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Solutions Manual

Engineering Mechanics: Statics 1st Edition

Michael E. Plesha University of Wisconsin–Madison

Gary L. Gray The Pennsylvania State University

Francesco Costanzo The Pennsylvania State University With the assistance of: Chris Punshon Andrew J. Miller Justin High Chris O’Brien Chandan Kumar Joseph Wyne Jonathan Fleischmann

Version: August 24, 2009

The McGraw-Hill Companies, Inc.

Copyright © 2002–2010 Michael E. Plesha, Gary L. Gray, and Francesco Costanzo

This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc. It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited.

3

Statics 1e

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Contact the Authors If you find any errors and/or have questions concerning a solution, please do not hesitate to contact the authors and editors via email at: [email protected] We welcome your input.

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Accuracy of Numbers in Calculations Throughout this solutions manual, we will generally assume that the data given for problems is accurate to 3 significant digits. When calculations are performed, all intermediate numerical results are reported to 4 significant digits. Final answers are usually reported with 3 significant digits. If you verify the calculations in this solutions manual using the rounded intermediate numerical results that are reported, you should obtain the final answers that are reported to 3 significant digits.

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Chapter 4 Solutions Problem 4.1 Compute the moment of force F about point B, using the following procedures. (a) Determine the moment arm d and then evaluate MB D F d . (b) Resolve force F into x and y components at point A and use the principle of moments. (c) Use the principle of moments with F positioned at point C . (d) Use the principle of moments with F positioned at point D. (e) Use a vector approach.

Solution Part (a) We first use geometry and trigonometry to find that tan 1 . 3=4/ D 36:9ı , tan 1 . 4=3/ D 53:1ı , and 180ı 60ı 53:1ı D 66:9ı as shown in the figure to the right. We then use the law of sines with triangle ABD to find that 12 in: b D ) b D 13:8 in: (1) ı sin 53:1 sin 66:9ı Through trigonometry, it follows that the moment arm d is d D b sin 53:1ı D 11:0 in:

(2)

Using Eq. (4.1) on page 182, and noting that positive moment is counterclockwise, MB is then given by MB D F d D .25 lb/.11:0 in:/ D 276 in.lb:

(3)

Part (b) Using the figure to the right, and noting that positive moment is counterclockwise, Eq. (4.1) provides ! ! 3 4 MB D 25 lb .12 in: sin 60ı / C 25 lb .12 in: cos 60ı / D 276 in.lb: 5 5 (4)

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Statics 1e Part (c) Using triangle ABC shown at the right, the law of sines provides 12 in: h D ı sin 36:9 sin 113:1ı

)

h D 18:4 in:

(5)

In the figure to the right, we see that the vertical force at C produces no moment about B. By using Eq. (4.1), and noting that positive moment is counterclockwise, MB is given by 3 MB D 25 lb .18:4 in:/ D 276 in.lb: 5

(6)

Part (d) In the figure to the right, we see that the horizontal force on D produces no moment about B. Using Eq. (4.1), and noting that positive moment is counterclockwise, MB is given by 4 MB D 25 lb .13:8 in:/ D 276 in.lb: 5

(7)

Part (e) We find that rEBA D 12 in: cos 60ı {O C sin 60ı |O 3 4 E F D 25 lb {O |O : 5 5

(8) (9)

E B is given by Using Eq. (4.2) on page 183, M ˇ ˇ ˇ E B D rEBA FE D .12 in:/.25 lb/ ˇ M ˇ ˇ

ˇ {O |O kO ˇˇ cos 60ı sin 60ı 0 ˇˇ D 276 kO in.lb: 3 4 0 ˇ 5 5

(10)

Note that instead of Eq. (8), we could have used rEBC or rEBD instead.

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Problem 4.2 Compute the moment of force F about point B, using the following procedures. (a) Determine the moment arm d and then evaluate MB D F d . (b) Resolve force F into x and y components at point A and use the principle of moments. (c) Use the principle of moments with F positioned at point C . (d) Use the principle of moments with F positioned at point D. (e) Use a vector approach.

Solution Part (a) Using the 30ı and 60ı angles given in the problem statement, we identify the additional angles shown in the figure at the right. We then use the law of sines for triangle ABD to determine b 12 in: D ı sin 30 sin 30ı

)

b D 12 in:

(1)

Through trigonometry, it follows that the moment arm d is d D b sin 30ı D 6 in:

(2)

Using Eq. (4.1) on page 182, and noting that positive moment is counterclockwise, MB is then given by MB D F d D

25 lb.6 in:/ D

150 in.lb:

(3)

Part (b) Using the figure to the right, and noting that positive moment is counterclockwise, Eq. (4.1) provides MB D D

25 lb cos 30ı .12 in: sin 60ı / C 25 lb sin 30ı .12 in: cos 60ı /

(4)

150 in.lb:

(5)

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Statics 1e Part (c) Using triangle ABC shown at the right, the law of sines provides h 12 in: D ı sin 120 sin 30ı

)

h D 6:93 in:

(6)

In the figure to the right, we see that the vertical force at C produces no moment about B. Using Eq. (4.1), and noting that positive moment is counterclockwise, MB is given by MB D

25 lb cos 30ı .6:93 in:/ D

150 in.lb:

(7)

Part (d) In the figure to the right, we see that the horizontal force at D produces no moment about B. Using Eq. (4.1), and noting that positive moment is counterclockwise, MB is given by MB D

25 lb sin 30ı .12 in:/ D

150 in.lb:

(8)

Part (e) We find that cos 60ı {O C sin 60ı |O FE D 25 lb cos 30ı {O sin 30ı |O :

rEBA D 12 in:

(9) (10)

E B is given by Using Eq. (4.2) on page 183, M ˇ ˇ {O ˇ E E MB D rEBA F D .12 in:/.25 lb/ ˇˇ cos 60ı ˇ cos 30ı

ˇ |O kO ˇˇ sin 60ı 0 ˇˇ D sin 30ı 0 ˇ

150 kO in.lb:

(11)

Rather than rEBA , we could have used rEBC or rEBD instead.

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Problem 4.3 The cover of a computer mouse is hinged at point B so that it may be “clicked.” Repeat Prob. 4.1 to determine the moment about point B.

Solution Part (a) Using trigonometry and the figure to the right, we determine that the moment arm d is d D 60 mm sin 45ı D 42:4 mm

(1)

Using Eq. (4.1) on page 182, and noting that positive moment is counterclockwise, MB is given by MB D F d D

3 N.42:4 mm/ D

127 Nmm:

(2)

Part (b) We first use geometry to determine h D 20 mm as shown in the figure to the right. Using Eq. (4.1), and noting that positive moment is counterclockwise, we determine that MB D D

3 N cos 45ı .20 mm/

3 N sin 45ı .40 mm/

127 Nmm:

(3) (4)

Part (c) In the figure to the right, we see that the vertical force at C produces no moment about B. Using Eq. (4.1), and noting that positive moment is counterclockwise, MB is given by MB D

3 N cos 45ı .60 mm/ D

127 Nmm:

(5)

Part (d) In the figure to the right, we see that the horizontal force at D produces no moment about B. Using Eq. (4.1), and noting that positive moment is counterclockwise, MB is given by MB D

3 N sin 45ı .60 mm/ D

127 Nmm

(6)

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Statics 1e Part (e) Using the value of h found in Part (b), we find that rEBA D .40 {O C 20 |O/ mm FE D 3 N sin 45ı {O

(7)

cos 45ı |O :

(8)

E B is given by Using Eq. (4.2) on page 183, M ˇ ˇ {O ˇ ˇ E E MB D rEBA F D 3 Nmm ˇ 40 ˇ sin 45ı

ˇ |O kO ˇˇ 20 0 ˇˇ D ı cos 45 0 ˇ

127 kO Nmm

(9)

Rather than rEBA , we could have used rEBC or rEBD instead.

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Problem 4.4 The cover of a computer mouse is hinged at point B so that it may be “clicked.” Repeat Prob. 4.1 to determine the moment about point B.

Solution Part (a) We use some elementary geometry to determine the dimensions shown in the figure to the right. The moment arm d is then d D 20 mm cos 45ı D 14:1 mm

(1)

Using Eq. (4.1) on page 182, and noting that positive moment is counterclockwise, MB is given by MB D F d D

3 N.14:1 mm/ D

42:4 Nmm:

(2)

Part (b) Using Eq. (4.1), and noting that positive moment is counterclockwise, we determine MB D 3 N cos 45ı .20 mm/ D

3 N sin 45ı .40 mm/

42:4 Nmm:

(3) (4)

Part (c) In the figure to the right, we see that the vertical force at C produces no moment about B. Using Eq. (4.1) again, MB is given by MB D

3 N cos 45ı .20 mm/ D

42:4 Nmm:

(5)

Part (d) We see that the horizontal force on D produces no moment about B in the figure to the right, then by using Eq. (4.1), and noting that positive moment is counterclockwise, MB is given by MB D

3 N sin 45ı .20 mm/ D

42:4 Nmm:

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(6)

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Statics 1e Part (e) We find that rEBA D .40 {O C 20 |O/ mm FE D 3 N

cos 45ı {O

(7) sin 45ı |O :

(8)

E B is given by Using Eq. (4.2) on page 183, M ˇ ˇ ˇ E E MB D rEBA F D 3 Nmm ˇˇ ˇ

{O 40 cos 45ı

ˇ |O kO ˇˇ 20 0 ˇˇ D ı sin 45 0 ˇ

42:4 kO Nmm

(9)

Rather than rEBA , we could have used rEBC or rEBD instead.

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Problem 4.5 An atomic force microscope (AFM) is a state-of-the-art device used to study the mechanical and topological properties of surfaces on length scales as small as the size of individual atoms. The device uses a flexible cantilever beam AB with a very sharp, stiff tip BC that is brought into contact with the surface to be studied. Due to contact forces at C , the cantilever beam deflects. If the tip of the AFM is subjected to the forces shown, determine the resultant moment of both forces about point A. Use both scalar and vector approaches.

Solution For the scalar approach we use Eq. (4.1) on page 182. Noting that positive moment is counterclockwise, MA is given by

MA D D

.13 nN/.12m/ C .5 nN/.1:5m/ D 1:4910

13

! ! 10 9 N 10 6 m 149 nNm nN m

Nm:

(1) (2)

For the vector approach we use Eq. (4.2) on page 183. It follows that rEAC D . 12 {O

1:5 |O/ m

FE D .5 {O C 13 |O/ nN

(3) (4)

EA is given by and M EA D rEAC M D D

ˇ ˇ ˇ FE D ˇˇ ˇ

{O 12 5

ˇ |O kO ˇˇ 1:5 0 ˇˇ nNm 13 0 ˇ

149 kO nNm 1:4910 13 kO Nm:

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(5) (6) (7)

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Problem 4.6 The door of an oven has 22 lb weight which acts vertically through point D. The door is supported by a hinge at point A and two springs that are symmetrically located on each side of the door. For the positions specified below, determine the force needed in each of the two springs if the resultant moment about point A of the weight and spring forces is to be zero. (a) ˛ D 45ı . (b) ˛ D 90ı . (c) Based on your answers to Parts (a) and (b), determine an appropriate stiffness for the springs.

Solution Part (a) We first determine some needed dimensions using trigonometry as shown in the figure at the left below. The vertical and horizontal distances between points A and C are .5 in:/ sin 45ı D .5 in:/ cos 45ı D 3:54 in. Then, using the Pythagorean theorem we determine the distance between points B and C as q (1) LBC D .10 in: C 3:54 in:/2 C .8 in: 3:54 in:/2 D 14:25 in: Let F be the force in one spring. The forces in the springs act along line BC , therefore the total horizontal and vertical forces at C from the springs are, respectively, Fhorizontal D 2F .13:54 in:=14:25 in:/;

Fvertical D 2F .4:46 in:=14:25 in:/;

(2)

and these forces are shown in the figure at the right below. The moment arm for the 22 lb weight is .12 in:/ cos 45ı D 8:49 in: as shown in the figure to the right below.

With positive moment being counterclockwise, the resultant moment of the two spring forces and the weight of the door about point A is MA D

.22 lb/.8:49 in:/ C 2F

13:54 in: 4:46 in: .3:54 in:/ C 2F .3:54 in:/: 14:25 in: 14:25 in:

(3)

Requiring MA D 0, as stated in the problem statement, we set Eq. (3) equal to zero to solve for F as F D 21:0 lb:

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(4)

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Part (b) When the door is horizontal (fully open), points A, B, and C have the positions shown in the figure at the left below. The Pythagorean theorem is used to determine the distance between points B and C as q LBC D .10 in: C 5 in:/2 C .8 in:/2 D 17 in: (5) Let F be the force in one spring. The forces in the springs act along line BC , therefore the total horizontal and vertical forces at C from the springs are, respectively Fhorizontal D 2F .15 in:=17 in:/;

Fvertical D 2F .8 in:=17 in:/:

(6)

and these forces are shown in the figure to the right below.

With positive moment being counterclockwise, and noting that the horizontal force at C produces no moment about point A, the resultant moment of the two spring forces and the weight of the door about point A is MA D

.22 lb/.12 in:/ C 2F

8 in: .5 in:/: 17:0 in:

(7)

Requiring MA D 0, as stated in the problem statement, we set Eq. (7) equal to zero to solve for F as F D 56:1 lb:

(8)

Part (c) We use the spring law to determine the necessary spring stiffness so that the springs will produce the forces determined in Parts (a) and (b) for ˛ D 45ı and ˛ D 90ı . Thus F D kı

)

kD

F ı

)

kD

change in force 56:1 lb D change in length 17:0 in:

21:0 lb D 13:0 lb=in: 14:3 in:

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(9)

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Problem 4.7 The ball of a trailer hitch is subjected to a force F . If failure occurs when the moment of F about point A reaches 10,000 in.lb, determine the largest value F may have and specify the value of ˛ for which the moment about A is the largest. Note that the value of F you determine must not produce a moment about A that exceeds 10,000 in.lb for any possible value of ˛.

Solution The maximum moment for a given value of force F occurs when the moment arm is at its maximum. Using the Pythagorean theorem in the figure to the right, the maximum value of the moment arm d is given by q d D .9 in:/2 C .6 in:/2 D 10:8 in:

(1)

We then use Eq. (4.1) on page 182 to determine the maximum force F , given that the moment of this force about point A may not exceed 10;000 in.lb. Thus, MA D F d D F .10:8 in:/ D 10;000 in.lb

)

F D 926 lb:

(2)

The angle ˛ at which this maximum force acts is then found to be d cos ˇ D 9 in: ˛ D 180ı

.180ı

) ˇ

ˇ D 33:6ı 90ı / D 123:6ı :

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(3) (4)

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Problem 4.8 Repeat Prob. 4.7 if the moment at point B may not exceed 5000 in.lb.

Solution The maximum moment for a given value of force F occurs when the moment arm is at its maximum. Using the Pythagorean theorem in the figure to the right, the maximum value of the moment arm d is given by q d D .5 in:/2 C .6 in:/2 D 7:81 in:

(1)

We then use Eq. (4.1) on page 182 to determine the maximum force F , given that the moment of this force about point B may not exceed 5000 in.lb. Thus, MB D F d D F .7:81 in:/ D 5000 in.lb

)

F D 640 lb:

(2)

The angle ˛ at which this maximum force acts is then found to be d cos ˇ D 5 in: ˛ D 180ı

.180ı

) ˇ

ˇ D 50:2ı

(3)

90ı / D 140ı :

(4)

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Problem 4.9 The port hull of a catamaran (top view shown) has cleats at points A, B, and C . A rope having 100 N force is to be attached to one of these cleats. If the force is to produce the largest possible counterclockwise moment about point O, determine the cleat to which the rope should be attached and the direction the rope should be pulled (measured positive counterclockwise from the positive x direction). Also, determine the value of MO produced. Assume all cleats, point O, and the rope lie in the same plane.

Solution Cleat C offers the largest possible moment arm, therefore Attaching the rope to cleat C will produce the largest moment.

(1)

As shown to the right, the Pythagorean theorem is used to determine the moment arm d as q (2) d D .1:5 m/2 C .3 m/2 D 3:35 m: The angle at which the force acts is given by d cos ˇ D 3 m

)

ˇ D 26:4ı

˛ D 90ı C ˇ D 116:4ı :

(3) (4)

The moment of the 100 N force about O is given by Eq. (4.1) on page 182 as MO D 100 N.3:35 m/ D 335 Nm:

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(5)

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Problem 4.10 Frame ABC has a frictionless pulley at C around which a cable is wrapped. Determine the resultant moment about point A produced by the cable forces if W D 5 kN and (a) ˛ D 0ı . (b) ˛ D 90ı . (c) ˛ D 30ı .

Solution Part (a) T D W D 5 kN for the pulley to be in equilibrium. We may determine the moment of the cable forces about point A by considering the forces supported by the cables, or by considering the forces the pulley applies to the structure at point C . Both of these options are illustrated in the figures below, to the left and right, respectively.

Use of either of the force systems shown above is convenient, and we will use the force system shown at the left. Noting that positive moment is counterclockwise, we then use Eq. (4.1) on page 182 to determine the moment about point A as MA D 5 kN.3 m C 0:5 m/ C 5 kN.3 m

0:5 m/

(1)

D 30:0 kNm:

(2)

Part (b) We use the same procedure as for Part (a), and use either of the following force systems.

Use of either of the force systems shown above is convenient, and we will use the force system shown at the left. Noting that positive moment is counterclockwise, we then use Eq. (4.1) on page 182 to determine the moment about point A as MA D 5 kN.4 m C 0:5 m/ C 5 kN.3 m

0:5 m/

D 35:0 kNm: This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc. It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited.

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Part (c) In principle, either of the force systems shown below may be used to determine the moment about point A.

However, if the force system at the left is used, it will be tedious to determine the moment arm for the cable force with the 30ı orientation. Thus, it will be considerably more convenient to use the force system shown at the right, which provides MA D 5 kN.3 m/ C 5 kN.cos 30ı /3 m C 5 kN.sin 30ı /4 m D 38:0 kNm:

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(5) (6)

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Solutions Manual

Problem 4.11 Frame ABC has a frictionless pulley at C around which a cable is wrapped. If the resultant moment about point A produced by the cable forces is not to exceed 20 kNm, determine the largest weight W that may be supported and the corresponding value of ˛.

Solution First we determine the angle ˛ that provides the largest moment arm for force T . Using the figure shown at the right, it follows that tan ˇ D

4m 3m

)

ˇ D ˛ D 53:1ı :

(1)

Sincepthe pulley is in equilibrium, T D W . The Pythagorean theorem provides d D .4 m/2 C .3 m/2 D 5 m. We then use Eq. (4.1) on page 182, to evaluate the moment of the cable forces about point A, noting that positive moment is counterclockwise MA D W .3 m/ C W cos 53:1ı .3 m/ C W sin 53:1ı .4 m/:

(2)

As given in the problem statement, MA is not to exceed 20 kNm. Thus we set Eq. (2) equal to 20 kNm and solve for W to obtain W D 2:5 kN: (3)

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Statics 1e

Problem 4.12 The load carrying capacity of the frame of Prob. 4.11 can be increased by placing a counterweight Q at point D as shown. The resultant moment about point A due to the cable forces and Q is not to exceed 20 kNm. (a) Determine the largest value of counterweight Q. (Hint: Let W D 0 and determine Q so that MA D 20 kNm.) (b) With the value of Q determined in Part (a), determine the largest weight W that may be supported and the corresponding value of ˛.

Solution Part (a) We begin by setting W D 0, and then use Eq. (4.1) on page 182 to evaluate the moment of Q about point A, noting that this moment may not exceed 20 kNm. With positive moment being counterclockwise, it follows that MA D

Q.2 m/ D

20 kNm

)

Q D 10 kN:

(1)

Part (b) With W ¤ 0, we know that T D W , since the pulley is in equilibrium. As determined in the solution to Prob. 4.11, the moment of T about point A is largest when ˛ D 53:1ı (the moment of T about point A is largest when its line of action is perpendicular to the line connecting points A and C , as shown above). Using the figure above, we evaluate the moment about point A, taking positive moment to be counterclockwise, and we require this moment to not exceed 20 kNm. Thus, MA D

10 kN.2 m/ C W .3 m/ C W cos 53:1ı .3 m/ C W sin 53:1ı .4 m/ D 20 kNm

W D 5 kN:

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(2) (3)

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Problem 4.13 A flat rectangular plate is subjected to the forces shown, where all forces are parallel to the x or y axis. If F D 200 N and P D 300 N, determine the resultant moment of all forces about the (a) ´ axis. (b) a axis, which is parallel to the ´ axis.

Solution Part (a) We use Eq. (4.1) on page 182 to find M´ D F .300 mm/ D

F .500 mm/ C P .200 mm/

5

1:610 Nmm D

P .600 mm/

160 Nm:

(1) (2)

In writing the above expression, positive moment is taken to be in the positive ´ direction, according to the right-hand rule. Part (b) We use Eq. (4.1) again to find Ma D D

F .300 mm/ C F .200 mm/ 5

1:710 Nmm D

P .500 mm/

170 Nm:

(3) (4)

In writing the above expression, positive moment is taken to be in the positive a direction, according to the right-hand rule.

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Problem 4.14 A flat rectangular plate is subjected to the forces shown, where all forces are parallel to the x or y axis. If P D 300 N, determine F when the resultant moment of all forces about the ´ axis is M´ D 100 Nm.

Solution Use Eq. (4.1) on page 182 to evaluate the moment of F and P D 300 N about the ´ axis M´ D F .300 mm/

F .500 mm/ C 300 N.200 mm/

300 N.600 mm/:

(1)

In writing the above expression, positive moment is taken to be in the positive ´ direction, according to the right-hand rule. As given in the problem statement, M´ D 100 Nm. Thus, Eq. (1) becomes mm 100 Nm 103 D F .300 mm/ F .500 mm/ C 300 N.200 mm/ 300 N.600 mm/; (2) m which may be solved for F D

100 N:

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(3)

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Problem 4.15 Repeat Prob. 4.14 if the resultant moment of all forces about the a axis is Ma D 100 Nm.

Solution Use Eq. (4.1) on page 182 to evaluate the moment of F and P D 300 N about the a axis Ma D

F .300 mm/ C F .200 mm/

300 N.500 mm/:

(1)

In writing the above expression, positive moment is taken to be in the positive a direction, according to the right-hand rule. As given in the problem statement, Ma D 100 Nm. Thus, Eq. (1) becomes mm 100 Nm 103 D F .300 mm/ C F .200 mm/ 300 N.500 mm/; (2) m which may be solved for F D

500 N:

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(3)

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Problem 4.16 Structure OBCD is built in at point O and supports a 50 lb cable force at point C and 100 and 200 lb vertical forces at points B and D, respectively. Using a vector approach, determine the moment of these forces about (a) point B. (b) point O.

Solution From the figure in the problem statement, the following vector expressions may be written rEBC D 20 kO in. rEBD D 15 {O C 20 kO in. rEOC D 36 kO in. rEOD D 15 {O C 36 kO in. FE D

200 |O lb

PE D

100 |O lb

12 {O C 18 |O TE D 50 lb 42

36 kO

:

Part (a) Observing that PE produces no moment about B, we use Eq. (4.2) on page 183 to evaluate the resultant moment of TE and FE about point B as E B D rEBC TE C rEBD FE M ˇ ˇ ˇ ˇ {O |O ˇ {O O ˇ k |O kO ˇ ˇ ˇ .20 in:/.50 lb/ ˇ ˇ ˇ C 200 in.lb D 1 ˇ ˇ 0 0 ˇ 15 0 20 42 ˇ 12 18 ˇ 0 36 ˇ 1 0 h i h 1000 in.lb O D {O. 18/ |O. 12/ C k.0/ C 200 in.lb {O.20/ 42 E B D 3570 {O C 286 |O 3000 kO in.lb: M

(1) ˇ ˇ ˇ ˇ ˇ ˇ

(2)

i O 15/ |O.0/ C k.

(3) (4)

Part (b) We use Eq. (4.2) on page 183 to evaluate the resultant moment of TE , FE , and PE about point O as E O D rEOC TE C rEOD FE C rEOB PE M ˇ ˇ ˇ ˇ {O ˇ {O |O O ˇ |O kO k ˇ ˇ .36 in:/.50 lb/ ˇˇ ˇ ˇ D 1 ˇ C 200 in.lb ˇ 15 0 36 ˇ 0 0 42 ˇ 0 ˇ 12 18 1 0 36 ˇ ˇ ˇ ˇ {O |O kO ˇ ˇ ˇ C .100 lb/.16 in:/ ˇˇ 0 0 1 ˇˇ ˇ 0 1 0 ˇ i h 1800 in.lb h O D {O. 18/ |O. 12/ C k.0/ C 200 in.lb {O.36/ 42 C 1600 {O in.lb E O D 8030 {O C 514 |O M

(5) ˇ ˇ ˇ ˇ ˇ ˇ

(6)

i O 15/ |O.0/ C k.

3000 kO in.lb:

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(7)

(8)

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Problem 4.17 Repeat Prob. 4.16, using a scalar approach.

Solution The vector expression for TE is 12 {O C 18 |O TE D 50 lb 42

36 kO

:

(1)

Using this expression, the figure at the right is drawn, where the force TE is shown in terms of its components. We then use E B in Eq. (4.1) on page 182 to determine the components of M each of the coordinate directions, taking positive moment components to be in the positive coordinate directions. Part (a) It follows that .MB /x D 200 lb.20 in:/

50 lb

18 .20 in:/ D 3570 in.lb 42

12 .20 in:/ D 286 in.lb 42 200 lb.15 in:/ D 3000 in.lb:

(2)

.MB /y D 50 lb

(3)

.MB /´ D

(4)

Part (b) It follows that .MO /x D 200 lb.36 in:/

50 lb

18 .36 in:/ C 100 lb.16 in:/ D 8030 in.lb 42

12 .36 in:/ D 514 in.lb 42 200 lb.15 in:/ D 3000 in.lb:

(5)

.MO /y D 50 lb

(6)

.MO /´ D

(7)

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Problem 4.18 Structure OAB is built in at point O and supports forces from two cables. Cable CAD passes through a frictionless ring at point A, and cable DBE passes through a frictionless ring at point B. If the force in cable CAD is 250 N and the force in cable DBE is 100 N, use a vector approach to determine (a) the moment of all cable forces about point A. (b) the moment of all cable forces about point O.

Solution From the figure provided in the problem statement, the following vector expressions may be written 150 {O 200 kO FEBD D 100 N 250 150 |O 200 kO FEBE D 100 N 250 60 | O 80 kO FEAC D 250 N 100 150 {O 80 kO : FEAD D 250 N 170

rEAB D 120 kO mm rEOB D 200 kO mm rEOA D 80 kO mm

These force vectors are shown in the figure at the right. Part (a) Noting that FEAC and FEAD produce no moment about point A, we use Eq. (4.2) on page 183 to determine

EA D rEAB M

kO 1 400

ˇ ˇ {O |O ˇ .120 mm/.100 N/ ˇ 0 FEBD C FEBE D 0 ˇ 250 ˇ 150 150

D 48 Nmm ŒO{ . 150/

ˇ ˇ ˇ ˇ ˇ ˇ

(1)

|O. 150/

(2)

D . 7200 {O C 7200 |O/ Nmm:

(3)

Part (b) We use Eq. (4.2) to determine E O D rEOB FEBD C FEBE C rEOA FEAC M ˇ ˇ |O kO .200 mm/.100 N/ ˇˇ {O D 0 1 ˇ 0 250 ˇ 150 150 400 D 80 Nmm ŒO{ . 150/

C FEAD ˇ ˇ ˇ ˇ {O ˇ ˇ ˇ C .80 mm/.250 N/ ˇ 0 ˇ ˇ ˇ 150 ˇ

|O. 150/ C 20;000 Nmm ŒO{ .0:6/

170

(4) kO 1

|O 0 60 100

|O. 0:882/

D .29;600 |O/ Nmm:

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80 100

80 170

ˇ ˇ ˇ ˇ ˇ ˇ

(5) (6) (7)

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Problem 4.19 Repeat Prob. 4.18, using a scalar approach.

Solution We begin by adding the two cable forces at point A to obtain FEA , and adding the two cable forces at point B to obtain FEB , as follows 60 |O 80 kO 150 |O 80 kO FEA D FEAC C FEAD D 250 N C 250 N D 221 {O 150 |O 318 kO N; 100 170 O 150 |O 200 kO 150 {O 200 k C 100 N D 60 {O C 60 |O 160 kO N: FEB D FEBD C FEBE D 100 N 250 250

(1) (2)

The components of FEA and FEB are shown in the figure below. Part (a) Taking positive moment components to be in the positive coordinate directions, we use Eq. (4.1) on page 182 to write .MA /x D

60 N.120 mm/ D

7200 Nmm

(3)

.MA /y D 60 N.120 mm/ D 7200 Nmm

(4)

.MA /´ D 0:

(5)

Part (b) Taking positive moment components to be in the positive coordinate directions, we use Eq. (4.1) on page 182 to write .MO /x D

60 N.200 mm/ C 150 N.80 mm/ D 0 Nmm

(6)

.MO /y D 60 N.200 mm/ C 221 N.80 mm/ D 29;700 Nmm

(7)

.MO /´ D 0:

(8)

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Problem 4.20 Structure OABC is built in at point O and supports forces from two cables. Cable EAD passes through a frictionless ring at point A, and cable OC G passes through a frictionless ring at point C . If the force in cable EAD is 800 lb and the force in cable OC G is 400 lb, determine (a) the moment of forces from cable OC G about point B. (b) the moment of all cable forces about point A. (c) the moment of all cable forces about point O.

Solution From the figure provided in the problem statement, the following vector expressions may be written 9 |O 12 kO 25 9 |O 12 kO D 400 lb 15 8 {O 6 kO D 800 lb 10 8 |O 6 kO D 800 lb 10

20 {O FEC G D 400 lb

rEBC D 9 |O ft

FECO

rEAC D 9 |O C 6 kO ft

FEAE FEAD

rEOC D 9 |O C 12 kO ft rEOA D 6 kO ft:

These force vectors are shown in the figure at the right. Part (a) We use Eq. (4.2) on page 183 to determine the moment about point B due to the forces in cable OC G as ˇ ˇ O ˇ ˇ {O | O k ˇ ˇ ˇ ˇ E E E MB D rEBC FC G C FCO D .400 ftlb/ ˇ 0 9 0 ˇ 9 12 20 9 12 ˇ ˇ 25 h i25 25 15 15 O 7:20/ D D 400 ftlb {O. 11:5/ |O.0/ C k. 4600 {O 2880 kO ftlb:

(1) (2)

Part (b) Noting that FEAE and FEAD produce no moment about point A, we use Eq. (4.2) on page 183 to determine ˇ ˇ O ˇ {O ˇ | O k ˇ ˇ ˇ ˇ E E E MA D rEAC FC G C FCO D .400 ftlb/ ˇ 0 9 6 ˇ 20 9 9 12 12 ˇ ˇ 25 15 h i25 15 25 O 7:20/ D D 400 ftlb {O. 5:76/ |O. 4:80/ C k. 2304 {O C 1920 |O

(3) 2880 kO ftlb:

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Solutions Manual

Part (c) We use Eq. (4.2) on page 183 to determine

E O D rEOC FEC G C FECO C rEOA FEAE C FEAD M ˇ ˇ ˇ ˇ ˇ ˇ {O |O kO ˇ .6 ft/.800 lb/ ˇ {O |O ˇ ˇ ˇ 0 0 ˇ D .400 ftlb/ ˇ 0 9 12 ˇ C ˇ 10 9 9 12 12 ˇ ˇ 8 8 ˇ 20 25 25 15 25 15 h i O 7:20/ C 480 ftlb ŒO{ . 8/ |O. 8/ D 400 ftlb |O. 9:60/ C k. D 3840 {O C 7680 |O 2880 kO ftlb:

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(5) ˇ kO ˇˇ 1 ˇˇ 12 ˇ

(6) (7) (8)

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Problem 4.21 Repeat Prob. 4.20, using a scalar approach.

Solution We begin by adding the two cable forces at point A to obtain FEA , and adding the two cable forces at point C to obtain FEC , as follows 6 kO 8 |O 6 kO O C 800 lb D 640 {O C 640 |O 960 k lb; (1) 10 10 20 {O 9 |O 12 kO 9 |O 12 kO D 400 lb C 400 lb D 320 {O 384 |O 512 kO lb: (2) 25 15

8 {O FEA D FEAE C FEAD D 800 lb FEC D FEC G C FECO

The components of FEA and FEC are shown in the figure below. Part (a) Taking positive moment components to be in the positive coordinate directions, we use Eq. (4.1) on page 182 to determine the moment about point B due to the forces in cable OC G as .MB /x D

512 lb.9 ft/ D

4610 ftlb

(3)

.MB /y D 0 .MB /´ D

(4) 320 lb.9 ft/ D

2880 ftlb:

(5)

Part (b) Noting that the forces at point A produce no moment about point A, and taking positive moment components to be in the positive coordinate directions, we use Eq. (4.1) on page 182 to write .MA /x D 384 lb.6 ft/

512 lb.9 ft/ D

2300 ftlb

(6)

.MA /y D 320 lb.6 ft/ D 1920 ftlb

(7)

.MA /´ D

(8)

320 lb.9 ft/ D

2880 ftlb:

Part (c) Taking positive moment components to be in the positive coordinate directions, we use Eq. (4.1) on page 182 to write .MO /x D 384 lb.12 ft/

512 lb.9 ft/

640 lb.6 ft/ D

3840 ftlb

(9)

.MO /y D 320 lb.12 ft/ C 640 lb.6 ft/ D 7680 ftlb

(10)

.MO /´ D

(11)

320 lb.9 ft/ D

2880 ftlb:

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Solutions Manual

Problem 4.22 Structure OAB is built in at point O and supports two forces of magnitude F parallel to the y and ´ axes. If the magnitude of the moment about point O cannot exceed 1:0 kNm, determine the largest value F may have.

Solution Either the vector approach or the scalar approach may be conveniently used to determine the moment of the the two forces about point O, and we will use the scalar approach. With the dimensions provided, we use Eq. (4.1) on page 182 to determine .MO /x D

F .90 mm/

(1)

.MO /y D

F .400 mm/

(2)

.MO /´ D

F .400 mm/

(3)

The magnitude of the moment at point O may not exceed 1:0 kNm, thus ! 1 103 mm 2 2 2 =2 .MO /x C .MO /y C .MO /´ D 1:0 kNm m 1=2 F . 90 mm/2 C . 400 mm/2 C . 400 mm/2 D 1000 kNmm

(4) (5)

which may be solved for F D 1:75 kN:

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(6)

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Problem 4.23 Repeat Prob. 4.22 if the x component of the moment (torsional component) at point O may not exceed 0:5 kNm and the resultant of qthe y and ´ components (bending components) may not exceed 0:8 kNm (i.e., My2 C M´2 0:8 kNm).

Solution Either the vector approach or the scalar approach may be conveniently used to determine the moment of the the two forces about point O, and we will use the scalar approach. Taking positive moment components to be in the positive coordinate directions, we use Eq. (4.1) on page 182 to determine .MO /x D

F .90 mm/

(1)

.MO /y D

F .400 mm/

(2)

.MO /´ D

F .400 mm/:

(3)

The absolute value of the torsional component of the moment, .MO /x , may not exceed 0:50 kNm, thus ! 3 mm ˇ ˇ 10 ˇ.MO /x ˇ D F .90 mm/ 0:50 kNm ) F 5:56 kN: (4) m The bending component of the moment, q

F

q

q

.MO /y2 C .MO /2´ , may not exceed 0:80 kNm, thus

.MO /y2 C .MO /2´ 0:8 kNm

. 400 mm/2 C . 400 mm/2 0:8 kNm

(5) 103 mm m

! )

F 1:41 kN:

(6)

Both Eqs. (4), and (6) are satisfied by F 1:41 kN:

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(7)

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Solutions Manual

Problem 4.24 Forces of 3 kN and 200 N are exerted at points B and C of the main rotor of a helicopter, and force F is exerted at point D on the tail rotor. The 3 kN forces are parallel to the ´ axis, the 200 N forces are perpendicular to the main rotor and are parallel to the xy plane, F is parallel to the y axis, and ˛ D 45ı . (a) Determine the value of F so that the ´ component of the moment about point O of all rotor forces is zero. (b) Using the value of F found in Part (a), determine the resultant moment of all rotor forces about point O. (c) If ˛ is different than 45ı , do your answers to Parts (a) and (b) change? Explain.

Solution Part (a) We use Eq. (4.1) on page 182 to evaluate the moment of the rotor forces about the ´ axis, and as instructed in the problem statement, this moment is required to be zero. Thus, we obtain .MO /´ D

200 N.3 m/

200 N.3 m/ C F .5 m/ D 0

) F D 240 N:

(1)

Part (b) Using the geometry provided in the problem statement, with the aid of the sketch below to determine x and y components of the main rotor forces, the following vector expressions may be written

cos 45ı |O C 3000 N kO FEC D 200 N sin 45ı {O C cos 45ı |O C 3000 N kO

(2)

FED D F . |O/

(4)

FEB D 200 N

cos 45ı {O C sin 45ı |O C 1:8 kO m D 3 m cos 45ı {O sin 45ı |O C 1:8 kO m D 5 {O m C 1:2 kO m

rEOC D 3 m rEOB rEOD

sin 45ı {O

(3)

(5) (6) (7)

Using Eq. (4.2) on page 183, the moment about point O of all rotor forces is E O D rEOB FEB C rEOC FEC C rEOD FED M ˇ ˇ ˇ {O |O kO ˇˇ ˇ D ˇˇ 3 cos 45ı 3 sin 45ı 1:8 ˇˇ Nm ı ˇ 200 sin 45ı 200 cos 45 3000 ˇ ˇ ˇ ˇ ˇ ˇ ˇ {O O ˇ O ˇ { O | O k | O k ˇ ˇ ˇ ˇ C ˇˇ 3 cos 45ı 3 sin 45ı 1:8 ˇˇ Nm C ˇˇ 5 0 1:2 ˇˇ Nm ˇ 200 sin 45ı 200 cos 45ı 3000 ˇ ˇ 0 240 0 ˇ h i h i O 600/ Nm C {O.6109/ |O. 6618/ C k. O 600/ Nm D {O. 6109/ |O.6618/ C k. h i O C {O.288/ |O.0/ C k.1200/ Nm; This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc. It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited.

(8)

(9)

(10)

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Statics 1e which simplifies to E O D .288 {O/ Nm: M

(11)

Part (c) If ˛ ¤ 45ı , the answer to Part (a) does not change, since the moment produced by the main rotor forces about the ´ axis is not a function of ˛. The answer to Part (b) does not change, for the following reason. The main rotor forces produce moments about the x and y axes that are functions of ˛, but the moment arms for components of force at points B and C are always opposite (negative of one another) so that the net x and y components are zero regardless of ˛.

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Problem 4.25 EA;1 D rEAB FE The moment of force FE about point A can be computed using M EA;2 D rEAC FE , where B and C are points on the line of action of FE . or using M EA;1 D M EA;2 . Noting that rEAC D rEAB C rEBC , show that M

Solution As specified in the problem statement, MA;2 D rEAC FE ;

and

rEAC D rEAB C rEBC :

(1) (2)

Equations (1) and (2) may be combined to write EA;2 D rEAB C rEBC FE : M

(3)

Using the distributive property of the cross product, Eq. (3) may be expanded to write EA;2 D rEAB FE C rEBC FE : M

(4)

E Hence, Eq. (4) becomes Because rEBC and FE are parallel, rEBC FE D 0. MA;2 D rEAB FE D MA;1 :

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(5)

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Problem 4.26 The steering wheel of a Ferrari sports car has circular shape with 190 mm radius, and it lies in a plane that is perpendicular to the steering column AB. Point C , where the driver’s hand applies force FE to the steering wheel, lies on the y axis with y coordinate yC D 190 mm. Point A is at the origin of the coordinate system, and point B has the coordinates B . 120; 0; 50/ mm. Determine the moment of FE about line AB of the steering column if (a) FE has 10 N magnitude and lies in the plane of the steering wheel and has orientation such that its moment arm to line AB is 190 mm. Also determine the vector expression for this force. 18 {O (b) FE D .10 N/

3 |O C 14 kO . 23

12 {O 5 kO . (c) FE D .10 N/ 13

Solution Part (a) The moment of force FE about line AB can be evaluated using the scalar approach as MAB D .10 N/.190 mm/ D 1900 Nmm:

(1)

To determine the vector expression for this force, note that FE is perpendicular to both rEAB and rEAC . Thus, FE is parallel to the direction nE where nE D rEAB rEAC D . 120 {O ˇ ˇ {O ˇ D ˇˇ 120 ˇ 0 D {O. 9500/

O mm . 190 |O/ mm 50 k/ ˇ |O kO ˇˇ 0 50 ˇˇ mm2 190 0 ˇ O |O.0/ C k.22800/ mm:

(2)

Using Eq. (2), a unit vector nO in the direction of nE is nO D

nE D jE nj

O 0:3846 {O C 0:9231 k:

(3)

Note that rather than using Eqs. (2) and (3), it is possible to write an expression for the unit normal vector by O inspection as nO D . 50 {O C 120 k/=130. Since the force FE is parallel to n, O the vector expression for the force is O D . 3:85 {O C 9:23 k/ O N: FE D .10 N/. 0:3846 {O C 0:9231 k/

(4)

Note that it is possible that the force FE could be in the direction opposite to n, O thus the answer FE D O .3:85 {O 9:23 k/ N is also acceptable. This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc. It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited.

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Solutions Manual

Part (b) Our strategy will be to determine the moment of force FE about point A, and then use the dot product to determine the component of this moment that acts in the direction of line AB. The moment of force FE about point A is ˇ ˇ ˇ {O |O kO ˇˇ 10 Nmm ˇ EA D rEAC FE D ˇ 0 M 190 0 ˇˇ ˇ 23 ˇ 18 3 14 ˇ 10 Nmm O D {O. 2660/ |O.0/ C k.3420/ : (5) 23 EA in the direction of line AB is The component of M EA MAB D M

rEAB jErAB j

10 Nmm 1 D . 2660/. 120/ C .0/.0/ C .3420/. 50/ 23 130

!

D 472 Nmm:

(6)

Part (c) Our strategy is the same as that used in Part (b). Thus, the moment of force FE about point A is ˇ ˇ ˇ {O |O kO ˇˇ 10 Nmm ˇ EA D rEAC FE D ˇ 0 M 190 0 ˇˇ ˇ 13 ˇ 12 0 5 ˇ O 2280/ 10 Nmm : D {O.950/ |O.0/ C k. (7) 13 EA in the direction of line AB is The component of M EA rEAB MAB D M jErAB j ! 10 Nmm 1 D .950/. 120/ C .0/.0/ C . 2280/. 50/ 13 130 D 0:

(8)

The reason that the force FE produces no moment about line AB is because FE is parallel to line AB.

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Problem 4.27 A rectangular piece of sheet metal is clamped along edge AB in a machine called a brake. The sheet is to be bent along line AB by applying a y direction force F . Determine the moment about line AB if F D 200 lb. Use both vector and scalar approaches.

Solution For the vector solution, we use the data given in the problem statement to write the following expressions rEAC D 36 {O in.;

(1)

FE D 200 |O lb; rEAB D 20 {O 15 kO in.

(2) (3)

We then use Eq. (4.2) on page 183 to determine the moment of FE about point A as ˇ ˇ ˇ {O |O kO ˇˇ ˇ O EA D rEAC FE D ˇ 36 0 0 ˇ in.lb D {O.0/ |O.0/ C k.7200 M in.lb/ D 7200 kO in.lb: ˇ ˇ ˇ 0 200 0 ˇ

(4)

EA that acts in the direction of line AB is given by The component of M rEAB .0 in.lb/.20 in:/ C .0 in.lb/.0 in:/ C .7200 in.lb/. 15 in:/ D rAB 25 in: 4320 in.lb;

EA MAB D M

(5)

D

(6)

where rAB is the magnitude of rEAB . The negative answer in Eq. (6) means that the moment of F about line AB is in the direction from B to A. Note: We could have also used the scalar triple product to obtain Eq. (6). For the scalar solution, we first determine the moment arm d as shown in the figure to the right. Through geometry and trigonometry, we find that 15 D 36:9ı ; 20 d D 36 in: sin D 21:6 in: D tan

1

(7) (8)

Then by applying Eq. (4.1) on page 182, taking positive moment to be in the direction from A to B according to the right-hand rule, it follows that MAB D

200 lb.21:6 in:/ D

4320 in.lb:

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Problem 4.28 In Prob. 4.27 determine F if the moment about line AB is to be 2000 in.lb.

Solution Using the data given in the problem statement, we write the following expressions rEAC D 36 {O in.; FE D F |O; rEAB D 20 {O

(1) (2)

15 kO in.

(3)

Then by using the scalar triple product, we determine the moment of force FE about line AB as ˇ ˇ ˇ 20 0 15 ˇˇ ˇ r E 1 AB ˇ 36 0 0 ˇˇ in. MAB D rEAC FE D rAB 25 ˇˇ 0 F 0 ˇ 1 D . 15/.36/F in. 25

(4) (5)

With MAB D 2000 in.lb, we solve Eq. (5) for F to obtain F D 25

2000 in.lb D . 15/.36/ in.

92:6 lb

)

F D 92:6 lb:

(6)

In writing our final answer for F , the negative sign is unimportant and is thus omitted.

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Problem 4.29 In the pipe assembly shown, points B and C lie in the xy plane and force F is parallel to the ´ axis. If F D 150 N, determine the moment of F about lines OA and AB. Use both vector and scalar approaches.

Solution Using the figure to the right, we write the following expressions rEOA D 350 |O mm;

(1) ı

ı

rEAB D 350 mm sin 30 {O C cos 30 |O D .175 {O C 303 |O/ mm; rEBC D 350 mm cos 30ı {O sin 30ı |O D .260 {O 150 |O/ mm; FE D 150 kO N:

(2) (3) (4)

For the vector approach, we use Eq. (4.2) on page 183 to determine the moment of the force about point A, EA . Then, M EA rEOA =rOA produces the moment of the force about line OA, and M EA rEAB =rAB namely M produces the moment of the force about line AB. Thus ˇ ˇ ˇ {O |O kO ˇˇ ˇ EA D rEAC FE D rEAB C rEBC FE D ˇ 0:435 0:153 M (5) 0 ˇˇ Nm ˇ ˇ 0 0 150 ˇ h i O D {O. 23:0/ |O. 65:2/ C k.0/ Nm D . 23:0 {O C 65:2 |O/ Nm: (6) EA rEOA D .65:2 Nm/.0:350 m/ D 65:2 Nm; MOA D M rOA 0:350 m . 23:0 Nm/.0:175 m/ C .65:2 Nm/.0:303 m/ r E EA AB D D 44:9 Nm: MAB D M rAB 0:350 m

(7) (8)

For the scalar approach, we use Eq. (4.1) on page 182 and the figure shown above, taking positive moment MOA to be in the direction from O to A, and positive moment MAB to be in the direction from A to B, according to the right-hand rule, to determine MOA D F d1 D 150 N .0:350 m/ sin 30ı C .0:300 m/ cos 30ı D 65:2 Nm; MAB D F d2 D 150 N .0:300 m/ D 45:0 Nm:

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Problem 4.30 In the pipe assembly shown, points B and C lie in the xy plane and force F is parallel to the ´ axis. If a twisting moment (torque) of 50 Nm will cause a pipe to begin twisting in the flange fitting at O or at either end of the elbow fitting at A, determine the first fitting that twists and the value of F that causes it.

Solution We use Eq. (4.1) on page 182, and the figure shown at the right for the determination of the moment arms d1 and d2 , to determine the moments of force F about lines OA and AB as MOA D F d1 D F .0:350 m/ sin 30ı C .0:300 m/ cos 30ı ; (1) MAB D F d2 D F .0:300 m/:

(2)

According to the failure criteria given in the problem statement, pipe OA will twist at O and A if MOA reaches 50 Nm, and pipe AB will twist at A and B if MAB reaches 50 Nm. Thus, If MOA D 50 Nm;

then

If MAB D 50 Nm;

then

50 Nm D 115 N; .0:350 m sin 30ı C 0:300 m cos 30ı / 50 Nm F D D 167 N: .0:300 m/ F D

(3) (4)

The smallest value of F among Eqs. (3) and (4) is the value of the force that will cause twisting to occur. Thus, the largest force F that may be applied is F D 115 N; and twisting occurs at both threads on pipe OA.

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(5)

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Problem 4.31 Determine the moment of force F about line AB as follows. EA , and then determine the (a) Determine the moment of F about point A, M component of this moment in the direction of line AB. E B , and then determine the (b) Determine the moment of F about point B, M component of this moment in the direction of line AB. EA , M E B , and the (c) Comment on differences and/or agreement between M moment about line AB found in Parts (a) and (b). Also comment on the meaning of the sign (positive or negative) found for the moment about line AB.

Solution With the information provided in the problem statement, the following position vectors may be written rEAB D .15 {O 20 |O/ in.; rEAC D 20 |O C 18 kO in.; rEBC D 15 {O C 18 kO in. (1) Part (a) We use Eq. (4.2) on page 183 to determine the moment of the force about point A as ˇ ˇ ˇ {O |O kO ˇˇ ˇ EA D rEAC FE D ˇ 0 M 20 18 ˇˇ in.lb; ˇ ˇ 10 80 40 ˇ h EA D {O. 640/ M

i O |O. 180/ C k.200/ in.lb D 640 {O C 180 |O C 200 kO in.lb:

(2)

(3)

EA in the direction of line AB is given by The component of M EA MAB D M

rEAB . 640/.15/ C .180/. 20/ C .200/.0/ D in.lb D rAB 25

528 in.lb:

Part (b) We use Eq. (4.2) on page 183 to determine the moment of the force about point B as ˇ ˇ ˇ {O O ˇ | O k ˇ ˇ E B D rEBC FE D ˇ 15 0 M 18 ˇˇ in.lb ˇ ˇ 10 80 40 ˇ

(4)

(5) (6)

h E B D {O. 1440/ M

i O 1200/ in.lb D |O.420/ C k. 1440 {O

420 |O

1200 kO in.lb:

(7)

E B in the direction of line AB is given by The component of M E B rEAB D . 1440/.15/ C . 420/. 20/ C . 1200/.0/ in.lb D MAB D M rAB 25 This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc. It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited.

528 in.lb:

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Part (c) EA and M E B are different, both in magnitude and direction. 1. M EA or M E B is used. 2. MAB is the same regardless if M 3. The negative result for MAB means that the twisting action of F is directed from B to A.

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Problem 4.32 Determine the moment of force F about line AB as follows. EA , and then determine the (a) Determine the moment of F about point A, M component of this moment in the direction of line AB. E B , and then determine the (b) Determine the moment of F about point B, M component of this moment in the direction of line AB. EA , M E B , and the (c) Comment on differences and/or agreement between M moment about line AB found in Parts (a) and (b). Also comment on the meaning of the sign (positive or negative) found for the moment about line AB.

Solution With the information provided in the problem statement, the following position vectors may be written rEAB D 4 {O 7 |O C 4 kO mm; rEAC D 4 {O C 4 kO mm; rEBC D 7 |O mm: (1) Part (a) We use Eq. (4.2) on page 183 to determine the moment of the force about point A as ˇ ˇ ˇ {O |O kO ˇˇ ˇ 2 EA D rEAC FE D ˇ 4 0 4 ˇ Nmm; M ˇ 7 ˇˇ 3 2 6 ˇ h EA D 2 {O.8/ M 7

i O |O. 36/ C k.8/ Nmm:

(2)

(3)

EA in the direction of line AB is given by The component of M EA rEAB D 2 Nmm Œ.8/. 4/ C .36/. 7/ C .8/.4/ 1 D MAB D M rAB 7 9

8:00 Nmm:

Part (b) We use Eq. (4.2) on page 183 to determine the moment of the force about point B as ˇ ˇ ˇ {O |O kO ˇ ˇ ˇ E B D rEBC FE D 2 ˇ 0 7 0 ˇ Nmm; M ˇ 7 ˇˇ 3 2 6 ˇ h E B D 2 {O.42/ M 7

i O 21/ Nmm: |O.0/ C k.

(4)

(5)

(6)

E B in the direction of line AB is given by The component of M E B rEAB D 2 Nmm Œ.42/. 4/ C .0/. 7/ C . 21/.4/ 1 D MAB D M rAB 7 9

8:00 Nmm:

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Solutions Manual

Part (c) EA and M E B are different, both in magnitude and direction. 1. M EA or M E B is used. 2. MAB is the same regardless if M 3. The negative result for MAB means that the twisting action of F is directed from B to A.

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Problem 4.33 The moment of force FE about line AB can be computed using MAB;1 or MAB;2 where rEAB MAB;1 D .ErAC FE / ; jErAB j

rEAB MAB;2 D .ErBC FE / ; jErAB j

where C is a point on the line of action of FE . Noting that rEAC D rEAB C rEBC , show that MAB;1 D MAB;2 .

Solution We summarize the information provided in the problem statement by writing the following expressions rE AB MAB;1 D rEAC FE ; rAB rE AB MAB;2 D rEBC FE ; rAB rEAC D rEAB C rEBC :

(1) (2) (3)

Substituting Eq. (3) into Eq. (1) provides rEAB MAB;1 D rEAB C rEBC FE rAB rE AB D rEAB FE C rEBC FE rAB rE rE AB AB D rEAB FE C rEBC FE : rAB rAB In Eq. (6), the result of rEAB FE is a vector that is orthogonal to rEAB , therefore, rEAB FE rEAB Thus, Eq. (6) reduces to EAB;1 D 0 C rEBC FE rEAB : M rAB

(4) (5) (6) D 0. (7)

Equation (7) is identical to Eq. (2), thus EAB;1 D M EAB;2 : M

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Problem 4.34 An automobile windshield wiper is actuated by a force FE D .120 {O Determine the moment of this force about shaft OB.

50 |O/ N.

Solution We use Eq. (4.2) on page 183 to determine the moment of force FE about point O as ˇ ˇ ˇ {O |O kO ˇˇ ˇ E O D rEOA FE D ˇ 0 M 60 80 ˇˇ Nmm ˇ ˇ 120 50 0 ˇ h i O D {O.4000/ |O. 9600/ C k.7200/ Nmm:

(1) (2)

E O about line OB is The component of M rEOB .4000/.20/ C .9600/.90/ C .7200/.60/ D Nmm rOB 110 D 12;500 Nmm D 12:5 Nm:

EO MOB D M

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(3) (4)

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Problem 4.35 In the windshield wiper of Prob. 4.34, if FE has 130 N magnitude, determine the direction in which it should be applied so that its moment about shaft OB is as large as possible, and determine the value of this moment.

Solution To maximize the moment of F about line OB, FE should be perpendicular to the plane containing lines OA and OB. Thus, we determine a vector nE , and unit vector n, O that are perpendicular to this plane as ˇ ˇ ˇ {O |O kO ˇˇ h i ˇ O 1200/ mm2 ; nE D rEOB rEOA D ˇˇ 20 90 60 ˇˇ mm2 D {O.10800/ |O.1600/ C k. (1) ˇ 0 ˇ 60 80 nO D

nE D 0:983 {O n

0:146 |O

O 0:109 k:

(2)

Using n, O we construct FE as FE D 130 N nO D 130 N 0:983 {O

0:146 |O

0:109 kO D 128 {O

19:0 |O

14:2 kO N:

(3)

We then use the scalar approach to determine the moment of this force about line OB as q MOB D F d D F rOA D .130 N/ . 60 mm/2 C .80 mm/2 D 13;000 Nmm D 13 Nm:

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Solutions Manual

Problem 4.36 Wrench AB is used to twist a nut on a threaded shaft. Point A is located 120 mm from point O, and line a has x, y, and ´ direction angles of 36ı , 60ı , and 72ı , respectively. Point B is located at .0; 300; 100/ mm, and the force is O N. FE D . 80 |O C 20 k/ EA , and then find the compo(a) Determine the moment of FE about point A, M nent of this moment about line a, Ma . E O , and then find the compo(b) Determine the moment of FE about point O, M nent of this moment about line a, Ma .

Solution Part (a)

Using the coordinates and direction angles provided, the following position vectors may be written rEOA D 120 mm cos 36ı {O C cos 60ı |O C cos 72ı kO D 97:1 {O C 60 |O C 37:1 kO mm; (1) h i O rEOB D {O.0 0/ C |O.300 0/ C k.100 0/ mm D 300 |O C 100 kO mm: (2)

It then follows that rEAB D rEAO C rEOB D rEOA C rEOB D 97:1 {O C 240 |O C 62:9 kO mm:

(3) (4)

We use Eq. (4.2) on page 183 to determine the moment of force FE about point A as

EA D rEAB M

ˇ |O kO ˇˇ 240 62:9 ˇˇ Nmm 80 20 ˇ i O |O. 1940/ C k.7770/ Nmm D 9830 {O C 1940 |O C 7770 kO Nmm:

ˇ ˇ ˇ E F D ˇˇ ˇ

h D {O.9830/

{O 97:1 0

(5) (6)

EA about line a is given by The component of M EA rOOA D .9830/.cos 36ı / C .1940/.cos 60ı / C .7770/.cos 72ı / Nmm Ma D M D 11;300 Nmm D 11:3 Nm:

(7) (8)

Part (b) We use Eq. (4.2) to determine the moment of force FE about point O as

E O D rEOB M

ˇ |O kO ˇˇ 300 100 ˇˇ Nmm 80 20 ˇ i O |O.0/ C k.0/ Nmm D 14000 {O Nmm:

ˇ ˇ {O ˇ E F D ˇˇ 0 ˇ 0

h D {O.14000/

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Statics 1e E O about line a is given by The component of M E O rOOA D .14000/.cos 36ı / C .0/.cos 60ı / C .0/.cos 72ı / Nmm Ma D M D 11;300 Nmm D 11:3 Nm:

(11) (12)

EA and M E O are different, yet Ma is the same. Observe that M

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Solutions Manual

Problem 4.37 A trailer has a rectangular door DEGH hinged about edge GH . If FE D O lb, determine the moment of F about edge GH . . 2 {O C 5 |O C 14 k/

Solution With the coordinates provided in the problem statement, the following position vector may be written rEHD D 16 |O 32 kO in. We then use Eq. (4.2) on page 183 to determine the moment of force FE about point H as ˇ ˇ ˇ {O |O kO ˇˇ ˇ E H D rEHD FE D ˇ 0 16 M 32 ˇˇ in.lb; ˇ ˇ 2 5 14 ˇ h i D 384 {O . 64/ |O C 32 kO in.lb:

(1)

(2) (3)

E H in the direction of line GH is simply the negative of the x component of M E H (or, The component of M E evaluate the dot product MGH D MH . {O/ to obtain the following result) MGH D

384 in.lb:

(4)

The negative sign for MGH means that the twisting action of the force is in the direction from H to G, according to the right-hand rule.

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Problem 4.38 In the trailer of Prob. 4.37, if F D 15 lb, determine the direction in which it should be applied so that its moment about edge GH of the rectangular door is as large as possible, and determine the value of this moment.

Solution To maximize the moment of FE about line GH , FE should be applied perpendicular (normal) to the door. From the information provided in the problem statement, the following position vectors may be written rOHG D {O; h rEHD D {O. 40 C 40/ C |O.24

(1) O 8/ C k.20

i 52/ in. D 16 |O

32 kO in.

We then determine a unit vector normal to these two vectors as ˇ ˇ ˇ {O |O kO ˇˇ h i ˇ O nE D rOHG rEHD D ˇˇ 1 0 in. 0 ˇˇ in. D |O. 32/ C k.16/ ˇ 0 16 32 ˇ O in. 32 | O C 16 k nE O nO D D p D 0:894 |O C 0:447 k: n .32 in:/2 C .16 in:/2

(2)

(3)

(4)

We then construct FE in this direction as FE D 15 lb nO D 15 lb 0:894 |O C 0:447 kO D 13:4 |O C 6:71 kO lb:

(5)

Using the scalar approach, the moment of FE about line GH is q MGH D F rHD D .15 lb/ .16 in:/2 C . 32 in:/2 D 537 in.lb:

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Problem 4.39 E D . {O C 4 |O C A trailer has a triangular door ABC hinged about edge BC . If Q O 8 k/ lb, determine the moment of Q about edge BC .

Solution With the coordinates provided in the problem statement, the following position vectors may be written h i O rEBA D {O.0 24/ C |O.28 4/ C k.12 12/ in. D . 24 {O C 24 |O/ in.; (1) h i O 12/ in. D 24 {O C 48 kO in. (2) rEBC D {O.0 24/ C |O.4 4/ C k.60 E about point B as We then use Eq. (4.2) on page 183 to determine the moment of force Q ˇ ˇ ˇ {O O ˇ | O k ˇ ˇ E B D rEBA Q E D ˇ 24 24 0 ˇ in.lb; M ˇ ˇ ˇ 1 4 8 ˇ h i D 192 {O . 192/ |O C . 120/ kO in.lb:

(3) (4)

E B in the direction of line BC is The component of M 2 E B rEBC D Œ.192/. 24/pC .192/.0/ C . 120/.48/ in. lb MBC D M rBC . 24 in:/2 C .48 in:/2 D 193 in.lb:

(5) (6)

The negative sign for MBC means that the twisting action of the force is in the direction from C to B, according to the right-hand rule.

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Problem 4.40 In the trailer of Prob. 4.39, if Q D 9 lb, determine the direction in which it should be applied so that its moment about edge BC of the triangular door is as large as possible, and determine the value of this moment.

Solution E about line BC , Q E should be applied perpendicular (normal) to the door. From To maximize the moment of Q the information provided in the problem statement, the following position vectors may be written rEBA D . 24 {O C 24 |O/ in. D . {O C |O/ 24 in:; rEBC D . 24 {O C 48 |O/ in. D {O C 2 kO 24 in:

(1) (2)

We then determine a unit vector normal to these two vectors as ˇ ˇ ˇ {O |O kO ˇ h i ˇ ˇ O nE D rEBA rEBC D ˇˇ 1 1 0 ˇˇ .24 in:/2 D {O.2/ |O. 2/ C k.1/ .24 in:/2 ˇ 1 0 2 ˇ 2 2 {O C 2 |O C kO .24 in:/ nE O nO D D D 0:667 {O C 0:667 |O C 0:333 k: p n .24 in:/2 .2/2 C .2/2 C .1/2

(3)

(4)

E in this direction as We then construct Q E D 9 lb 0:667 {O C 0:667 |O C 0:333 kO D 6 {O C 6 |O C 3 kO lb: Q E about point B is Using Eq. (4.2) on page 183, the moment of Q ˇ ˇ ˇ {O |O kO ˇ ˇ ˇ E B D rEBA Q E D ˇ 1 1 0 ˇ 24 in.lb M ˇ ˇ ˇ 6 6 3 ˇ h i O 12/ 24 in.lb D 3 {O C 3 |O D {O.3/ |O. 3/ C k.

(5)

(6) 12 kO 24 in.lb:

(7)

E B about line BC is The component of M C 2. 12/ .24 in.lb/.24 in:/ E B rEBC D Œ 1.3/ C 0.3/ MBC D M D p rBC . 1/2 C .2/2 .24 in:/

290 in.lb:

(8)

The negative sign for MBC means that the twisting action of the force is in the direction from C to B, according to the right-hand rule.

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Solutions Manual

Problem 4.41 A poorly leveled crane rotates about line a, which lies in the xy plane and has a y direction angle of 5ı . Line AB lies in the x´ plane. If the crane supports a weight W D 5000lb and ˛ D 45ı , determine the moment of W about line a.

Solution From the information provided in the problem statement, the following vector expressions may be written WE D rEAB

5000 |O lb; D 40 ft cos 45ı {O C sin 45ı kO :

(1) (2)

Using Eq. (4.2) on page 183, the moment of WE about point A is ˇ ˇ ˇ ˇ {O |O kO ˇ ˇ ı ı ˇ E E MA D rEAB W D ˇ cos 45 0 sin 45 ˇˇ .40 ft/. 5000 lb/ ˇ ˇ 0 1 0 h i O 0:707/ D 1:41105 ftlb {O C kO : D 2105 ftlb {O. 0:707/ |O.0/ C k.

(3) (4)

Noting that the a direction lies in the xy plane and has a y direction angle of 5ı , the unit vector uO in the a direction is uO D sin 5ı {O C cos 5ı |O: (5) EA in the a direction is given by The component of M EA uO D M EA Ma D M sin 5ı {O C cos 5ı |O D 1:41105 ftlb .1/. sin 5ı / C .0/.cos 5ı / C .1/.0/ D

12;300 ftlb:

(6) (7) (8)

The negative sign for Ma means that the twisting action of the weight is in the a direction, according to the right-hand rule.

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Problem 4.42 If the poorly leveled crane of Prob. 4.41 can have any position ˛, where 0 ˛ 135ı , determine the largest moment of the weight W D 5000 lb about line a and the position ˛ that produces this moment. Assume h D 40 ft for any position ˛ (this requires the operator to slightly change the boom’s height and/or length as the crane rotates about line a).

Solution From the information provided in the problem statement, the following vector expressions may be written WE D rEAB

5000 |O lb; D 40 ft cos ˛ {O C sin ˛ kO :

Using Eq. (4.2) on page 183, the moment of WE ˇ ˇ ˇ E E MA D rEAB W D ˇˇ ˇ h D 2105 ftlb

(1) (2)

about point A is {O |O kO cos ˛ 0 sin ˛ 0 1 0

(3)

sin ˛ {O

(4)

ˇ ˇ ˇ ˇ .40 ft/. 5000 lb/ ˇ ˇ i cos ˛ kO :

Noting that the a direction lies in the xy plane and has a y direction angle of 5ı , the unit vector uO in the a direction is uO D sin 5ı {O C cos 5ı |O: (5) EA in the a direction is given by The component of M EA uO D M EA Ma D M sin 5ı {O C cos 5ı |O D 2105 ftlb . sin ˛/. sin 5ı / C .0/.cos 5ı / C . cos ˛/.0/ D

17;400 ftlb.sin ˛/:

(6) (7) (8)

From this we see that Ma is maximum when sin ˛ D 1, which occurs when ˛ D 90ı . Therefore the maximum moment about line a is Ma D

17;400 ftlb

at ˛ D 90ı :

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Problem 4.43 Three-wheel and four-wheel all-terrain vehicles (ATVs) are shown. For both ATVs, the combined weight of the driver and vehicle is W D 2600 N. During a hard turn, the tendency for the ATV to tip is modeled by the force F which acts in the negative y direction. Determine the value of F such that the resultant moment of F and the weight about line AB is zero. Comment on which ATV is more prone to tip during hard turns. Note: One of these models of ATV are no longer manufactured because of their poor safety record.

Solution We first consider the three-wheel ATV. With point O being the origin of the coordinate system, the following vector expressions may be written rEAO D 0:9 {O C 0:8 kO m; (1) E D FE C WE D Q

F |O

O 2600 N k:

(2)

Note that the lines of action of both FE and WE intersect at point O. Using Eq. (4.2) on page 183, the moment E about point A is of force Q ˇ ˇ ˇ ˇ h {O |O kO i ˇ ˇ O ˇ E E MA D rEAO Q D ˇ 0:9 m 0 mF / : (3) 0:8 m ˇˇ D {O.0:8 mF / |O.2340 Nm/ C k.0:9 ˇ 0 F 2600 N ˇ Noting that the position vector from point A to point B is rEAB D . 1:6 {O C 0:6 |O/ m;

(4)

EA about line AB is given by the component of M EA rEAB MAB D M rAB .0:8 mF /. 1:6 m/ C . 2340 Nm/.0:6 m/ C .0:9 mF /.0 m/ D : p . 1:6 m/2 C .0:6 m/2

(5) (6)

As instructed in the problem statement, MAB D 0, hence we set Eq. (6) equal to zero and solve for F to obtain 1404 Nm F D D 1096 N ) F D 1096 N: (7) 1:28 m In Eq. (7), the negative sign for F is irrelevant and is thus ignored. For the four-wheel ATV, we use the scalar approach (Eq. (4.1) on page 182) to determine the moment of the forces about line AB as MAB D F .0:8 m/ 2600 N.0:6 m/: (8) This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc. It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited.

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325

As instructed in the problem statement, MAB D 0, hence we set Eq. (8) equal to zero and solve for F to obtain F D 1950 N: (9) Comparing Eqs. (7) and (9), we see that the force required to tip the four-wheel ATV is 78% larger than that for the three-wheel ATV. Note that three-wheel ATVs are no longer manufactured because of their poor safety record. Photo credit: © Richard Hamilton Smith/CORBIS.

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Problem 4.44 An open-end wrench applies the forces F to the head of a bolt to produce the moment M , where each force F is normal to the surface on which it acts. Determine M if F D 400 lb.

Solution Using the sketch shown at the right, we determine the dimension d to be d cos 30ı D

5 in. 16

)

d D 0:361 in:

(1)

Using Eq. (4.12) on page 209, we determine the moment of the couple to be M D F d D 400 lb.0:361 in:/ D 144 in.lb:

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(2)

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Problem 4.45 A box-end wrench applies the forces F to the head of a bolt to produce the moment M , where each force F is normal to the surface on which it acts. Determine F if M D 20 ftlb.

Solution Using the sketch shown at the right, we determine the dimension d to be d cos 30ı D

5 in. 16

)

d D 0:361 in:

(1)

Using Eq. (4.12) on page 209, we determine the moment of the three couples to be ! 12 in: M D 3F d D 20 ftlb D 3F .0:361 in:/ 1 ft

)

F D 222 lb:

(2)

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Problem 4.46 The top view of a workpiece that fits loosely in a fixture for drilling is shown. The drill bit has two edges that apply in-plane cutting forces F to the workpiece. (a) If F D 600 N, determine the forces Q between the workpiece and fixture so that the resultant couple moment is zero when ˛ D 30ı . (b) Does your answer for Q from Part (a) change if ˛ has different value? If yes, then repeat Part (a) with ˛ D 60ı .

Solution Part (a) The workpiece is subjected to two couples. Equation (4.12) on page 209 is used, with positive moment being counterclockwise, to determine the resultant moment of the two couples as M D Q.150 mm/

F .30 mm/:

(1)

As instructed in the problem statement, M D 0. Thus, we set Eq. (1) equal to zero, and using F D 600 N, we solve for Q to obtain Q D 120 N:

(2)

Part (b) The answer obtained in Part (a) is valid for any value of ˛.

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Problem 4.47 Three tugboats are used to turn a barge in a narrow channel. To avoid producing any net translation of the barge, the forces applied should be couples. The tugboat at point A applies a 400 lb force. (a) Determine FB and FC so that only couples are applied. (b) Using your answers to Part (a), determine the resultant couple moment that is produced. (c) Resolve the forces at A and B into x and y components, and identify the pairs of forces that constitute couples.

Solution Part (a) In order to have loading due to couple forces only, the resultant of all forces must be zero. Hence, X FRx D Fx D 0 W .400 lb/ sin 30ı FB sin 45ı D 0; (1) X FRy D Fy D 0 W .400 lb/ cos 30ı FB cos 45ı FC D 0: (2) Solving these equations provides FB D 282:8 lb;

(3)

FC D 146:4 lb:

(4)

Part (b) With the values given in Eqs. (3) and (4), the system of forces are couples, and the moment produced by these, taking positive moment to be counterclockwise, is X MR D MB D .400 lb/ cos 30ı .50 ft/ .400 lb/ sin 30ı .44 ft/ FC .40 ft/; (5) which provides MR D

31;980 ftlb:

(6)

Note that the same value of MR would be obtained regardless of the summation point used in Eq. (5), because the forces are couples. Part (c) This sketch below illustrates two force systems that are equivalent. The original force system is shown on the left, with forces resolved into x and y directions. The force system on the right shows more explicitly the three couples that are applied to the barge.

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Solutions Manual

Problem 4.48 A bracket with pulleys for positioning magnetic tape in an electronic data storage device is shown. The pulleys are frictionless, the tape supports a force T D 2 N, and the thickness of the tape can be neglected. (a) Use the two tape forces in Fig. p4.48(a) to compute the resultant couple moment. (b) Replace the forces on the two pulleys with point forces on the pinions as shown in Fig. p4.48(b), and compute the resultant couple moment (replacement of pulley forces by bearing forces is discussed in Chapter 3 in connection with Fig. 3.9).

Solution Part (a) Using the geometry shown in the figure to the right, we determine the moment arm d for the couple to be q d D 10 mm C .20 mm/2 C .20 mm/2 C 10 mm D 48:3 mm: (1) We then use Eq. (4.12) on page 209, taking positive moment to be counterclockwise, to determine the moment of the couple as M D T d D .2 N/.48:3 mm/ D 96:6 Nmm:

(2)

Part (b) Using Fig. (b) in the problem statement, the loading from the tape consists of two couples. Using Eq. (4.12), taking positive moment to be counterclockwise, the moment of the couples is M DT

q

.20 mm/2 C .20 mm/2 C T .20 mm/ D 96:6 Nmm:

(3)

As expected, the results of Parts (a) and (b) agree.

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Problem 4.49 Consider an object with forces FEA and FEB applied. The first column of the following table lists resultant moments at various points due to both of these forces. For each statement select True or False.

E C D rECA FEA C rECB FEB M E C D rEAB FEB M E C D rEBA FEA M EA D rEAB FEB M E D D rECA FEA C rECB FEB M

If FEA and FEB are not a couple

If FEA and FEB are a couple

T or F? T or F? T or F? T or F? T or F?

T or F? T or F? T or F? T or F? T or F?

Solution

E C D rECA FEA C rECB FEB M E C D rEAB FEB M E C D rEBA FEA M EA D rEAB FEB M E D D rECA FEA C rECB FEB M

If FA and FB are not a couple

If FA and FB are a couple

T F F T F

T T T T T

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Solutions Manual

Problem 4.50 A structure built in at point O supports 300 and 400 N couples. Determine the resultant couple moment vector, using both scalar and vector approaches.

Solution For the scalar approach, we take positive moments to be in the positive coordinate directions, according to the right-hand rule. Using Eq. (4.12) on page 209, the moment of the 300 N couple is M1 D 300 N.2 m 4 m/ D E 1 D 600 kO Nm: M

600 Nm;

(1) (2)

Using Eq. (4.12) again, the moment of the 400 N couple is M2 D 400 N.2 m

1:5 m/ D 200 Nm;

(3)

E 2 D 200 {O Nm: M

(4)

The resultant couple moment is the sum of Eqs. (2) and (4), which provides E DM E1 C M E 2 D 200 {O M

600 kO Nm:

(5)

For the vector approach, we use Eq. (4.11) on page 209, to determine the resultant couple moment is ˇ ˇ ˇ ˇ {O |O kO |O kO ˇˇ ˇˇ {O ˇ ˇ ˇ ˇ E E E M D rEBC FC C rEAB FB D ˇ 0 0 2 m 0 ˇ C ˇ 0 0:5 m ˇ 300 N 0 0 ˇ ˇ 0 0 400 N O 600 Nm/ C {O.200 Nm/ D 200 {O 600 kO Nm: E D k. M

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ˇ ˇ ˇ ˇ ˇ ˇ

(6) (7)

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Problem 4.51 A structure built in at point O supports 70 and 85 N couples and a tip moment. Determine the resultant couple moment, using both scalar and vector approaches.

Solution From the geometry provided in the problem statement, we write vector expressions for the forces applied at points B and D as 1 O FEB D .70 N/ . 6 {O C 3 |O 2 k/; 7 1 O FED D .85 N/ . 8 {O 12 |O C 9 k/: 17

(1) (2)

For the scalar approach, we evaluate the moments of each of the x, y, and ´ components of the 70 N and 85 N forces, taking positive moments to be in the positive coordinate directions according to the right-hand rule. Thus, using Eq. (4.12) on page 209, we obtain ! ! 3 6 E D 70 N M 2 m {O C 70 N 2 m |O C 0 kO 7 7 ! ! 9 8 C 85 N 2 m {O C 0 |O C 85 N 2 m kO 200 {O Nm; (3) 17 17 which provides E D M

50 {O C 120 |O C 80 kO Nm:

(4)

For the vector approach, the following position vectors are needed rEAB D

2 kO m;

rECD D 2 |O m:

(5)

We use Eq. (4.11) on page 209 to write E D rEAB FEB C rECD FED C M EE M ˇ ˇ ˇ ˇ ˇ O ˇ |O kO 70 Nm ˇˇ {O |O k ˇˇ 85 Nm ˇˇ {O D 2 ˇC 0 2 0 ˇ 0 0 7 17 ˇˇ ˇ 6 3 2 ˇ 8 12 9 i 85 Nm h 70 Nm h O D {O.6/ |O. 12/ C k.0/ C {O.18/ 7 17

(6) ˇ ˇ ˇ ˇ ˇ ˇ

200 {O Nm O |O.0/ C k.16/

(7) i

200 {O Nm;

(8)

which provides E D M

50 {O C 120 |O C 80 kO Nm:

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Solutions Manual

Problem 4.52 Determine the distance between the lines of action of the 70 N forces. Hint: Use the vector approach to determine the moment of this couple. Then note that the magnitude of this result is equal to .70 N/d; where d is the distance between the lines of action for the two forces.

Solution From the geometry provided in the problem statement, we write the following expressions rEAB D

1 FEB D .70 N/ . 6 {O C 3 |O 7

2 kO m;

O 2 k/;

(1)

where FEB is the force applied at point B. The moment due to the 70 N couple is obtained using Eq. (4.11) on page 209 as ˇ ˇ ˇ {O |O kO ˇ ˇ ˇ E D rEAB FEB D 70 Nm ˇ 0 0 M (2) 2 ˇˇ ˇ 7 ˇ 6 3 2 ˇ i 70 Nm h O D {O.6/ |O. 12/ C k.0/ ; (3) 7 E is and the magnitude of M M D 134 Nm:

(4)

We then use the scalar approach given by Eq. (4.12) on page 209 to solve for d as follows M D Fd

(5)

134 Nm D .70 N/d;

(6)

which is solved to obtain dD

134 Nm D 1:92 m: 70 N

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Problem 4.53 Determine the distance between the lines of action of the 85 N forces. See the hint in Prob. 4.52.

Solution From the geometry provided in the problem statement, we write the following expressions rECD D 2 |O m;

1 FED D .85 N/ . 8 {O 17

O 12 |O C 9 k/;

(1)

where FED is the force applied at point D. The moment due to the 85 N couple is obtained using Eq. (4.11) on page 209 as ˇ ˇ ˇ {O O ˇ | O k ˇ ˇ E D rECD FED D 85 Nm ˇ 0 M (2) 2 0 ˇˇ ˇ 17 ˇ 8 12 9 ˇ i 85 Nm h O D {O.18/ |O.0/ C k.16/ ; (3) 17 E is and the magnitude of M M D 120 Nm:

(4)

We then use the scalar approach given by Eq. (4.12) on page 209 to solve for d as follows M D Fd

(5)

120 Nm D .85 N/d;

(6)

which is solved to obtain dD

120 Nm D 1:42 m: 85 N

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Problem 4.54 The input shaft of a speed reducer supports a 200 Nm moment, and the output shafts support 300 and 500 Nm moments. (a) Determine the resultant moment applied by the shafts to the speed reducer. (b) If the speed reducer is bolted to a surface that lies in the xy plane, speculate on the characteristics of the forces these bolts apply to the speed reducer. In other words, do you expect these forces to constitute couples only, or may they be more general? Remark: Problems such as this are discussed in detail in Chapter 5.

Solution Part (a) The resultant moment applied to the speed reducer by the three shafts is simply the sum of the three moment vectors that are applied E D . 300 {O M

500 {O

200 |O/ Nm D . 800 {O

200 |O/ Nm:

(1)

Part (b) Since the input and output shafts apply moments only, there is no net force applied by the shafts to the speed reducer and hence we would expect the base to also apply no net force. The speed reducer is subjected to a resultant moment by the shafts, so we would expect the forces on the base, whatever they are, to be a system of couples with a resultant of .800 {O C 200 |O/ Nm so that the total moment applied to the speed reducer is zero.

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Problem 4.55 Satellites and other spacecraft perform attitude positioning using thrusters that are fired in pairs so as to produce couples. If thrusters at points A, B, C , and D each produce 3 N forces, determine the resultant moment and hence the axis through the center of mass about which an initially nonrotating satellite will begin to rotate.

Solution From the information provided in the problem statement, the following vector expressions may be written rEAC D .1:2 |O

O 1:6 k/m; O D . 1:2 |O 1:6 k/m;

FEC D 3 kO N;

(1)

rEBD

FED D

(2)

3 {O N:

Using Eq. (4.11) on page 209, the resultant moment of the couples is given by E D rEAC FEC C rEBD FED M ˇ ˇ ˇ ˇ {O ˇ ˇ {O |O kO |O kO ˇ ˇ ˇ D ˇˇ 0 1:2 m 1:6 m ˇˇ C ˇˇ 0 1:2 m 1:6 m ˇ 0 ˇ ˇ 0 3N 3N 0 0 h i O 3:6/ Nm; D 3:6 Nm {O C {O.0/ |O. 4:8/ C k.

(3) ˇ ˇ ˇ ˇ ˇ ˇ

(4) (5)

which provides E D 3:6 {O C 4:8 |O M

3:6 kO Nm:

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Problem 4.56 Forces F and T exerted by air on a rotating airplane propeller are shown. Forces F lie in the xy plane and are normal to the axis of each propeller blade, and thrust forces T act in the ´ direction. Show that the forces F can be represented as a couple or system of couples.

Solution The figures shown at the right are views looking down the ´ axis toward the propeller; forces T are not shown and these are irrelevant to the question being asked. We resolve the forces F into horizontal and vertical components, as shown in the second figure at the right. We then observe that the vertical force F cos 30ı in the positive y direction and the vertical force F cos 30ı in the negative y direction constitute a couple. The sum of the two horizontal forces F sin 30ı in the negative x direction and the force F in the positive x direction constitute another couple.

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Problem 4.57 If the structure is subjected to couple forces applied at points A and B and the O lb, determine the moment of the force applied at A is FE D .8 {O C 10 |O 40 k/ couple about line a. Line a has direction angles x D 72ı , y D 36ı , and ´ D 60ı .

Solution E of the couple is is given by Using Eq. (4.11) on page 209, the moment M E D rEBA FEA D 24 {O C 12 |O 8 kO in. 8 {O C 10 |O M ˇ ˇ ˇ {O |O kO ˇˇ ˇ D ˇˇ 24 12 8 ˇˇ in.lb ˇ 8 10 40 ˇ D 400 {O C 896 |O C 144 kO in.lb:

40 kO lb

(1) (2) (3)

E in the direction of line a is then given by The component of M E aO D Ma D M

400 {O C 896 |O C 144 kO in.lb cos 72ı {O C cos 36ı |O C cos 60ı kO

D 673 in.lb:

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(4) (5)

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Problem 4.58 The structure shown has two forces of magnitude F applied. If F D 42 lb and the forces are to be a couple, determine the orientation they should have so that the moment of the couple about line a is as large as possible. Line a has direction angles x D 72ı , y D 36ı , and ´ D 60ı .

Solution Two solutions will be provided. Solution 1 does not require careful visualization, is straightforward in concept, but will require the use of a computer mathematics program to solve a system of nonlinear equations. Solution 2 is straightforward in terms of the mathematics involved, but the optimal direction for the forces must be identified by inspection. Solution 1 From the information given in the problem statement, the position vector from point B to point A, and a unit vector in the a direction, may be written as O in.; 8 k/

rEBA D .24 {O C 12 |O

O uO a D cos 72ı {O C cos 36ı |O C cos 60ı k:

Let the force applied at point A be expressed as FEA D 42 lb c1 {O C c2 |O C c3 kO

where c12 C c22 C c32 D 1:

(1)

(2)

In the above expression, c1 , c2 , and c3 are the direction cosines for FE . The moment of the couple about line a can be determined using the scalar triple product as ˇ ˇ ˇ cos 72ı cos 36ı cos 60ı ˇ ˇ ˇ 12 8 ˇˇ 42 in.lb Ma D rEBA FEA uO a D ˇˇ 24 (3) ˇ c1 ˇ c2 c3 D cos 72ı .12c3 C 8c2 / cos 36ı .24c3 C 8c1 / C cos 60ı .24c2 12c1 / 42 in.lb: (4) Now determine c1 , c2 , and c3 so that Ma is maximized subject to the constraint that c12 C c22 C c32 D 1. We will use Mathematica to follow through on this strategy: eq1 = c12 + c22 + c32

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Engineering Mechanics: Statics 1st Edition

Michael E. Plesha University of Wisconsin–Madison

Gary L. Gray The Pennsylvania State University

Francesco Costanzo The Pennsylvania State University With the assistance of: Chris Punshon Andrew J. Miller Justin High Chris O’Brien Chandan Kumar Joseph Wyne Jonathan Fleischmann

Version: August 24, 2009

The McGraw-Hill Companies, Inc.

Copyright © 2002–2010 Michael E. Plesha, Gary L. Gray, and Francesco Costanzo

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Statics 1e

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Accuracy of Numbers in Calculations Throughout this solutions manual, we will generally assume that the data given for problems is accurate to 3 significant digits. When calculations are performed, all intermediate numerical results are reported to 4 significant digits. Final answers are usually reported with 3 significant digits. If you verify the calculations in this solutions manual using the rounded intermediate numerical results that are reported, you should obtain the final answers that are reported to 3 significant digits.

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Chapter 4 Solutions Problem 4.1 Compute the moment of force F about point B, using the following procedures. (a) Determine the moment arm d and then evaluate MB D F d . (b) Resolve force F into x and y components at point A and use the principle of moments. (c) Use the principle of moments with F positioned at point C . (d) Use the principle of moments with F positioned at point D. (e) Use a vector approach.

Solution Part (a) We first use geometry and trigonometry to find that tan 1 . 3=4/ D 36:9ı , tan 1 . 4=3/ D 53:1ı , and 180ı 60ı 53:1ı D 66:9ı as shown in the figure to the right. We then use the law of sines with triangle ABD to find that 12 in: b D ) b D 13:8 in: (1) ı sin 53:1 sin 66:9ı Through trigonometry, it follows that the moment arm d is d D b sin 53:1ı D 11:0 in:

(2)

Using Eq. (4.1) on page 182, and noting that positive moment is counterclockwise, MB is then given by MB D F d D .25 lb/.11:0 in:/ D 276 in.lb:

(3)

Part (b) Using the figure to the right, and noting that positive moment is counterclockwise, Eq. (4.1) provides ! ! 3 4 MB D 25 lb .12 in: sin 60ı / C 25 lb .12 in: cos 60ı / D 276 in.lb: 5 5 (4)

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Statics 1e Part (c) Using triangle ABC shown at the right, the law of sines provides 12 in: h D ı sin 36:9 sin 113:1ı

)

h D 18:4 in:

(5)

In the figure to the right, we see that the vertical force at C produces no moment about B. By using Eq. (4.1), and noting that positive moment is counterclockwise, MB is given by 3 MB D 25 lb .18:4 in:/ D 276 in.lb: 5

(6)

Part (d) In the figure to the right, we see that the horizontal force on D produces no moment about B. Using Eq. (4.1), and noting that positive moment is counterclockwise, MB is given by 4 MB D 25 lb .13:8 in:/ D 276 in.lb: 5

(7)

Part (e) We find that rEBA D 12 in: cos 60ı {O C sin 60ı |O 3 4 E F D 25 lb {O |O : 5 5

(8) (9)

E B is given by Using Eq. (4.2) on page 183, M ˇ ˇ ˇ E B D rEBA FE D .12 in:/.25 lb/ ˇ M ˇ ˇ

ˇ {O |O kO ˇˇ cos 60ı sin 60ı 0 ˇˇ D 276 kO in.lb: 3 4 0 ˇ 5 5

(10)

Note that instead of Eq. (8), we could have used rEBC or rEBD instead.

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Problem 4.2 Compute the moment of force F about point B, using the following procedures. (a) Determine the moment arm d and then evaluate MB D F d . (b) Resolve force F into x and y components at point A and use the principle of moments. (c) Use the principle of moments with F positioned at point C . (d) Use the principle of moments with F positioned at point D. (e) Use a vector approach.

Solution Part (a) Using the 30ı and 60ı angles given in the problem statement, we identify the additional angles shown in the figure at the right. We then use the law of sines for triangle ABD to determine b 12 in: D ı sin 30 sin 30ı

)

b D 12 in:

(1)

Through trigonometry, it follows that the moment arm d is d D b sin 30ı D 6 in:

(2)

Using Eq. (4.1) on page 182, and noting that positive moment is counterclockwise, MB is then given by MB D F d D

25 lb.6 in:/ D

150 in.lb:

(3)

Part (b) Using the figure to the right, and noting that positive moment is counterclockwise, Eq. (4.1) provides MB D D

25 lb cos 30ı .12 in: sin 60ı / C 25 lb sin 30ı .12 in: cos 60ı /

(4)

150 in.lb:

(5)

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Statics 1e Part (c) Using triangle ABC shown at the right, the law of sines provides h 12 in: D ı sin 120 sin 30ı

)

h D 6:93 in:

(6)

In the figure to the right, we see that the vertical force at C produces no moment about B. Using Eq. (4.1), and noting that positive moment is counterclockwise, MB is given by MB D

25 lb cos 30ı .6:93 in:/ D

150 in.lb:

(7)

Part (d) In the figure to the right, we see that the horizontal force at D produces no moment about B. Using Eq. (4.1), and noting that positive moment is counterclockwise, MB is given by MB D

25 lb sin 30ı .12 in:/ D

150 in.lb:

(8)

Part (e) We find that cos 60ı {O C sin 60ı |O FE D 25 lb cos 30ı {O sin 30ı |O :

rEBA D 12 in:

(9) (10)

E B is given by Using Eq. (4.2) on page 183, M ˇ ˇ {O ˇ E E MB D rEBA F D .12 in:/.25 lb/ ˇˇ cos 60ı ˇ cos 30ı

ˇ |O kO ˇˇ sin 60ı 0 ˇˇ D sin 30ı 0 ˇ

150 kO in.lb:

(11)

Rather than rEBA , we could have used rEBC or rEBD instead.

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Problem 4.3 The cover of a computer mouse is hinged at point B so that it may be “clicked.” Repeat Prob. 4.1 to determine the moment about point B.

Solution Part (a) Using trigonometry and the figure to the right, we determine that the moment arm d is d D 60 mm sin 45ı D 42:4 mm

(1)

Using Eq. (4.1) on page 182, and noting that positive moment is counterclockwise, MB is given by MB D F d D

3 N.42:4 mm/ D

127 Nmm:

(2)

Part (b) We first use geometry to determine h D 20 mm as shown in the figure to the right. Using Eq. (4.1), and noting that positive moment is counterclockwise, we determine that MB D D

3 N cos 45ı .20 mm/

3 N sin 45ı .40 mm/

127 Nmm:

(3) (4)

Part (c) In the figure to the right, we see that the vertical force at C produces no moment about B. Using Eq. (4.1), and noting that positive moment is counterclockwise, MB is given by MB D

3 N cos 45ı .60 mm/ D

127 Nmm:

(5)

Part (d) In the figure to the right, we see that the horizontal force at D produces no moment about B. Using Eq. (4.1), and noting that positive moment is counterclockwise, MB is given by MB D

3 N sin 45ı .60 mm/ D

127 Nmm

(6)

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Statics 1e Part (e) Using the value of h found in Part (b), we find that rEBA D .40 {O C 20 |O/ mm FE D 3 N sin 45ı {O

(7)

cos 45ı |O :

(8)

E B is given by Using Eq. (4.2) on page 183, M ˇ ˇ {O ˇ ˇ E E MB D rEBA F D 3 Nmm ˇ 40 ˇ sin 45ı

ˇ |O kO ˇˇ 20 0 ˇˇ D ı cos 45 0 ˇ

127 kO Nmm

(9)

Rather than rEBA , we could have used rEBC or rEBD instead.

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Problem 4.4 The cover of a computer mouse is hinged at point B so that it may be “clicked.” Repeat Prob. 4.1 to determine the moment about point B.

Solution Part (a) We use some elementary geometry to determine the dimensions shown in the figure to the right. The moment arm d is then d D 20 mm cos 45ı D 14:1 mm

(1)

Using Eq. (4.1) on page 182, and noting that positive moment is counterclockwise, MB is given by MB D F d D

3 N.14:1 mm/ D

42:4 Nmm:

(2)

Part (b) Using Eq. (4.1), and noting that positive moment is counterclockwise, we determine MB D 3 N cos 45ı .20 mm/ D

3 N sin 45ı .40 mm/

42:4 Nmm:

(3) (4)

Part (c) In the figure to the right, we see that the vertical force at C produces no moment about B. Using Eq. (4.1) again, MB is given by MB D

3 N cos 45ı .20 mm/ D

42:4 Nmm:

(5)

Part (d) We see that the horizontal force on D produces no moment about B in the figure to the right, then by using Eq. (4.1), and noting that positive moment is counterclockwise, MB is given by MB D

3 N sin 45ı .20 mm/ D

42:4 Nmm:

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(6)

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Statics 1e Part (e) We find that rEBA D .40 {O C 20 |O/ mm FE D 3 N

cos 45ı {O

(7) sin 45ı |O :

(8)

E B is given by Using Eq. (4.2) on page 183, M ˇ ˇ ˇ E E MB D rEBA F D 3 Nmm ˇˇ ˇ

{O 40 cos 45ı

ˇ |O kO ˇˇ 20 0 ˇˇ D ı sin 45 0 ˇ

42:4 kO Nmm

(9)

Rather than rEBA , we could have used rEBC or rEBD instead.

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Solutions Manual

Problem 4.5 An atomic force microscope (AFM) is a state-of-the-art device used to study the mechanical and topological properties of surfaces on length scales as small as the size of individual atoms. The device uses a flexible cantilever beam AB with a very sharp, stiff tip BC that is brought into contact with the surface to be studied. Due to contact forces at C , the cantilever beam deflects. If the tip of the AFM is subjected to the forces shown, determine the resultant moment of both forces about point A. Use both scalar and vector approaches.

Solution For the scalar approach we use Eq. (4.1) on page 182. Noting that positive moment is counterclockwise, MA is given by

MA D D

.13 nN/.12m/ C .5 nN/.1:5m/ D 1:4910

13

! ! 10 9 N 10 6 m 149 nNm nN m

Nm:

(1) (2)

For the vector approach we use Eq. (4.2) on page 183. It follows that rEAC D . 12 {O

1:5 |O/ m

FE D .5 {O C 13 |O/ nN

(3) (4)

EA is given by and M EA D rEAC M D D

ˇ ˇ ˇ FE D ˇˇ ˇ

{O 12 5

ˇ |O kO ˇˇ 1:5 0 ˇˇ nNm 13 0 ˇ

149 kO nNm 1:4910 13 kO Nm:

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(5) (6) (7)

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Problem 4.6 The door of an oven has 22 lb weight which acts vertically through point D. The door is supported by a hinge at point A and two springs that are symmetrically located on each side of the door. For the positions specified below, determine the force needed in each of the two springs if the resultant moment about point A of the weight and spring forces is to be zero. (a) ˛ D 45ı . (b) ˛ D 90ı . (c) Based on your answers to Parts (a) and (b), determine an appropriate stiffness for the springs.

Solution Part (a) We first determine some needed dimensions using trigonometry as shown in the figure at the left below. The vertical and horizontal distances between points A and C are .5 in:/ sin 45ı D .5 in:/ cos 45ı D 3:54 in. Then, using the Pythagorean theorem we determine the distance between points B and C as q (1) LBC D .10 in: C 3:54 in:/2 C .8 in: 3:54 in:/2 D 14:25 in: Let F be the force in one spring. The forces in the springs act along line BC , therefore the total horizontal and vertical forces at C from the springs are, respectively, Fhorizontal D 2F .13:54 in:=14:25 in:/;

Fvertical D 2F .4:46 in:=14:25 in:/;

(2)

and these forces are shown in the figure at the right below. The moment arm for the 22 lb weight is .12 in:/ cos 45ı D 8:49 in: as shown in the figure to the right below.

With positive moment being counterclockwise, the resultant moment of the two spring forces and the weight of the door about point A is MA D

.22 lb/.8:49 in:/ C 2F

13:54 in: 4:46 in: .3:54 in:/ C 2F .3:54 in:/: 14:25 in: 14:25 in:

(3)

Requiring MA D 0, as stated in the problem statement, we set Eq. (3) equal to zero to solve for F as F D 21:0 lb:

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(4)

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Part (b) When the door is horizontal (fully open), points A, B, and C have the positions shown in the figure at the left below. The Pythagorean theorem is used to determine the distance between points B and C as q LBC D .10 in: C 5 in:/2 C .8 in:/2 D 17 in: (5) Let F be the force in one spring. The forces in the springs act along line BC , therefore the total horizontal and vertical forces at C from the springs are, respectively Fhorizontal D 2F .15 in:=17 in:/;

Fvertical D 2F .8 in:=17 in:/:

(6)

and these forces are shown in the figure to the right below.

With positive moment being counterclockwise, and noting that the horizontal force at C produces no moment about point A, the resultant moment of the two spring forces and the weight of the door about point A is MA D

.22 lb/.12 in:/ C 2F

8 in: .5 in:/: 17:0 in:

(7)

Requiring MA D 0, as stated in the problem statement, we set Eq. (7) equal to zero to solve for F as F D 56:1 lb:

(8)

Part (c) We use the spring law to determine the necessary spring stiffness so that the springs will produce the forces determined in Parts (a) and (b) for ˛ D 45ı and ˛ D 90ı . Thus F D kı

)

kD

F ı

)

kD

change in force 56:1 lb D change in length 17:0 in:

21:0 lb D 13:0 lb=in: 14:3 in:

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(9)

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Problem 4.7 The ball of a trailer hitch is subjected to a force F . If failure occurs when the moment of F about point A reaches 10,000 in.lb, determine the largest value F may have and specify the value of ˛ for which the moment about A is the largest. Note that the value of F you determine must not produce a moment about A that exceeds 10,000 in.lb for any possible value of ˛.

Solution The maximum moment for a given value of force F occurs when the moment arm is at its maximum. Using the Pythagorean theorem in the figure to the right, the maximum value of the moment arm d is given by q d D .9 in:/2 C .6 in:/2 D 10:8 in:

(1)

We then use Eq. (4.1) on page 182 to determine the maximum force F , given that the moment of this force about point A may not exceed 10;000 in.lb. Thus, MA D F d D F .10:8 in:/ D 10;000 in.lb

)

F D 926 lb:

(2)

The angle ˛ at which this maximum force acts is then found to be d cos ˇ D 9 in: ˛ D 180ı

.180ı

) ˇ

ˇ D 33:6ı 90ı / D 123:6ı :

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(3) (4)

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Problem 4.8 Repeat Prob. 4.7 if the moment at point B may not exceed 5000 in.lb.

Solution The maximum moment for a given value of force F occurs when the moment arm is at its maximum. Using the Pythagorean theorem in the figure to the right, the maximum value of the moment arm d is given by q d D .5 in:/2 C .6 in:/2 D 7:81 in:

(1)

We then use Eq. (4.1) on page 182 to determine the maximum force F , given that the moment of this force about point B may not exceed 5000 in.lb. Thus, MB D F d D F .7:81 in:/ D 5000 in.lb

)

F D 640 lb:

(2)

The angle ˛ at which this maximum force acts is then found to be d cos ˇ D 5 in: ˛ D 180ı

.180ı

) ˇ

ˇ D 50:2ı

(3)

90ı / D 140ı :

(4)

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Problem 4.9 The port hull of a catamaran (top view shown) has cleats at points A, B, and C . A rope having 100 N force is to be attached to one of these cleats. If the force is to produce the largest possible counterclockwise moment about point O, determine the cleat to which the rope should be attached and the direction the rope should be pulled (measured positive counterclockwise from the positive x direction). Also, determine the value of MO produced. Assume all cleats, point O, and the rope lie in the same plane.

Solution Cleat C offers the largest possible moment arm, therefore Attaching the rope to cleat C will produce the largest moment.

(1)

As shown to the right, the Pythagorean theorem is used to determine the moment arm d as q (2) d D .1:5 m/2 C .3 m/2 D 3:35 m: The angle at which the force acts is given by d cos ˇ D 3 m

)

ˇ D 26:4ı

˛ D 90ı C ˇ D 116:4ı :

(3) (4)

The moment of the 100 N force about O is given by Eq. (4.1) on page 182 as MO D 100 N.3:35 m/ D 335 Nm:

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(5)

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Problem 4.10 Frame ABC has a frictionless pulley at C around which a cable is wrapped. Determine the resultant moment about point A produced by the cable forces if W D 5 kN and (a) ˛ D 0ı . (b) ˛ D 90ı . (c) ˛ D 30ı .

Solution Part (a) T D W D 5 kN for the pulley to be in equilibrium. We may determine the moment of the cable forces about point A by considering the forces supported by the cables, or by considering the forces the pulley applies to the structure at point C . Both of these options are illustrated in the figures below, to the left and right, respectively.

Use of either of the force systems shown above is convenient, and we will use the force system shown at the left. Noting that positive moment is counterclockwise, we then use Eq. (4.1) on page 182 to determine the moment about point A as MA D 5 kN.3 m C 0:5 m/ C 5 kN.3 m

0:5 m/

(1)

D 30:0 kNm:

(2)

Part (b) We use the same procedure as for Part (a), and use either of the following force systems.

Use of either of the force systems shown above is convenient, and we will use the force system shown at the left. Noting that positive moment is counterclockwise, we then use Eq. (4.1) on page 182 to determine the moment about point A as MA D 5 kN.4 m C 0:5 m/ C 5 kN.3 m

0:5 m/

D 35:0 kNm: This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc. It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited.

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Part (c) In principle, either of the force systems shown below may be used to determine the moment about point A.

However, if the force system at the left is used, it will be tedious to determine the moment arm for the cable force with the 30ı orientation. Thus, it will be considerably more convenient to use the force system shown at the right, which provides MA D 5 kN.3 m/ C 5 kN.cos 30ı /3 m C 5 kN.sin 30ı /4 m D 38:0 kNm:

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(5) (6)

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Problem 4.11 Frame ABC has a frictionless pulley at C around which a cable is wrapped. If the resultant moment about point A produced by the cable forces is not to exceed 20 kNm, determine the largest weight W that may be supported and the corresponding value of ˛.

Solution First we determine the angle ˛ that provides the largest moment arm for force T . Using the figure shown at the right, it follows that tan ˇ D

4m 3m

)

ˇ D ˛ D 53:1ı :

(1)

Sincepthe pulley is in equilibrium, T D W . The Pythagorean theorem provides d D .4 m/2 C .3 m/2 D 5 m. We then use Eq. (4.1) on page 182, to evaluate the moment of the cable forces about point A, noting that positive moment is counterclockwise MA D W .3 m/ C W cos 53:1ı .3 m/ C W sin 53:1ı .4 m/:

(2)

As given in the problem statement, MA is not to exceed 20 kNm. Thus we set Eq. (2) equal to 20 kNm and solve for W to obtain W D 2:5 kN: (3)

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Statics 1e

Problem 4.12 The load carrying capacity of the frame of Prob. 4.11 can be increased by placing a counterweight Q at point D as shown. The resultant moment about point A due to the cable forces and Q is not to exceed 20 kNm. (a) Determine the largest value of counterweight Q. (Hint: Let W D 0 and determine Q so that MA D 20 kNm.) (b) With the value of Q determined in Part (a), determine the largest weight W that may be supported and the corresponding value of ˛.

Solution Part (a) We begin by setting W D 0, and then use Eq. (4.1) on page 182 to evaluate the moment of Q about point A, noting that this moment may not exceed 20 kNm. With positive moment being counterclockwise, it follows that MA D

Q.2 m/ D

20 kNm

)

Q D 10 kN:

(1)

Part (b) With W ¤ 0, we know that T D W , since the pulley is in equilibrium. As determined in the solution to Prob. 4.11, the moment of T about point A is largest when ˛ D 53:1ı (the moment of T about point A is largest when its line of action is perpendicular to the line connecting points A and C , as shown above). Using the figure above, we evaluate the moment about point A, taking positive moment to be counterclockwise, and we require this moment to not exceed 20 kNm. Thus, MA D

10 kN.2 m/ C W .3 m/ C W cos 53:1ı .3 m/ C W sin 53:1ı .4 m/ D 20 kNm

W D 5 kN:

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(2) (3)

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Problem 4.13 A flat rectangular plate is subjected to the forces shown, where all forces are parallel to the x or y axis. If F D 200 N and P D 300 N, determine the resultant moment of all forces about the (a) ´ axis. (b) a axis, which is parallel to the ´ axis.

Solution Part (a) We use Eq. (4.1) on page 182 to find M´ D F .300 mm/ D

F .500 mm/ C P .200 mm/

5

1:610 Nmm D

P .600 mm/

160 Nm:

(1) (2)

In writing the above expression, positive moment is taken to be in the positive ´ direction, according to the right-hand rule. Part (b) We use Eq. (4.1) again to find Ma D D

F .300 mm/ C F .200 mm/ 5

1:710 Nmm D

P .500 mm/

170 Nm:

(3) (4)

In writing the above expression, positive moment is taken to be in the positive a direction, according to the right-hand rule.

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Problem 4.14 A flat rectangular plate is subjected to the forces shown, where all forces are parallel to the x or y axis. If P D 300 N, determine F when the resultant moment of all forces about the ´ axis is M´ D 100 Nm.

Solution Use Eq. (4.1) on page 182 to evaluate the moment of F and P D 300 N about the ´ axis M´ D F .300 mm/

F .500 mm/ C 300 N.200 mm/

300 N.600 mm/:

(1)

In writing the above expression, positive moment is taken to be in the positive ´ direction, according to the right-hand rule. As given in the problem statement, M´ D 100 Nm. Thus, Eq. (1) becomes mm 100 Nm 103 D F .300 mm/ F .500 mm/ C 300 N.200 mm/ 300 N.600 mm/; (2) m which may be solved for F D

100 N:

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(3)

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Problem 4.15 Repeat Prob. 4.14 if the resultant moment of all forces about the a axis is Ma D 100 Nm.

Solution Use Eq. (4.1) on page 182 to evaluate the moment of F and P D 300 N about the a axis Ma D

F .300 mm/ C F .200 mm/

300 N.500 mm/:

(1)

In writing the above expression, positive moment is taken to be in the positive a direction, according to the right-hand rule. As given in the problem statement, Ma D 100 Nm. Thus, Eq. (1) becomes mm 100 Nm 103 D F .300 mm/ C F .200 mm/ 300 N.500 mm/; (2) m which may be solved for F D

500 N:

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(3)

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Problem 4.16 Structure OBCD is built in at point O and supports a 50 lb cable force at point C and 100 and 200 lb vertical forces at points B and D, respectively. Using a vector approach, determine the moment of these forces about (a) point B. (b) point O.

Solution From the figure in the problem statement, the following vector expressions may be written rEBC D 20 kO in. rEBD D 15 {O C 20 kO in. rEOC D 36 kO in. rEOD D 15 {O C 36 kO in. FE D

200 |O lb

PE D

100 |O lb

12 {O C 18 |O TE D 50 lb 42

36 kO

:

Part (a) Observing that PE produces no moment about B, we use Eq. (4.2) on page 183 to evaluate the resultant moment of TE and FE about point B as E B D rEBC TE C rEBD FE M ˇ ˇ ˇ ˇ {O |O ˇ {O O ˇ k |O kO ˇ ˇ ˇ .20 in:/.50 lb/ ˇ ˇ ˇ C 200 in.lb D 1 ˇ ˇ 0 0 ˇ 15 0 20 42 ˇ 12 18 ˇ 0 36 ˇ 1 0 h i h 1000 in.lb O D {O. 18/ |O. 12/ C k.0/ C 200 in.lb {O.20/ 42 E B D 3570 {O C 286 |O 3000 kO in.lb: M

(1) ˇ ˇ ˇ ˇ ˇ ˇ

(2)

i O 15/ |O.0/ C k.

(3) (4)

Part (b) We use Eq. (4.2) on page 183 to evaluate the resultant moment of TE , FE , and PE about point O as E O D rEOC TE C rEOD FE C rEOB PE M ˇ ˇ ˇ ˇ {O ˇ {O |O O ˇ |O kO k ˇ ˇ .36 in:/.50 lb/ ˇˇ ˇ ˇ D 1 ˇ C 200 in.lb ˇ 15 0 36 ˇ 0 0 42 ˇ 0 ˇ 12 18 1 0 36 ˇ ˇ ˇ ˇ {O |O kO ˇ ˇ ˇ C .100 lb/.16 in:/ ˇˇ 0 0 1 ˇˇ ˇ 0 1 0 ˇ i h 1800 in.lb h O D {O. 18/ |O. 12/ C k.0/ C 200 in.lb {O.36/ 42 C 1600 {O in.lb E O D 8030 {O C 514 |O M

(5) ˇ ˇ ˇ ˇ ˇ ˇ

(6)

i O 15/ |O.0/ C k.

3000 kO in.lb:

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(7)

(8)

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Problem 4.17 Repeat Prob. 4.16, using a scalar approach.

Solution The vector expression for TE is 12 {O C 18 |O TE D 50 lb 42

36 kO

:

(1)

Using this expression, the figure at the right is drawn, where the force TE is shown in terms of its components. We then use E B in Eq. (4.1) on page 182 to determine the components of M each of the coordinate directions, taking positive moment components to be in the positive coordinate directions. Part (a) It follows that .MB /x D 200 lb.20 in:/

50 lb

18 .20 in:/ D 3570 in.lb 42

12 .20 in:/ D 286 in.lb 42 200 lb.15 in:/ D 3000 in.lb:

(2)

.MB /y D 50 lb

(3)

.MB /´ D

(4)

Part (b) It follows that .MO /x D 200 lb.36 in:/

50 lb

18 .36 in:/ C 100 lb.16 in:/ D 8030 in.lb 42

12 .36 in:/ D 514 in.lb 42 200 lb.15 in:/ D 3000 in.lb:

(5)

.MO /y D 50 lb

(6)

.MO /´ D

(7)

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Problem 4.18 Structure OAB is built in at point O and supports forces from two cables. Cable CAD passes through a frictionless ring at point A, and cable DBE passes through a frictionless ring at point B. If the force in cable CAD is 250 N and the force in cable DBE is 100 N, use a vector approach to determine (a) the moment of all cable forces about point A. (b) the moment of all cable forces about point O.

Solution From the figure provided in the problem statement, the following vector expressions may be written 150 {O 200 kO FEBD D 100 N 250 150 |O 200 kO FEBE D 100 N 250 60 | O 80 kO FEAC D 250 N 100 150 {O 80 kO : FEAD D 250 N 170

rEAB D 120 kO mm rEOB D 200 kO mm rEOA D 80 kO mm

These force vectors are shown in the figure at the right. Part (a) Noting that FEAC and FEAD produce no moment about point A, we use Eq. (4.2) on page 183 to determine

EA D rEAB M

kO 1 400

ˇ ˇ {O |O ˇ .120 mm/.100 N/ ˇ 0 FEBD C FEBE D 0 ˇ 250 ˇ 150 150

D 48 Nmm ŒO{ . 150/

ˇ ˇ ˇ ˇ ˇ ˇ

(1)

|O. 150/

(2)

D . 7200 {O C 7200 |O/ Nmm:

(3)

Part (b) We use Eq. (4.2) to determine E O D rEOB FEBD C FEBE C rEOA FEAC M ˇ ˇ |O kO .200 mm/.100 N/ ˇˇ {O D 0 1 ˇ 0 250 ˇ 150 150 400 D 80 Nmm ŒO{ . 150/

C FEAD ˇ ˇ ˇ ˇ {O ˇ ˇ ˇ C .80 mm/.250 N/ ˇ 0 ˇ ˇ ˇ 150 ˇ

|O. 150/ C 20;000 Nmm ŒO{ .0:6/

170

(4) kO 1

|O 0 60 100

|O. 0:882/

D .29;600 |O/ Nmm:

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80 100

80 170

ˇ ˇ ˇ ˇ ˇ ˇ

(5) (6) (7)

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Problem 4.19 Repeat Prob. 4.18, using a scalar approach.

Solution We begin by adding the two cable forces at point A to obtain FEA , and adding the two cable forces at point B to obtain FEB , as follows 60 |O 80 kO 150 |O 80 kO FEA D FEAC C FEAD D 250 N C 250 N D 221 {O 150 |O 318 kO N; 100 170 O 150 |O 200 kO 150 {O 200 k C 100 N D 60 {O C 60 |O 160 kO N: FEB D FEBD C FEBE D 100 N 250 250

(1) (2)

The components of FEA and FEB are shown in the figure below. Part (a) Taking positive moment components to be in the positive coordinate directions, we use Eq. (4.1) on page 182 to write .MA /x D

60 N.120 mm/ D

7200 Nmm

(3)

.MA /y D 60 N.120 mm/ D 7200 Nmm

(4)

.MA /´ D 0:

(5)

Part (b) Taking positive moment components to be in the positive coordinate directions, we use Eq. (4.1) on page 182 to write .MO /x D

60 N.200 mm/ C 150 N.80 mm/ D 0 Nmm

(6)

.MO /y D 60 N.200 mm/ C 221 N.80 mm/ D 29;700 Nmm

(7)

.MO /´ D 0:

(8)

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Problem 4.20 Structure OABC is built in at point O and supports forces from two cables. Cable EAD passes through a frictionless ring at point A, and cable OC G passes through a frictionless ring at point C . If the force in cable EAD is 800 lb and the force in cable OC G is 400 lb, determine (a) the moment of forces from cable OC G about point B. (b) the moment of all cable forces about point A. (c) the moment of all cable forces about point O.

Solution From the figure provided in the problem statement, the following vector expressions may be written 9 |O 12 kO 25 9 |O 12 kO D 400 lb 15 8 {O 6 kO D 800 lb 10 8 |O 6 kO D 800 lb 10

20 {O FEC G D 400 lb

rEBC D 9 |O ft

FECO

rEAC D 9 |O C 6 kO ft

FEAE FEAD

rEOC D 9 |O C 12 kO ft rEOA D 6 kO ft:

These force vectors are shown in the figure at the right. Part (a) We use Eq. (4.2) on page 183 to determine the moment about point B due to the forces in cable OC G as ˇ ˇ O ˇ ˇ {O | O k ˇ ˇ ˇ ˇ E E E MB D rEBC FC G C FCO D .400 ftlb/ ˇ 0 9 0 ˇ 9 12 20 9 12 ˇ ˇ 25 h i25 25 15 15 O 7:20/ D D 400 ftlb {O. 11:5/ |O.0/ C k. 4600 {O 2880 kO ftlb:

(1) (2)

Part (b) Noting that FEAE and FEAD produce no moment about point A, we use Eq. (4.2) on page 183 to determine ˇ ˇ O ˇ {O ˇ | O k ˇ ˇ ˇ ˇ E E E MA D rEAC FC G C FCO D .400 ftlb/ ˇ 0 9 6 ˇ 20 9 9 12 12 ˇ ˇ 25 15 h i25 15 25 O 7:20/ D D 400 ftlb {O. 5:76/ |O. 4:80/ C k. 2304 {O C 1920 |O

(3) 2880 kO ftlb:

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(4)

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Solutions Manual

Part (c) We use Eq. (4.2) on page 183 to determine

E O D rEOC FEC G C FECO C rEOA FEAE C FEAD M ˇ ˇ ˇ ˇ ˇ ˇ {O |O kO ˇ .6 ft/.800 lb/ ˇ {O |O ˇ ˇ ˇ 0 0 ˇ D .400 ftlb/ ˇ 0 9 12 ˇ C ˇ 10 9 9 12 12 ˇ ˇ 8 8 ˇ 20 25 25 15 25 15 h i O 7:20/ C 480 ftlb ŒO{ . 8/ |O. 8/ D 400 ftlb |O. 9:60/ C k. D 3840 {O C 7680 |O 2880 kO ftlb:

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(5) ˇ kO ˇˇ 1 ˇˇ 12 ˇ

(6) (7) (8)

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Problem 4.21 Repeat Prob. 4.20, using a scalar approach.

Solution We begin by adding the two cable forces at point A to obtain FEA , and adding the two cable forces at point C to obtain FEC , as follows 6 kO 8 |O 6 kO O C 800 lb D 640 {O C 640 |O 960 k lb; (1) 10 10 20 {O 9 |O 12 kO 9 |O 12 kO D 400 lb C 400 lb D 320 {O 384 |O 512 kO lb: (2) 25 15

8 {O FEA D FEAE C FEAD D 800 lb FEC D FEC G C FECO

The components of FEA and FEC are shown in the figure below. Part (a) Taking positive moment components to be in the positive coordinate directions, we use Eq. (4.1) on page 182 to determine the moment about point B due to the forces in cable OC G as .MB /x D

512 lb.9 ft/ D

4610 ftlb

(3)

.MB /y D 0 .MB /´ D

(4) 320 lb.9 ft/ D

2880 ftlb:

(5)

Part (b) Noting that the forces at point A produce no moment about point A, and taking positive moment components to be in the positive coordinate directions, we use Eq. (4.1) on page 182 to write .MA /x D 384 lb.6 ft/

512 lb.9 ft/ D

2300 ftlb

(6)

.MA /y D 320 lb.6 ft/ D 1920 ftlb

(7)

.MA /´ D

(8)

320 lb.9 ft/ D

2880 ftlb:

Part (c) Taking positive moment components to be in the positive coordinate directions, we use Eq. (4.1) on page 182 to write .MO /x D 384 lb.12 ft/

512 lb.9 ft/

640 lb.6 ft/ D

3840 ftlb

(9)

.MO /y D 320 lb.12 ft/ C 640 lb.6 ft/ D 7680 ftlb

(10)

.MO /´ D

(11)

320 lb.9 ft/ D

2880 ftlb:

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Solutions Manual

Problem 4.22 Structure OAB is built in at point O and supports two forces of magnitude F parallel to the y and ´ axes. If the magnitude of the moment about point O cannot exceed 1:0 kNm, determine the largest value F may have.

Solution Either the vector approach or the scalar approach may be conveniently used to determine the moment of the the two forces about point O, and we will use the scalar approach. With the dimensions provided, we use Eq. (4.1) on page 182 to determine .MO /x D

F .90 mm/

(1)

.MO /y D

F .400 mm/

(2)

.MO /´ D

F .400 mm/

(3)

The magnitude of the moment at point O may not exceed 1:0 kNm, thus ! 1 103 mm 2 2 2 =2 .MO /x C .MO /y C .MO /´ D 1:0 kNm m 1=2 F . 90 mm/2 C . 400 mm/2 C . 400 mm/2 D 1000 kNmm

(4) (5)

which may be solved for F D 1:75 kN:

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Problem 4.23 Repeat Prob. 4.22 if the x component of the moment (torsional component) at point O may not exceed 0:5 kNm and the resultant of qthe y and ´ components (bending components) may not exceed 0:8 kNm (i.e., My2 C M´2 0:8 kNm).

Solution Either the vector approach or the scalar approach may be conveniently used to determine the moment of the the two forces about point O, and we will use the scalar approach. Taking positive moment components to be in the positive coordinate directions, we use Eq. (4.1) on page 182 to determine .MO /x D

F .90 mm/

(1)

.MO /y D

F .400 mm/

(2)

.MO /´ D

F .400 mm/:

(3)

The absolute value of the torsional component of the moment, .MO /x , may not exceed 0:50 kNm, thus ! 3 mm ˇ ˇ 10 ˇ.MO /x ˇ D F .90 mm/ 0:50 kNm ) F 5:56 kN: (4) m The bending component of the moment, q

F

q

q

.MO /y2 C .MO /2´ , may not exceed 0:80 kNm, thus

.MO /y2 C .MO /2´ 0:8 kNm

. 400 mm/2 C . 400 mm/2 0:8 kNm

(5) 103 mm m

! )

F 1:41 kN:

(6)

Both Eqs. (4), and (6) are satisfied by F 1:41 kN:

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(7)

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Solutions Manual

Problem 4.24 Forces of 3 kN and 200 N are exerted at points B and C of the main rotor of a helicopter, and force F is exerted at point D on the tail rotor. The 3 kN forces are parallel to the ´ axis, the 200 N forces are perpendicular to the main rotor and are parallel to the xy plane, F is parallel to the y axis, and ˛ D 45ı . (a) Determine the value of F so that the ´ component of the moment about point O of all rotor forces is zero. (b) Using the value of F found in Part (a), determine the resultant moment of all rotor forces about point O. (c) If ˛ is different than 45ı , do your answers to Parts (a) and (b) change? Explain.

Solution Part (a) We use Eq. (4.1) on page 182 to evaluate the moment of the rotor forces about the ´ axis, and as instructed in the problem statement, this moment is required to be zero. Thus, we obtain .MO /´ D

200 N.3 m/

200 N.3 m/ C F .5 m/ D 0

) F D 240 N:

(1)

Part (b) Using the geometry provided in the problem statement, with the aid of the sketch below to determine x and y components of the main rotor forces, the following vector expressions may be written

cos 45ı |O C 3000 N kO FEC D 200 N sin 45ı {O C cos 45ı |O C 3000 N kO

(2)

FED D F . |O/

(4)

FEB D 200 N

cos 45ı {O C sin 45ı |O C 1:8 kO m D 3 m cos 45ı {O sin 45ı |O C 1:8 kO m D 5 {O m C 1:2 kO m

rEOC D 3 m rEOB rEOD

sin 45ı {O

(3)

(5) (6) (7)

Using Eq. (4.2) on page 183, the moment about point O of all rotor forces is E O D rEOB FEB C rEOC FEC C rEOD FED M ˇ ˇ ˇ {O |O kO ˇˇ ˇ D ˇˇ 3 cos 45ı 3 sin 45ı 1:8 ˇˇ Nm ı ˇ 200 sin 45ı 200 cos 45 3000 ˇ ˇ ˇ ˇ ˇ ˇ ˇ {O O ˇ O ˇ { O | O k | O k ˇ ˇ ˇ ˇ C ˇˇ 3 cos 45ı 3 sin 45ı 1:8 ˇˇ Nm C ˇˇ 5 0 1:2 ˇˇ Nm ˇ 200 sin 45ı 200 cos 45ı 3000 ˇ ˇ 0 240 0 ˇ h i h i O 600/ Nm C {O.6109/ |O. 6618/ C k. O 600/ Nm D {O. 6109/ |O.6618/ C k. h i O C {O.288/ |O.0/ C k.1200/ Nm; This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc. It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited.

(8)

(9)

(10)

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Statics 1e which simplifies to E O D .288 {O/ Nm: M

(11)

Part (c) If ˛ ¤ 45ı , the answer to Part (a) does not change, since the moment produced by the main rotor forces about the ´ axis is not a function of ˛. The answer to Part (b) does not change, for the following reason. The main rotor forces produce moments about the x and y axes that are functions of ˛, but the moment arms for components of force at points B and C are always opposite (negative of one another) so that the net x and y components are zero regardless of ˛.

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Problem 4.25 EA;1 D rEAB FE The moment of force FE about point A can be computed using M EA;2 D rEAC FE , where B and C are points on the line of action of FE . or using M EA;1 D M EA;2 . Noting that rEAC D rEAB C rEBC , show that M

Solution As specified in the problem statement, MA;2 D rEAC FE ;

and

rEAC D rEAB C rEBC :

(1) (2)

Equations (1) and (2) may be combined to write EA;2 D rEAB C rEBC FE : M

(3)

Using the distributive property of the cross product, Eq. (3) may be expanded to write EA;2 D rEAB FE C rEBC FE : M

(4)

E Hence, Eq. (4) becomes Because rEBC and FE are parallel, rEBC FE D 0. MA;2 D rEAB FE D MA;1 :

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(5)

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Problem 4.26 The steering wheel of a Ferrari sports car has circular shape with 190 mm radius, and it lies in a plane that is perpendicular to the steering column AB. Point C , where the driver’s hand applies force FE to the steering wheel, lies on the y axis with y coordinate yC D 190 mm. Point A is at the origin of the coordinate system, and point B has the coordinates B . 120; 0; 50/ mm. Determine the moment of FE about line AB of the steering column if (a) FE has 10 N magnitude and lies in the plane of the steering wheel and has orientation such that its moment arm to line AB is 190 mm. Also determine the vector expression for this force. 18 {O (b) FE D .10 N/

3 |O C 14 kO . 23

12 {O 5 kO . (c) FE D .10 N/ 13

Solution Part (a) The moment of force FE about line AB can be evaluated using the scalar approach as MAB D .10 N/.190 mm/ D 1900 Nmm:

(1)

To determine the vector expression for this force, note that FE is perpendicular to both rEAB and rEAC . Thus, FE is parallel to the direction nE where nE D rEAB rEAC D . 120 {O ˇ ˇ {O ˇ D ˇˇ 120 ˇ 0 D {O. 9500/

O mm . 190 |O/ mm 50 k/ ˇ |O kO ˇˇ 0 50 ˇˇ mm2 190 0 ˇ O |O.0/ C k.22800/ mm:

(2)

Using Eq. (2), a unit vector nO in the direction of nE is nO D

nE D jE nj

O 0:3846 {O C 0:9231 k:

(3)

Note that rather than using Eqs. (2) and (3), it is possible to write an expression for the unit normal vector by O inspection as nO D . 50 {O C 120 k/=130. Since the force FE is parallel to n, O the vector expression for the force is O D . 3:85 {O C 9:23 k/ O N: FE D .10 N/. 0:3846 {O C 0:9231 k/

(4)

Note that it is possible that the force FE could be in the direction opposite to n, O thus the answer FE D O .3:85 {O 9:23 k/ N is also acceptable. This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc. It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited.

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Part (b) Our strategy will be to determine the moment of force FE about point A, and then use the dot product to determine the component of this moment that acts in the direction of line AB. The moment of force FE about point A is ˇ ˇ ˇ {O |O kO ˇˇ 10 Nmm ˇ EA D rEAC FE D ˇ 0 M 190 0 ˇˇ ˇ 23 ˇ 18 3 14 ˇ 10 Nmm O D {O. 2660/ |O.0/ C k.3420/ : (5) 23 EA in the direction of line AB is The component of M EA MAB D M

rEAB jErAB j

10 Nmm 1 D . 2660/. 120/ C .0/.0/ C .3420/. 50/ 23 130

!

D 472 Nmm:

(6)

Part (c) Our strategy is the same as that used in Part (b). Thus, the moment of force FE about point A is ˇ ˇ ˇ {O |O kO ˇˇ 10 Nmm ˇ EA D rEAC FE D ˇ 0 M 190 0 ˇˇ ˇ 13 ˇ 12 0 5 ˇ O 2280/ 10 Nmm : D {O.950/ |O.0/ C k. (7) 13 EA in the direction of line AB is The component of M EA rEAB MAB D M jErAB j ! 10 Nmm 1 D .950/. 120/ C .0/.0/ C . 2280/. 50/ 13 130 D 0:

(8)

The reason that the force FE produces no moment about line AB is because FE is parallel to line AB.

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Problem 4.27 A rectangular piece of sheet metal is clamped along edge AB in a machine called a brake. The sheet is to be bent along line AB by applying a y direction force F . Determine the moment about line AB if F D 200 lb. Use both vector and scalar approaches.

Solution For the vector solution, we use the data given in the problem statement to write the following expressions rEAC D 36 {O in.;

(1)

FE D 200 |O lb; rEAB D 20 {O 15 kO in.

(2) (3)

We then use Eq. (4.2) on page 183 to determine the moment of FE about point A as ˇ ˇ ˇ {O |O kO ˇˇ ˇ O EA D rEAC FE D ˇ 36 0 0 ˇ in.lb D {O.0/ |O.0/ C k.7200 M in.lb/ D 7200 kO in.lb: ˇ ˇ ˇ 0 200 0 ˇ

(4)

EA that acts in the direction of line AB is given by The component of M rEAB .0 in.lb/.20 in:/ C .0 in.lb/.0 in:/ C .7200 in.lb/. 15 in:/ D rAB 25 in: 4320 in.lb;

EA MAB D M

(5)

D

(6)

where rAB is the magnitude of rEAB . The negative answer in Eq. (6) means that the moment of F about line AB is in the direction from B to A. Note: We could have also used the scalar triple product to obtain Eq. (6). For the scalar solution, we first determine the moment arm d as shown in the figure to the right. Through geometry and trigonometry, we find that 15 D 36:9ı ; 20 d D 36 in: sin D 21:6 in: D tan

1

(7) (8)

Then by applying Eq. (4.1) on page 182, taking positive moment to be in the direction from A to B according to the right-hand rule, it follows that MAB D

200 lb.21:6 in:/ D

4320 in.lb:

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Problem 4.28 In Prob. 4.27 determine F if the moment about line AB is to be 2000 in.lb.

Solution Using the data given in the problem statement, we write the following expressions rEAC D 36 {O in.; FE D F |O; rEAB D 20 {O

(1) (2)

15 kO in.

(3)

Then by using the scalar triple product, we determine the moment of force FE about line AB as ˇ ˇ ˇ 20 0 15 ˇˇ ˇ r E 1 AB ˇ 36 0 0 ˇˇ in. MAB D rEAC FE D rAB 25 ˇˇ 0 F 0 ˇ 1 D . 15/.36/F in. 25

(4) (5)

With MAB D 2000 in.lb, we solve Eq. (5) for F to obtain F D 25

2000 in.lb D . 15/.36/ in.

92:6 lb

)

F D 92:6 lb:

(6)

In writing our final answer for F , the negative sign is unimportant and is thus omitted.

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Problem 4.29 In the pipe assembly shown, points B and C lie in the xy plane and force F is parallel to the ´ axis. If F D 150 N, determine the moment of F about lines OA and AB. Use both vector and scalar approaches.

Solution Using the figure to the right, we write the following expressions rEOA D 350 |O mm;

(1) ı

ı

rEAB D 350 mm sin 30 {O C cos 30 |O D .175 {O C 303 |O/ mm; rEBC D 350 mm cos 30ı {O sin 30ı |O D .260 {O 150 |O/ mm; FE D 150 kO N:

(2) (3) (4)

For the vector approach, we use Eq. (4.2) on page 183 to determine the moment of the force about point A, EA . Then, M EA rEOA =rOA produces the moment of the force about line OA, and M EA rEAB =rAB namely M produces the moment of the force about line AB. Thus ˇ ˇ ˇ {O |O kO ˇˇ ˇ EA D rEAC FE D rEAB C rEBC FE D ˇ 0:435 0:153 M (5) 0 ˇˇ Nm ˇ ˇ 0 0 150 ˇ h i O D {O. 23:0/ |O. 65:2/ C k.0/ Nm D . 23:0 {O C 65:2 |O/ Nm: (6) EA rEOA D .65:2 Nm/.0:350 m/ D 65:2 Nm; MOA D M rOA 0:350 m . 23:0 Nm/.0:175 m/ C .65:2 Nm/.0:303 m/ r E EA AB D D 44:9 Nm: MAB D M rAB 0:350 m

(7) (8)

For the scalar approach, we use Eq. (4.1) on page 182 and the figure shown above, taking positive moment MOA to be in the direction from O to A, and positive moment MAB to be in the direction from A to B, according to the right-hand rule, to determine MOA D F d1 D 150 N .0:350 m/ sin 30ı C .0:300 m/ cos 30ı D 65:2 Nm; MAB D F d2 D 150 N .0:300 m/ D 45:0 Nm:

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Solutions Manual

Problem 4.30 In the pipe assembly shown, points B and C lie in the xy plane and force F is parallel to the ´ axis. If a twisting moment (torque) of 50 Nm will cause a pipe to begin twisting in the flange fitting at O or at either end of the elbow fitting at A, determine the first fitting that twists and the value of F that causes it.

Solution We use Eq. (4.1) on page 182, and the figure shown at the right for the determination of the moment arms d1 and d2 , to determine the moments of force F about lines OA and AB as MOA D F d1 D F .0:350 m/ sin 30ı C .0:300 m/ cos 30ı ; (1) MAB D F d2 D F .0:300 m/:

(2)

According to the failure criteria given in the problem statement, pipe OA will twist at O and A if MOA reaches 50 Nm, and pipe AB will twist at A and B if MAB reaches 50 Nm. Thus, If MOA D 50 Nm;

then

If MAB D 50 Nm;

then

50 Nm D 115 N; .0:350 m sin 30ı C 0:300 m cos 30ı / 50 Nm F D D 167 N: .0:300 m/ F D

(3) (4)

The smallest value of F among Eqs. (3) and (4) is the value of the force that will cause twisting to occur. Thus, the largest force F that may be applied is F D 115 N; and twisting occurs at both threads on pipe OA.

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(5)

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Problem 4.31 Determine the moment of force F about line AB as follows. EA , and then determine the (a) Determine the moment of F about point A, M component of this moment in the direction of line AB. E B , and then determine the (b) Determine the moment of F about point B, M component of this moment in the direction of line AB. EA , M E B , and the (c) Comment on differences and/or agreement between M moment about line AB found in Parts (a) and (b). Also comment on the meaning of the sign (positive or negative) found for the moment about line AB.

Solution With the information provided in the problem statement, the following position vectors may be written rEAB D .15 {O 20 |O/ in.; rEAC D 20 |O C 18 kO in.; rEBC D 15 {O C 18 kO in. (1) Part (a) We use Eq. (4.2) on page 183 to determine the moment of the force about point A as ˇ ˇ ˇ {O |O kO ˇˇ ˇ EA D rEAC FE D ˇ 0 M 20 18 ˇˇ in.lb; ˇ ˇ 10 80 40 ˇ h EA D {O. 640/ M

i O |O. 180/ C k.200/ in.lb D 640 {O C 180 |O C 200 kO in.lb:

(2)

(3)

EA in the direction of line AB is given by The component of M EA MAB D M

rEAB . 640/.15/ C .180/. 20/ C .200/.0/ D in.lb D rAB 25

528 in.lb:

Part (b) We use Eq. (4.2) on page 183 to determine the moment of the force about point B as ˇ ˇ ˇ {O O ˇ | O k ˇ ˇ E B D rEBC FE D ˇ 15 0 M 18 ˇˇ in.lb ˇ ˇ 10 80 40 ˇ

(4)

(5) (6)

h E B D {O. 1440/ M

i O 1200/ in.lb D |O.420/ C k. 1440 {O

420 |O

1200 kO in.lb:

(7)

E B in the direction of line AB is given by The component of M E B rEAB D . 1440/.15/ C . 420/. 20/ C . 1200/.0/ in.lb D MAB D M rAB 25 This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc. It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited.

528 in.lb:

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Part (c) EA and M E B are different, both in magnitude and direction. 1. M EA or M E B is used. 2. MAB is the same regardless if M 3. The negative result for MAB means that the twisting action of F is directed from B to A.

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Problem 4.32 Determine the moment of force F about line AB as follows. EA , and then determine the (a) Determine the moment of F about point A, M component of this moment in the direction of line AB. E B , and then determine the (b) Determine the moment of F about point B, M component of this moment in the direction of line AB. EA , M E B , and the (c) Comment on differences and/or agreement between M moment about line AB found in Parts (a) and (b). Also comment on the meaning of the sign (positive or negative) found for the moment about line AB.

Solution With the information provided in the problem statement, the following position vectors may be written rEAB D 4 {O 7 |O C 4 kO mm; rEAC D 4 {O C 4 kO mm; rEBC D 7 |O mm: (1) Part (a) We use Eq. (4.2) on page 183 to determine the moment of the force about point A as ˇ ˇ ˇ {O |O kO ˇˇ ˇ 2 EA D rEAC FE D ˇ 4 0 4 ˇ Nmm; M ˇ 7 ˇˇ 3 2 6 ˇ h EA D 2 {O.8/ M 7

i O |O. 36/ C k.8/ Nmm:

(2)

(3)

EA in the direction of line AB is given by The component of M EA rEAB D 2 Nmm Œ.8/. 4/ C .36/. 7/ C .8/.4/ 1 D MAB D M rAB 7 9

8:00 Nmm:

Part (b) We use Eq. (4.2) on page 183 to determine the moment of the force about point B as ˇ ˇ ˇ {O |O kO ˇ ˇ ˇ E B D rEBC FE D 2 ˇ 0 7 0 ˇ Nmm; M ˇ 7 ˇˇ 3 2 6 ˇ h E B D 2 {O.42/ M 7

i O 21/ Nmm: |O.0/ C k.

(4)

(5)

(6)

E B in the direction of line AB is given by The component of M E B rEAB D 2 Nmm Œ.42/. 4/ C .0/. 7/ C . 21/.4/ 1 D MAB D M rAB 7 9

8:00 Nmm:

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Solutions Manual

Part (c) EA and M E B are different, both in magnitude and direction. 1. M EA or M E B is used. 2. MAB is the same regardless if M 3. The negative result for MAB means that the twisting action of F is directed from B to A.

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Problem 4.33 The moment of force FE about line AB can be computed using MAB;1 or MAB;2 where rEAB MAB;1 D .ErAC FE / ; jErAB j

rEAB MAB;2 D .ErBC FE / ; jErAB j

where C is a point on the line of action of FE . Noting that rEAC D rEAB C rEBC , show that MAB;1 D MAB;2 .

Solution We summarize the information provided in the problem statement by writing the following expressions rE AB MAB;1 D rEAC FE ; rAB rE AB MAB;2 D rEBC FE ; rAB rEAC D rEAB C rEBC :

(1) (2) (3)

Substituting Eq. (3) into Eq. (1) provides rEAB MAB;1 D rEAB C rEBC FE rAB rE AB D rEAB FE C rEBC FE rAB rE rE AB AB D rEAB FE C rEBC FE : rAB rAB In Eq. (6), the result of rEAB FE is a vector that is orthogonal to rEAB , therefore, rEAB FE rEAB Thus, Eq. (6) reduces to EAB;1 D 0 C rEBC FE rEAB : M rAB

(4) (5) (6) D 0. (7)

Equation (7) is identical to Eq. (2), thus EAB;1 D M EAB;2 : M

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Problem 4.34 An automobile windshield wiper is actuated by a force FE D .120 {O Determine the moment of this force about shaft OB.

50 |O/ N.

Solution We use Eq. (4.2) on page 183 to determine the moment of force FE about point O as ˇ ˇ ˇ {O |O kO ˇˇ ˇ E O D rEOA FE D ˇ 0 M 60 80 ˇˇ Nmm ˇ ˇ 120 50 0 ˇ h i O D {O.4000/ |O. 9600/ C k.7200/ Nmm:

(1) (2)

E O about line OB is The component of M rEOB .4000/.20/ C .9600/.90/ C .7200/.60/ D Nmm rOB 110 D 12;500 Nmm D 12:5 Nm:

EO MOB D M

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(3) (4)

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Problem 4.35 In the windshield wiper of Prob. 4.34, if FE has 130 N magnitude, determine the direction in which it should be applied so that its moment about shaft OB is as large as possible, and determine the value of this moment.

Solution To maximize the moment of F about line OB, FE should be perpendicular to the plane containing lines OA and OB. Thus, we determine a vector nE , and unit vector n, O that are perpendicular to this plane as ˇ ˇ ˇ {O |O kO ˇˇ h i ˇ O 1200/ mm2 ; nE D rEOB rEOA D ˇˇ 20 90 60 ˇˇ mm2 D {O.10800/ |O.1600/ C k. (1) ˇ 0 ˇ 60 80 nO D

nE D 0:983 {O n

0:146 |O

O 0:109 k:

(2)

Using n, O we construct FE as FE D 130 N nO D 130 N 0:983 {O

0:146 |O

0:109 kO D 128 {O

19:0 |O

14:2 kO N:

(3)

We then use the scalar approach to determine the moment of this force about line OB as q MOB D F d D F rOA D .130 N/ . 60 mm/2 C .80 mm/2 D 13;000 Nmm D 13 Nm:

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Solutions Manual

Problem 4.36 Wrench AB is used to twist a nut on a threaded shaft. Point A is located 120 mm from point O, and line a has x, y, and ´ direction angles of 36ı , 60ı , and 72ı , respectively. Point B is located at .0; 300; 100/ mm, and the force is O N. FE D . 80 |O C 20 k/ EA , and then find the compo(a) Determine the moment of FE about point A, M nent of this moment about line a, Ma . E O , and then find the compo(b) Determine the moment of FE about point O, M nent of this moment about line a, Ma .

Solution Part (a)

Using the coordinates and direction angles provided, the following position vectors may be written rEOA D 120 mm cos 36ı {O C cos 60ı |O C cos 72ı kO D 97:1 {O C 60 |O C 37:1 kO mm; (1) h i O rEOB D {O.0 0/ C |O.300 0/ C k.100 0/ mm D 300 |O C 100 kO mm: (2)

It then follows that rEAB D rEAO C rEOB D rEOA C rEOB D 97:1 {O C 240 |O C 62:9 kO mm:

(3) (4)

We use Eq. (4.2) on page 183 to determine the moment of force FE about point A as

EA D rEAB M

ˇ |O kO ˇˇ 240 62:9 ˇˇ Nmm 80 20 ˇ i O |O. 1940/ C k.7770/ Nmm D 9830 {O C 1940 |O C 7770 kO Nmm:

ˇ ˇ ˇ E F D ˇˇ ˇ

h D {O.9830/

{O 97:1 0

(5) (6)

EA about line a is given by The component of M EA rOOA D .9830/.cos 36ı / C .1940/.cos 60ı / C .7770/.cos 72ı / Nmm Ma D M D 11;300 Nmm D 11:3 Nm:

(7) (8)

Part (b) We use Eq. (4.2) to determine the moment of force FE about point O as

E O D rEOB M

ˇ |O kO ˇˇ 300 100 ˇˇ Nmm 80 20 ˇ i O |O.0/ C k.0/ Nmm D 14000 {O Nmm:

ˇ ˇ {O ˇ E F D ˇˇ 0 ˇ 0

h D {O.14000/

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Statics 1e E O about line a is given by The component of M E O rOOA D .14000/.cos 36ı / C .0/.cos 60ı / C .0/.cos 72ı / Nmm Ma D M D 11;300 Nmm D 11:3 Nm:

(11) (12)

EA and M E O are different, yet Ma is the same. Observe that M

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Solutions Manual

Problem 4.37 A trailer has a rectangular door DEGH hinged about edge GH . If FE D O lb, determine the moment of F about edge GH . . 2 {O C 5 |O C 14 k/

Solution With the coordinates provided in the problem statement, the following position vector may be written rEHD D 16 |O 32 kO in. We then use Eq. (4.2) on page 183 to determine the moment of force FE about point H as ˇ ˇ ˇ {O |O kO ˇˇ ˇ E H D rEHD FE D ˇ 0 16 M 32 ˇˇ in.lb; ˇ ˇ 2 5 14 ˇ h i D 384 {O . 64/ |O C 32 kO in.lb:

(1)

(2) (3)

E H in the direction of line GH is simply the negative of the x component of M E H (or, The component of M E evaluate the dot product MGH D MH . {O/ to obtain the following result) MGH D

384 in.lb:

(4)

The negative sign for MGH means that the twisting action of the force is in the direction from H to G, according to the right-hand rule.

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Problem 4.38 In the trailer of Prob. 4.37, if F D 15 lb, determine the direction in which it should be applied so that its moment about edge GH of the rectangular door is as large as possible, and determine the value of this moment.

Solution To maximize the moment of FE about line GH , FE should be applied perpendicular (normal) to the door. From the information provided in the problem statement, the following position vectors may be written rOHG D {O; h rEHD D {O. 40 C 40/ C |O.24

(1) O 8/ C k.20

i 52/ in. D 16 |O

32 kO in.

We then determine a unit vector normal to these two vectors as ˇ ˇ ˇ {O |O kO ˇˇ h i ˇ O nE D rOHG rEHD D ˇˇ 1 0 in. 0 ˇˇ in. D |O. 32/ C k.16/ ˇ 0 16 32 ˇ O in. 32 | O C 16 k nE O nO D D p D 0:894 |O C 0:447 k: n .32 in:/2 C .16 in:/2

(2)

(3)

(4)

We then construct FE in this direction as FE D 15 lb nO D 15 lb 0:894 |O C 0:447 kO D 13:4 |O C 6:71 kO lb:

(5)

Using the scalar approach, the moment of FE about line GH is q MGH D F rHD D .15 lb/ .16 in:/2 C . 32 in:/2 D 537 in.lb:

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Problem 4.39 E D . {O C 4 |O C A trailer has a triangular door ABC hinged about edge BC . If Q O 8 k/ lb, determine the moment of Q about edge BC .

Solution With the coordinates provided in the problem statement, the following position vectors may be written h i O rEBA D {O.0 24/ C |O.28 4/ C k.12 12/ in. D . 24 {O C 24 |O/ in.; (1) h i O 12/ in. D 24 {O C 48 kO in. (2) rEBC D {O.0 24/ C |O.4 4/ C k.60 E about point B as We then use Eq. (4.2) on page 183 to determine the moment of force Q ˇ ˇ ˇ {O O ˇ | O k ˇ ˇ E B D rEBA Q E D ˇ 24 24 0 ˇ in.lb; M ˇ ˇ ˇ 1 4 8 ˇ h i D 192 {O . 192/ |O C . 120/ kO in.lb:

(3) (4)

E B in the direction of line BC is The component of M 2 E B rEBC D Œ.192/. 24/pC .192/.0/ C . 120/.48/ in. lb MBC D M rBC . 24 in:/2 C .48 in:/2 D 193 in.lb:

(5) (6)

The negative sign for MBC means that the twisting action of the force is in the direction from C to B, according to the right-hand rule.

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Problem 4.40 In the trailer of Prob. 4.39, if Q D 9 lb, determine the direction in which it should be applied so that its moment about edge BC of the triangular door is as large as possible, and determine the value of this moment.

Solution E about line BC , Q E should be applied perpendicular (normal) to the door. From To maximize the moment of Q the information provided in the problem statement, the following position vectors may be written rEBA D . 24 {O C 24 |O/ in. D . {O C |O/ 24 in:; rEBC D . 24 {O C 48 |O/ in. D {O C 2 kO 24 in:

(1) (2)

We then determine a unit vector normal to these two vectors as ˇ ˇ ˇ {O |O kO ˇ h i ˇ ˇ O nE D rEBA rEBC D ˇˇ 1 1 0 ˇˇ .24 in:/2 D {O.2/ |O. 2/ C k.1/ .24 in:/2 ˇ 1 0 2 ˇ 2 2 {O C 2 |O C kO .24 in:/ nE O nO D D D 0:667 {O C 0:667 |O C 0:333 k: p n .24 in:/2 .2/2 C .2/2 C .1/2

(3)

(4)

E in this direction as We then construct Q E D 9 lb 0:667 {O C 0:667 |O C 0:333 kO D 6 {O C 6 |O C 3 kO lb: Q E about point B is Using Eq. (4.2) on page 183, the moment of Q ˇ ˇ ˇ {O |O kO ˇ ˇ ˇ E B D rEBA Q E D ˇ 1 1 0 ˇ 24 in.lb M ˇ ˇ ˇ 6 6 3 ˇ h i O 12/ 24 in.lb D 3 {O C 3 |O D {O.3/ |O. 3/ C k.

(5)

(6) 12 kO 24 in.lb:

(7)

E B about line BC is The component of M C 2. 12/ .24 in.lb/.24 in:/ E B rEBC D Œ 1.3/ C 0.3/ MBC D M D p rBC . 1/2 C .2/2 .24 in:/

290 in.lb:

(8)

The negative sign for MBC means that the twisting action of the force is in the direction from C to B, according to the right-hand rule.

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Solutions Manual

Problem 4.41 A poorly leveled crane rotates about line a, which lies in the xy plane and has a y direction angle of 5ı . Line AB lies in the x´ plane. If the crane supports a weight W D 5000lb and ˛ D 45ı , determine the moment of W about line a.

Solution From the information provided in the problem statement, the following vector expressions may be written WE D rEAB

5000 |O lb; D 40 ft cos 45ı {O C sin 45ı kO :

(1) (2)

Using Eq. (4.2) on page 183, the moment of WE about point A is ˇ ˇ ˇ ˇ {O |O kO ˇ ˇ ı ı ˇ E E MA D rEAB W D ˇ cos 45 0 sin 45 ˇˇ .40 ft/. 5000 lb/ ˇ ˇ 0 1 0 h i O 0:707/ D 1:41105 ftlb {O C kO : D 2105 ftlb {O. 0:707/ |O.0/ C k.

(3) (4)

Noting that the a direction lies in the xy plane and has a y direction angle of 5ı , the unit vector uO in the a direction is uO D sin 5ı {O C cos 5ı |O: (5) EA in the a direction is given by The component of M EA uO D M EA Ma D M sin 5ı {O C cos 5ı |O D 1:41105 ftlb .1/. sin 5ı / C .0/.cos 5ı / C .1/.0/ D

12;300 ftlb:

(6) (7) (8)

The negative sign for Ma means that the twisting action of the weight is in the a direction, according to the right-hand rule.

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Problem 4.42 If the poorly leveled crane of Prob. 4.41 can have any position ˛, where 0 ˛ 135ı , determine the largest moment of the weight W D 5000 lb about line a and the position ˛ that produces this moment. Assume h D 40 ft for any position ˛ (this requires the operator to slightly change the boom’s height and/or length as the crane rotates about line a).

Solution From the information provided in the problem statement, the following vector expressions may be written WE D rEAB

5000 |O lb; D 40 ft cos ˛ {O C sin ˛ kO :

Using Eq. (4.2) on page 183, the moment of WE ˇ ˇ ˇ E E MA D rEAB W D ˇˇ ˇ h D 2105 ftlb

(1) (2)

about point A is {O |O kO cos ˛ 0 sin ˛ 0 1 0

(3)

sin ˛ {O

(4)

ˇ ˇ ˇ ˇ .40 ft/. 5000 lb/ ˇ ˇ i cos ˛ kO :

Noting that the a direction lies in the xy plane and has a y direction angle of 5ı , the unit vector uO in the a direction is uO D sin 5ı {O C cos 5ı |O: (5) EA in the a direction is given by The component of M EA uO D M EA Ma D M sin 5ı {O C cos 5ı |O D 2105 ftlb . sin ˛/. sin 5ı / C .0/.cos 5ı / C . cos ˛/.0/ D

17;400 ftlb.sin ˛/:

(6) (7) (8)

From this we see that Ma is maximum when sin ˛ D 1, which occurs when ˛ D 90ı . Therefore the maximum moment about line a is Ma D

17;400 ftlb

at ˛ D 90ı :

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Problem 4.43 Three-wheel and four-wheel all-terrain vehicles (ATVs) are shown. For both ATVs, the combined weight of the driver and vehicle is W D 2600 N. During a hard turn, the tendency for the ATV to tip is modeled by the force F which acts in the negative y direction. Determine the value of F such that the resultant moment of F and the weight about line AB is zero. Comment on which ATV is more prone to tip during hard turns. Note: One of these models of ATV are no longer manufactured because of their poor safety record.

Solution We first consider the three-wheel ATV. With point O being the origin of the coordinate system, the following vector expressions may be written rEAO D 0:9 {O C 0:8 kO m; (1) E D FE C WE D Q

F |O

O 2600 N k:

(2)

Note that the lines of action of both FE and WE intersect at point O. Using Eq. (4.2) on page 183, the moment E about point A is of force Q ˇ ˇ ˇ ˇ h {O |O kO i ˇ ˇ O ˇ E E MA D rEAO Q D ˇ 0:9 m 0 mF / : (3) 0:8 m ˇˇ D {O.0:8 mF / |O.2340 Nm/ C k.0:9 ˇ 0 F 2600 N ˇ Noting that the position vector from point A to point B is rEAB D . 1:6 {O C 0:6 |O/ m;

(4)

EA about line AB is given by the component of M EA rEAB MAB D M rAB .0:8 mF /. 1:6 m/ C . 2340 Nm/.0:6 m/ C .0:9 mF /.0 m/ D : p . 1:6 m/2 C .0:6 m/2

(5) (6)

As instructed in the problem statement, MAB D 0, hence we set Eq. (6) equal to zero and solve for F to obtain 1404 Nm F D D 1096 N ) F D 1096 N: (7) 1:28 m In Eq. (7), the negative sign for F is irrelevant and is thus ignored. For the four-wheel ATV, we use the scalar approach (Eq. (4.1) on page 182) to determine the moment of the forces about line AB as MAB D F .0:8 m/ 2600 N.0:6 m/: (8) This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc. It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited.

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325

As instructed in the problem statement, MAB D 0, hence we set Eq. (8) equal to zero and solve for F to obtain F D 1950 N: (9) Comparing Eqs. (7) and (9), we see that the force required to tip the four-wheel ATV is 78% larger than that for the three-wheel ATV. Note that three-wheel ATVs are no longer manufactured because of their poor safety record. Photo credit: © Richard Hamilton Smith/CORBIS.

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Solutions Manual

Problem 4.44 An open-end wrench applies the forces F to the head of a bolt to produce the moment M , where each force F is normal to the surface on which it acts. Determine M if F D 400 lb.

Solution Using the sketch shown at the right, we determine the dimension d to be d cos 30ı D

5 in. 16

)

d D 0:361 in:

(1)

Using Eq. (4.12) on page 209, we determine the moment of the couple to be M D F d D 400 lb.0:361 in:/ D 144 in.lb:

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(2)

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Problem 4.45 A box-end wrench applies the forces F to the head of a bolt to produce the moment M , where each force F is normal to the surface on which it acts. Determine F if M D 20 ftlb.

Solution Using the sketch shown at the right, we determine the dimension d to be d cos 30ı D

5 in. 16

)

d D 0:361 in:

(1)

Using Eq. (4.12) on page 209, we determine the moment of the three couples to be ! 12 in: M D 3F d D 20 ftlb D 3F .0:361 in:/ 1 ft

)

F D 222 lb:

(2)

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Solutions Manual

Problem 4.46 The top view of a workpiece that fits loosely in a fixture for drilling is shown. The drill bit has two edges that apply in-plane cutting forces F to the workpiece. (a) If F D 600 N, determine the forces Q between the workpiece and fixture so that the resultant couple moment is zero when ˛ D 30ı . (b) Does your answer for Q from Part (a) change if ˛ has different value? If yes, then repeat Part (a) with ˛ D 60ı .

Solution Part (a) The workpiece is subjected to two couples. Equation (4.12) on page 209 is used, with positive moment being counterclockwise, to determine the resultant moment of the two couples as M D Q.150 mm/

F .30 mm/:

(1)

As instructed in the problem statement, M D 0. Thus, we set Eq. (1) equal to zero, and using F D 600 N, we solve for Q to obtain Q D 120 N:

(2)

Part (b) The answer obtained in Part (a) is valid for any value of ˛.

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Problem 4.47 Three tugboats are used to turn a barge in a narrow channel. To avoid producing any net translation of the barge, the forces applied should be couples. The tugboat at point A applies a 400 lb force. (a) Determine FB and FC so that only couples are applied. (b) Using your answers to Part (a), determine the resultant couple moment that is produced. (c) Resolve the forces at A and B into x and y components, and identify the pairs of forces that constitute couples.

Solution Part (a) In order to have loading due to couple forces only, the resultant of all forces must be zero. Hence, X FRx D Fx D 0 W .400 lb/ sin 30ı FB sin 45ı D 0; (1) X FRy D Fy D 0 W .400 lb/ cos 30ı FB cos 45ı FC D 0: (2) Solving these equations provides FB D 282:8 lb;

(3)

FC D 146:4 lb:

(4)

Part (b) With the values given in Eqs. (3) and (4), the system of forces are couples, and the moment produced by these, taking positive moment to be counterclockwise, is X MR D MB D .400 lb/ cos 30ı .50 ft/ .400 lb/ sin 30ı .44 ft/ FC .40 ft/; (5) which provides MR D

31;980 ftlb:

(6)

Note that the same value of MR would be obtained regardless of the summation point used in Eq. (5), because the forces are couples. Part (c) This sketch below illustrates two force systems that are equivalent. The original force system is shown on the left, with forces resolved into x and y directions. The force system on the right shows more explicitly the three couples that are applied to the barge.

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Solutions Manual

Problem 4.48 A bracket with pulleys for positioning magnetic tape in an electronic data storage device is shown. The pulleys are frictionless, the tape supports a force T D 2 N, and the thickness of the tape can be neglected. (a) Use the two tape forces in Fig. p4.48(a) to compute the resultant couple moment. (b) Replace the forces on the two pulleys with point forces on the pinions as shown in Fig. p4.48(b), and compute the resultant couple moment (replacement of pulley forces by bearing forces is discussed in Chapter 3 in connection with Fig. 3.9).

Solution Part (a) Using the geometry shown in the figure to the right, we determine the moment arm d for the couple to be q d D 10 mm C .20 mm/2 C .20 mm/2 C 10 mm D 48:3 mm: (1) We then use Eq. (4.12) on page 209, taking positive moment to be counterclockwise, to determine the moment of the couple as M D T d D .2 N/.48:3 mm/ D 96:6 Nmm:

(2)

Part (b) Using Fig. (b) in the problem statement, the loading from the tape consists of two couples. Using Eq. (4.12), taking positive moment to be counterclockwise, the moment of the couples is M DT

q

.20 mm/2 C .20 mm/2 C T .20 mm/ D 96:6 Nmm:

(3)

As expected, the results of Parts (a) and (b) agree.

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Problem 4.49 Consider an object with forces FEA and FEB applied. The first column of the following table lists resultant moments at various points due to both of these forces. For each statement select True or False.

E C D rECA FEA C rECB FEB M E C D rEAB FEB M E C D rEBA FEA M EA D rEAB FEB M E D D rECA FEA C rECB FEB M

If FEA and FEB are not a couple

If FEA and FEB are a couple

T or F? T or F? T or F? T or F? T or F?

T or F? T or F? T or F? T or F? T or F?

Solution

E C D rECA FEA C rECB FEB M E C D rEAB FEB M E C D rEBA FEA M EA D rEAB FEB M E D D rECA FEA C rECB FEB M

If FA and FB are not a couple

If FA and FB are a couple

T F F T F

T T T T T

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Solutions Manual

Problem 4.50 A structure built in at point O supports 300 and 400 N couples. Determine the resultant couple moment vector, using both scalar and vector approaches.

Solution For the scalar approach, we take positive moments to be in the positive coordinate directions, according to the right-hand rule. Using Eq. (4.12) on page 209, the moment of the 300 N couple is M1 D 300 N.2 m 4 m/ D E 1 D 600 kO Nm: M

600 Nm;

(1) (2)

Using Eq. (4.12) again, the moment of the 400 N couple is M2 D 400 N.2 m

1:5 m/ D 200 Nm;

(3)

E 2 D 200 {O Nm: M

(4)

The resultant couple moment is the sum of Eqs. (2) and (4), which provides E DM E1 C M E 2 D 200 {O M

600 kO Nm:

(5)

For the vector approach, we use Eq. (4.11) on page 209, to determine the resultant couple moment is ˇ ˇ ˇ ˇ {O |O kO |O kO ˇˇ ˇˇ {O ˇ ˇ ˇ ˇ E E E M D rEBC FC C rEAB FB D ˇ 0 0 2 m 0 ˇ C ˇ 0 0:5 m ˇ 300 N 0 0 ˇ ˇ 0 0 400 N O 600 Nm/ C {O.200 Nm/ D 200 {O 600 kO Nm: E D k. M

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ˇ ˇ ˇ ˇ ˇ ˇ

(6) (7)

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Problem 4.51 A structure built in at point O supports 70 and 85 N couples and a tip moment. Determine the resultant couple moment, using both scalar and vector approaches.

Solution From the geometry provided in the problem statement, we write vector expressions for the forces applied at points B and D as 1 O FEB D .70 N/ . 6 {O C 3 |O 2 k/; 7 1 O FED D .85 N/ . 8 {O 12 |O C 9 k/: 17

(1) (2)

For the scalar approach, we evaluate the moments of each of the x, y, and ´ components of the 70 N and 85 N forces, taking positive moments to be in the positive coordinate directions according to the right-hand rule. Thus, using Eq. (4.12) on page 209, we obtain ! ! 3 6 E D 70 N M 2 m {O C 70 N 2 m |O C 0 kO 7 7 ! ! 9 8 C 85 N 2 m {O C 0 |O C 85 N 2 m kO 200 {O Nm; (3) 17 17 which provides E D M

50 {O C 120 |O C 80 kO Nm:

(4)

For the vector approach, the following position vectors are needed rEAB D

2 kO m;

rECD D 2 |O m:

(5)

We use Eq. (4.11) on page 209 to write E D rEAB FEB C rECD FED C M EE M ˇ ˇ ˇ ˇ ˇ O ˇ |O kO 70 Nm ˇˇ {O |O k ˇˇ 85 Nm ˇˇ {O D 2 ˇC 0 2 0 ˇ 0 0 7 17 ˇˇ ˇ 6 3 2 ˇ 8 12 9 i 85 Nm h 70 Nm h O D {O.6/ |O. 12/ C k.0/ C {O.18/ 7 17

(6) ˇ ˇ ˇ ˇ ˇ ˇ

200 {O Nm O |O.0/ C k.16/

(7) i

200 {O Nm;

(8)

which provides E D M

50 {O C 120 |O C 80 kO Nm:

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Solutions Manual

Problem 4.52 Determine the distance between the lines of action of the 70 N forces. Hint: Use the vector approach to determine the moment of this couple. Then note that the magnitude of this result is equal to .70 N/d; where d is the distance between the lines of action for the two forces.

Solution From the geometry provided in the problem statement, we write the following expressions rEAB D

1 FEB D .70 N/ . 6 {O C 3 |O 7

2 kO m;

O 2 k/;

(1)

where FEB is the force applied at point B. The moment due to the 70 N couple is obtained using Eq. (4.11) on page 209 as ˇ ˇ ˇ {O |O kO ˇ ˇ ˇ E D rEAB FEB D 70 Nm ˇ 0 0 M (2) 2 ˇˇ ˇ 7 ˇ 6 3 2 ˇ i 70 Nm h O D {O.6/ |O. 12/ C k.0/ ; (3) 7 E is and the magnitude of M M D 134 Nm:

(4)

We then use the scalar approach given by Eq. (4.12) on page 209 to solve for d as follows M D Fd

(5)

134 Nm D .70 N/d;

(6)

which is solved to obtain dD

134 Nm D 1:92 m: 70 N

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Statics 1e

Problem 4.53 Determine the distance between the lines of action of the 85 N forces. See the hint in Prob. 4.52.

Solution From the geometry provided in the problem statement, we write the following expressions rECD D 2 |O m;

1 FED D .85 N/ . 8 {O 17

O 12 |O C 9 k/;

(1)

where FED is the force applied at point D. The moment due to the 85 N couple is obtained using Eq. (4.11) on page 209 as ˇ ˇ ˇ {O O ˇ | O k ˇ ˇ E D rECD FED D 85 Nm ˇ 0 M (2) 2 0 ˇˇ ˇ 17 ˇ 8 12 9 ˇ i 85 Nm h O D {O.18/ |O.0/ C k.16/ ; (3) 17 E is and the magnitude of M M D 120 Nm:

(4)

We then use the scalar approach given by Eq. (4.12) on page 209 to solve for d as follows M D Fd

(5)

120 Nm D .85 N/d;

(6)

which is solved to obtain dD

120 Nm D 1:42 m: 85 N

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Solutions Manual

Problem 4.54 The input shaft of a speed reducer supports a 200 Nm moment, and the output shafts support 300 and 500 Nm moments. (a) Determine the resultant moment applied by the shafts to the speed reducer. (b) If the speed reducer is bolted to a surface that lies in the xy plane, speculate on the characteristics of the forces these bolts apply to the speed reducer. In other words, do you expect these forces to constitute couples only, or may they be more general? Remark: Problems such as this are discussed in detail in Chapter 5.

Solution Part (a) The resultant moment applied to the speed reducer by the three shafts is simply the sum of the three moment vectors that are applied E D . 300 {O M

500 {O

200 |O/ Nm D . 800 {O

200 |O/ Nm:

(1)

Part (b) Since the input and output shafts apply moments only, there is no net force applied by the shafts to the speed reducer and hence we would expect the base to also apply no net force. The speed reducer is subjected to a resultant moment by the shafts, so we would expect the forces on the base, whatever they are, to be a system of couples with a resultant of .800 {O C 200 |O/ Nm so that the total moment applied to the speed reducer is zero.

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Problem 4.55 Satellites and other spacecraft perform attitude positioning using thrusters that are fired in pairs so as to produce couples. If thrusters at points A, B, C , and D each produce 3 N forces, determine the resultant moment and hence the axis through the center of mass about which an initially nonrotating satellite will begin to rotate.

Solution From the information provided in the problem statement, the following vector expressions may be written rEAC D .1:2 |O

O 1:6 k/m; O D . 1:2 |O 1:6 k/m;

FEC D 3 kO N;

(1)

rEBD

FED D

(2)

3 {O N:

Using Eq. (4.11) on page 209, the resultant moment of the couples is given by E D rEAC FEC C rEBD FED M ˇ ˇ ˇ ˇ {O ˇ ˇ {O |O kO |O kO ˇ ˇ ˇ D ˇˇ 0 1:2 m 1:6 m ˇˇ C ˇˇ 0 1:2 m 1:6 m ˇ 0 ˇ ˇ 0 3N 3N 0 0 h i O 3:6/ Nm; D 3:6 Nm {O C {O.0/ |O. 4:8/ C k.

(3) ˇ ˇ ˇ ˇ ˇ ˇ

(4) (5)

which provides E D 3:6 {O C 4:8 |O M

3:6 kO Nm:

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Problem 4.56 Forces F and T exerted by air on a rotating airplane propeller are shown. Forces F lie in the xy plane and are normal to the axis of each propeller blade, and thrust forces T act in the ´ direction. Show that the forces F can be represented as a couple or system of couples.

Solution The figures shown at the right are views looking down the ´ axis toward the propeller; forces T are not shown and these are irrelevant to the question being asked. We resolve the forces F into horizontal and vertical components, as shown in the second figure at the right. We then observe that the vertical force F cos 30ı in the positive y direction and the vertical force F cos 30ı in the negative y direction constitute a couple. The sum of the two horizontal forces F sin 30ı in the negative x direction and the force F in the positive x direction constitute another couple.

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Problem 4.57 If the structure is subjected to couple forces applied at points A and B and the O lb, determine the moment of the force applied at A is FE D .8 {O C 10 |O 40 k/ couple about line a. Line a has direction angles x D 72ı , y D 36ı , and ´ D 60ı .

Solution E of the couple is is given by Using Eq. (4.11) on page 209, the moment M E D rEBA FEA D 24 {O C 12 |O 8 kO in. 8 {O C 10 |O M ˇ ˇ ˇ {O |O kO ˇˇ ˇ D ˇˇ 24 12 8 ˇˇ in.lb ˇ 8 10 40 ˇ D 400 {O C 896 |O C 144 kO in.lb:

40 kO lb

(1) (2) (3)

E in the direction of line a is then given by The component of M E aO D Ma D M

400 {O C 896 |O C 144 kO in.lb cos 72ı {O C cos 36ı |O C cos 60ı kO

D 673 in.lb:

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(4) (5)

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Problem 4.58 The structure shown has two forces of magnitude F applied. If F D 42 lb and the forces are to be a couple, determine the orientation they should have so that the moment of the couple about line a is as large as possible. Line a has direction angles x D 72ı , y D 36ı , and ´ D 60ı .

Solution Two solutions will be provided. Solution 1 does not require careful visualization, is straightforward in concept, but will require the use of a computer mathematics program to solve a system of nonlinear equations. Solution 2 is straightforward in terms of the mathematics involved, but the optimal direction for the forces must be identified by inspection. Solution 1 From the information given in the problem statement, the position vector from point B to point A, and a unit vector in the a direction, may be written as O in.; 8 k/

rEBA D .24 {O C 12 |O

O uO a D cos 72ı {O C cos 36ı |O C cos 60ı k:

Let the force applied at point A be expressed as FEA D 42 lb c1 {O C c2 |O C c3 kO

where c12 C c22 C c32 D 1:

(1)

(2)

In the above expression, c1 , c2 , and c3 are the direction cosines for FE . The moment of the couple about line a can be determined using the scalar triple product as ˇ ˇ ˇ cos 72ı cos 36ı cos 60ı ˇ ˇ ˇ 12 8 ˇˇ 42 in.lb Ma D rEBA FEA uO a D ˇˇ 24 (3) ˇ c1 ˇ c2 c3 D cos 72ı .12c3 C 8c2 / cos 36ı .24c3 C 8c1 / C cos 60ı .24c2 12c1 / 42 in.lb: (4) Now determine c1 , c2 , and c3 so that Ma is maximized subject to the constraint that c12 C c22 C c32 D 1. We will use Mathematica to follow through on this strategy: eq1 = c12 + c22 + c32

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