staticsChapter 3

PROBLEM 3.1 A 13.2-N force P is applied to the lever which controls the auger of a snowblower. Determine the moment of P about A when α is equal to 30°.

SOLUTION First note Px = P sin α = (13.2 N ) sin 30° = 6.60 N

Py = P cos α = (13.2 N ) cos 30° = 11.4315 N Noting that the direction of the moment of each force component about A is counterclockwise, M A = xB/ A Py + yB/ A Px = ( 0.086 m )(11.4315 N ) + ( 0.122 m )( 6.60 N ) = 1.78831 N ⋅ m or M A = 1.788 N ⋅ m

W

PROBLEM 3.2 The force P is applied to the lever which controls the auger of a snowblower. Determine the magnitude and the direction of the smallest force P which has a 2.20- N ⋅ m counterclockwise moment about A.

SOLUTION For P to be a minimum, it must be perpendicular to the line joining points A and B. rAB =

(86 mm )2 + (122 mm )2  y

= 149.265 mm

 122 mm 

α = θ = tan −1   = tan −1   = 54.819° x  86 mm  M A = rAB Pmin

Then or

Pmin = =

MA rAB 2.20 N ⋅ m  1000 mm    149.265 mm  1 m 

= 14.7389 N ∴ Pmin = 14.74 N

54.8° or Pmin = 14.74 N

35.2° W

PROBLEM 3.3 A 13.1-N force P is applied to the lever which controls the auger of a snowblower. Determine the value of α knowing that the moment of P about A is counterclockwise and has a magnitude of 1.95 N ⋅ m.

SOLUTION M A = rB/ A P sin θ

By definition

θ = φ + ( 90° − α )

where

 122 mm 

φ = tan −1   = 54.819°  86 mm 

and

Also Then

rB/ A =

(86 mm )2 + (122 mm )2

= 149.265 mm

1.95 N ⋅ m = ( 0.149265 m )(13.1 N ) sin ( 54.819° + 90° − α )

or

sin (144.819° − α ) = 0.99725

or

144.819° − α = 85.752°

and

144.819° − α = 94.248° ∴ α = 50.6°, 59.1° W

PROBLEM 3.4 A foot valve for a pneumatic system is hinged at B. Knowing that α = 28°, determine the moment of the 4-lb force about point B by resolving the force into horizontal and vertical components.

SOLUTION Note that

θ = α − 20° = 28° − 20° = 8°

and

Fx = ( 4 lb ) cos8° = 3.9611 lb Fy = ( 4 lb ) sin 8° = 0.55669 lb

Also

x = ( 6.5 in.) cos 20° = 6.1080 in. y = ( 6.5 in.) sin 20° = 2.2231 in.

Noting that the direction of the moment of each force component about B is counterclockwise, M B = xFy + yFx

= ( 6.1080 in.)( 0.55669 lb ) + ( 2.2231 in.)( 3.9611 lb ) = 12.2062 lb ⋅ in. or M B = 12.21 lb ⋅ in. W

PROBLEM 3.5 A foot valve for a pneumatic system is hinged at B. Knowing that α = 28°, determine the moment of the 4-lb force about point B by resolving the force into components along ABC and in a direction perpendicular to ABC.

SOLUTION First resolve the 4-lb force into components P and Q, where Q = ( 4.0 lb ) sin 28° = 1.87787 lb

Then

M B = rA/BQ

= ( 6.5 in.)(1.87787 lb ) = 12.2063 lb ⋅ in. or M B = 12.21 lb ⋅ in. W

PROBLEM 3.6 It is known that a vertical force of 800 N is required to remove the nail at C from the board. As the nail first starts moving, determine (a) the moment about B of the force exerted on the nail, (b) the magnitude of the force P which creates the same moment about B if α = 10°, (c) the smallest force P which creates the same moment about B.

SOLUTION (a) Have

M B = rC/B FN

= ( 0.1 m )( 800 N ) = 80.0 N ⋅ m or M B = 80.0 N ⋅ m

W

(b) By definition M B = rA/B P sin θ

where

θ = 90° − ( 90° − 70° ) − α = 90° − 20° − 10° = 60° ∴ 80.0 N ⋅ m = ( 0.45 m ) P sin 60° P = 205.28 N or P = 205 N W

(c) For P to be minimum, it must be perpendicular to the line joining points A and B. Thus, P must be directed as shown. Thus or

M B = dPmin = rA/B Pmin 80.0 N ⋅ m = ( 0.45 m ) Pmin ∴ Pmin = 177.778 N or Pmin = 177.8 N

20° W

PROBLEM 3.7 A sign is suspended from two chains AE and BF. Knowing that the tension in BF is 45 lb, determine (a) the moment about A of the force exert by the chain at B, (b) the smallest force applied at C which creates the same moment about A.

SOLUTION M A = rB/ A × TBF

(a) Have

Noting that the direction of the moment of each force component about A is counterclockwise, M A = xTBFy + yTBFx = ( 6.5 ft )( 45 lb ) sin 60° + ( 4.4 ft − 3.1 ft )( 45 lb ) cos 60° = 282.56 lb ⋅ ft or M A = 283 lb ⋅ ft (b) Have

W

M A = rC/ A × ( FC )min

For FC to be minimum, it must be perpendicular to the line joining points A and C. ∴ M A = d ( FC )min where

d = rC/ A =

( 6.5 ft )2 + ( 4.4 ft )2

= 7.8492 ft

∴ 282.56 lb ⋅ ft = ( 7.8492 ft ) ( FC )min

( FC )min

= 35.999 lb

 4.4 ft 

φ = tan −1   = 34.095°  6.5 ft  θ = 90° − φ = 90° − 34.095° = 55.905° or

( FC )min

= 36.0 lb

55.9° W

PROBLEM 3.8 A sign is suspended from two chains AE and BF. Knowing that the tension in BF is 45 lb, determine (a) the moment about A of the force exerted by the chain at B, (b) the magnitude and sense of the vertical force applied at C which creates the same moment about A, (c) the smallest force applied at B which creates the same moment about A.

SOLUTION M A = rB/ A × TBF

(a) Have

Noting that the direction of the moment of each force component about A is counterclockwise, M A = xTBFy + yTBFx = ( 6.5 ft )( 45 lb ) sin 60° + ( 4.4 ft − 3.1 ft )( 45 lb ) cos 60° = 282.56 lb ⋅ ft or M A = 283 lb ⋅ ft

W

M A = rC/ A × FC

(b) Have

M A = xFC

or ∴ FC =

MA 282.56 lb ⋅ ft = = 43.471 lb x 6.5 ft or FC = 43.5 lb W

M A = rB/ A × ( FB )min

(c) Have

For FB to be minimum, it must be perpendicular to the line joining points A and B. ∴ M A = d ( FB )min d =

where ∴ and

( 6.5 ft )2 + ( 4.4 ft

( FB )min

=

− 3.1 ft ) = 6.6287 ft 2

MA 282.56 lb ⋅ ft = = 42.627 lb d 6.6287 ft 

6.5 ft



θ = tan −1   = 78.690°  4.4 ft − 3.1 ft  or

( FB )min

= 42.6 lb

78.7° W

PROBLEM 3.9 The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts a 125-N force directed along its center line on the ball and socket at B, determine the moment of the force about A.

SOLUTION First note

dCB =

( 240 mm )2 + ( 46.6 mm )2

= 244.48 mm Then

and

where

240 mm 244.48 mm

sin θ =

46.6 mm 244.48 mm

FCB = FCB cosθ i − FCB sin θ j =

Now

cosθ =

125 N ( 240 mm ) i − ( 46.6 mm ) j 244.48 mm  M A = rB/ A × FCB

rB/ A = ( 306 mm ) i − ( 240 mm + 46.6 mm ) j = ( 306 mm ) i − ( 286.6 mm ) j

Then

125 N M A = ( 306 mm ) i − ( 286.6 mm ) j × ( 240i − 46.6 j) 244.48 = ( 27878 N ⋅ mm ) k = ( 27.878 N ⋅ m ) k or M A = 27.9 N ⋅ m

W

PROBLEM 3.10 The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts a 125-N force directed along its center line on the ball and socket at B, determine the moment of the force about A.

SOLUTION First note Then and

dCB =

( 344 mm )2 + (152.4 mm )2

cosθ =

344 mm 376.25 mm

152.4 mm 376.25 mm

FCB = ( FCB cosθ ) i − ( FCB sin θ ) j =

Now

sin θ =

= 376.25 mm

125 N ( 344 mm ) i + (152.4 mm ) j 376.25 mm  M A = rB/ A × FCB

where

rB/ A = ( 410 mm ) i − ( 87.6 mm ) j

Then

125 N M A = ( 410 mm ) i − ( 87.6 mm ) j × ( 344i − 152.4 j) 376.25 = ( 30770 N ⋅ mm ) k = ( 30.770 N ⋅ m ) k or M A = 30.8 N ⋅ m

W

PROBLEM 3.11 A winch puller AB is used to straighten a fence post. Knowing that the tension in cable BC is 260 lb, length a is 8 in., length b is 35 in., and length d is 76 in., determine the moment about D of the force exerted by the cable at C by resolving that force into horizontal and vertical components applied (a) at point C, (b) at point E.

SOLUTION Slope of line EC =

(a)

Then

and Then

TABx =

35 in. 5 = 76 in. + 8 in. 12

12 (TAB ) 13

=

12 ( 260 lb ) = 240 lb 13

TABy =

5 ( 260 lb ) = 100 lb 13

M D = TABx ( 35 in.) − TABy ( 8 in.) = ( 240 lb )( 35 in.) − (100 lb )( 8 in.) = 7600 lb ⋅ in.

or M D = 7600 lb ⋅ in.

(b) Have

M D = TABx ( y ) + TABy ( x )

= ( 240 lb )( 0 ) + (100 lb )( 76 in.) = 7600 lb ⋅ in. or M D = 7600 lb ⋅ in.

PROBLEM 3.12 It is known that a force with a moment of 7840 lb ⋅ in. about D is required to straighten the fence post CD. If a = 8 in., b = 35 in., and d = 112 in., determine the tension that must be developed in the cable of winch puller AB to create the required moment about point D.

SOLUTION

Slope of line EC =

35 in. 7 = 112 in. + 8 in. 24

Then

TABx =

24 TAB 25

and

TABy =

7 TAB 25

Have

M D = TABx ( y ) + TABy ( x ) ∴ 7840 lb ⋅ in. =

24 7 TAB ( 0 ) + TAB (112 in.) 25 25 TAB = 250 lb

or TAB = 250 lb

PROBLEM 3.13 It is known that a force with a moment of 1152 N ⋅ m about D is required to straighten the fence post CD. If the capacity of the winch puller AB is 2880 N, determine the minimum value of distance d to create the specified moment about point D knowing that a = 0.24 m and b = 1.05 m.

SOLUTION

The minimum value of d can be found based on the equation relating the moment of the force TAB about D: M D = (TAB max ) y ( d ) M D = 1152 N ⋅ m

where

(TAB max ) y

= TAB max sin θ = ( 2880 N ) sin θ 1.05 m

sin θ =

Now

(d

 ∴ 1152 N ⋅ m = 2880 N   

or or or

+ 0.24 ) + (1.05 ) m 2

1.05

( d + 0.24 )2 + (1.05)2

( d + 0.24 )2 + (1.05)2 (d

2

  (d )  

= 2.625d

+ 0.24 ) + (1.05 ) = 6.8906d 2 2

2

5.8906d 2 − 0.48d − 1.1601 = 0

Using the quadratic equation, the minimum values of d are 0.48639 m and −0.40490 m. Since only the positive value applies here, d = 0.48639 m or d = 486 mm

PROBLEM 3.14 A mechanic uses a piece of pipe AB as a lever when tightening an alternator belt. When he pushes down at A, a force of 580 N is exerted on the alternator B. Determine the moment of that force about bolt C if its line of action passes through O.

SOLUTION M C = rB/C × FB

Have

Noting the direction of the moment of each force component about C is clockwise, M C = xFBy + yFBx where

x = 144 mm − 78 mm = 66 mm y = 86 mm + 108 mm = 194 mm

and

FBx =

FBy =

78

( 78)

2

+ ( 86 )

2

86

( 78) + (86 ) 2

2

( 580 N ) = 389.65 N ( 580 N ) = 429.62 N

∴ M C = ( 66 mm )( 429.62 N ) + (194 mm )( 389.65 N ) = 103947 N ⋅ mm = 103.947 N ⋅ m or M C = 103.9 N ⋅ m

PROBLEM 3.15 Form the vector products B × C and B′ × C, where B = B′, and use the results obtained to prove the identity sin α cos β =

SOLUTION

1 sin 2

(α + β ) +

1 sin 2

(α − β ) .

B = B ( cos β i + sin β j)

First note

B′ = B ( cos β i − sin β j) C = C ( cos α i + sin α j) By definition

Now

B × C = BC sin (α − β )

(1)

B′ × C = BC sin (α + β )

(2)

B × C = B ( cos β i + sin β j) × C ( cos α i + sin α j) = BC ( cos β sin α − sin β cos α ) k

(3)

B × C = B ( cos β i − sin β j) × C ( cos α i + sin α j) = BC ( cos β sin α + sin β cos α ) k Equating magnitudes of B × C from Equations (1) and (3),

(4) (5)

sin (α − β ) = cos β sin α − sin β cos α Similarly, equating magnitudes of B′ × C from Equations (2) and (4), sin (α + β ) = cos β sin α + sin β cos α

(6)

Adding Equations (5) and (6) sin (α − β ) + sin (α + β ) = 2cos β sin α ∴ sin α cos β =

1 1 sin (α + β ) + sin (α − β ) 2 2

PROBLEM 3.16 A line passes through the points (420 mm, −150 mm) and (−140 mm, 180 mm). Determine the perpendicular distance d from the line to the origin O of the system of coordinates.

SOLUTION d = λ AB × rO/ A

Have

λ AB =

where

rB/ A rB/ A

rB/ A = ( −140 mm − 420 mm ) i + 180 mm − ( −150 mm )  j

and

= − ( 560 mm ) i + ( 330 mm ) j rB/ A =

∴ λ AB =

( −560 )2 + ( 330 )2 mm = 650 mm − ( 560 mm ) i + ( 330 mm ) j 1 = ( −56i + 33j) 650 mm 65

rO/ A = ( 0 − x A ) i + ( 0 − y A ) j = − ( 420 mm ) i + (150 mm ) j ∴ d =

1 ( −56i − 33j) ×  − ( 420 mm ) i + (150 mm ) j = 84.0 mm 65 d = 84.0 mm

PROBLEM 3.17 A plane contains the vectors A and B. Determine the unit vector normal to the plane when A and B are equal to, respectively, (a) 4i − 2j + 3k and −2i + 6j − 5k, (b) 7i + j − 4k and −6i − 3k + 2k.

SOLUTION λ =

(a) Have

A×B A×B

A = 4i − 2 j + 3k

where

B = −2i + 6 j − 5k

Then

i j k A × B = 4 −2 3 = (10 − 18 ) i + ( −6 + 20 ) j + ( 24 − 4 ) k = 2 ( −4i + 7 j + 10k ) −2 6 −5 A×B = 2

and ∴ λ =

( −4 )2 + ( 7 )2 + (10 )2

2 ( −4i + 7 j + 10k )

or λ =

2 165 λ =

(b) Have

= 2 165 1 ( −4i + 7 j + 10k ) 165

A×B A×B

A = 7i + j − 4k

where

B = −6i − 3j + 2k

Then

and

i j k A × B = 7 1 −4 = ( 2 − 12 ) i + ( 24 − 14 ) j + ( −21 + 6 ) k = 5 ( −2i + 2 j − 3k ) −6 −3 2 A×B = 5 ∴ λ =

( −2 )2 + ( 2 )2 + ( −3)2

5 ( −2i + 2 j − 3k ) 5 17

= 5 17 or λ =

1 ( −2i + 2 j − 3k ) 17

PROBLEM 3.18 The vectors P and Q are two adjacent sides of a parallelogram. Determine the area of the parallelogram when (a) P = (8 in.)i + (2 in.)j − (1 in.)k and Q = −(3 in.)i + (4 in.)j + (2 in.)k, (b) P = −(3 in.)i + (6 in.)j + (4 in.)k and Q = (2 in.)i + (5 in.)j − (3 in.)k.

SOLUTION A = P×Q

(a) Have

P = ( 8 in.) i + ( 2 in.) j − (1 in.) k

where

Q = − ( 3 in.) i + ( 4 in.) j + ( 2 in.) k i j k P × Q = 8 2 −1 in 2 = ( 4 + 4 ) i + ( 3 − 16 ) j + ( 32 + 6 ) k  in 2 −3 4 2

Then

(

) (

) (

)

= 8 in 2 i − 13 in 2 j + 38 in 2 k ∴ Α=

(8)2 + ( −13)2 + ( 38)2 in 2

= 40.951 in 2

or A = 41.0 in 2

A = P×Q

(b) Have

P = − ( 3 in.) i + ( 6 in.) j + ( 4 in.) k

where

Q = ( 2 in.) i + ( 5 in.) j − ( 3 in.) k

Then

i j k P × Q = −3 6 4 in 2 = ( −18 − 20 ) i + ( 8 − 9 ) j + ( −15 − 12 ) k  in 2 2 5 −3

(

) (

) (

)

= − 38 in 2 i − 1 in 2 j − 27 in 2 k ∴ Α=

( −38)2 + ( −1)2 + ( −27 )2 in 2

= 46.626 in 2

or A = 46.6 in 2

PROBLEM 3.19 Determine the moment about the origin O of the force F = −(5 N)i − (2 N)j + (3 N)k which acts at a point A. Assume that the position vector of A is (a) r = (4 m)i − (2 m)j − (1 m)k, (b) r = −(8 m)i + (3 m)j + (4 m)k, (c) r = (7.5 m)i + (3 m)j − (4.5 m)k.

SOLUTION MO = r × F

(a) Have

F = − (5 N ) i − ( 2 N ) j + (3 N ) k

where

r = ( 4 m ) i − ( 2 m ) j − (1 m ) k

∴ MO

i j k = 4 −2 −1 N ⋅ m = ( −6 − 2 ) i + ( 5 − 12 ) j + ( −8 − 10 ) k  N ⋅ m −5 −2 3 = ( −8i − 7 j − 18k ) N ⋅ m or M O = − ( 8 N ⋅ m ) i − ( 7 N ⋅ m ) j − (18 N ⋅ m ) k MO = r × F

(b) Have

F = − (5 N ) i − ( 2 N ) j + (3 N ) k

where

r = − (8 m ) i + ( 3 m ) j − ( 4 m ) k

∴ MO

i j k = −8 3 4 N ⋅ m = ( 9 + 8 ) i + ( −20 + 24 ) j + (16 + 15 ) k  N ⋅ m −5 −2 3 = (17i + 4 j + 31k ) N ⋅ m or M O = (17 N ⋅ m ) i + ( 4 N ⋅ m ) j + ( 31 N ⋅ m ) k

(c) Have where

MO = r × F F = − (5 N ) i − ( 2 N ) j + (3 N ) k r = ( 7.5 m ) i + ( 3 m ) j − ( 4.5 m ) k

PROBLEM 3.19 CONTINUED ∴ MO

i j k = 7.5 3 −4.5 N ⋅ m = ( 9 − 9 ) i + ( 22.5 − 22.5) j + ( −15 + 15 ) k  N ⋅ m −5 −2 3 or M O = 0

−2   r  . Therefore, vector F has a line of action This answer is expected since r and F are proportional  F = 3   passing through the origin at O.

PROBLEM 3.20 Determine the moment about the origin O of the force F = −(1.5 lb)i + (3 lb)j − (2 lb)k which acts at a point A. Assume that the position vector of A is (a) r = (2.5 ft)i − (1 ft)j + (2 ft)k, (b) r = (4.5 ft)i − (9 ft)j + (6 ft)k, (c) r = (4 ft)i − (1 ft)j + (7 ft)k.

SOLUTION MO = r × F

(a) Have

F = − (1.5 lb ) i + ( 3 lb ) j + ( 2 lb ) k

where

r = ( 2.5 ft ) i − (1 ft ) j + ( 2 ft ) k Then

MO

i j k = 2.5 −1 2 lb ⋅ ft = ( 2 − 6 ) i + ( −3 + 5 ) j + ( 7.5 − 1.5 ) k  lb ⋅ ft −1.5 3 −2 or M O = − ( 4 lb ⋅ ft ) i + ( 2 lb ⋅ ft ) j + ( 6 lb ⋅ ft ) k MO = r × F

(b) Have

F = − (1.5 lb ) i + ( 3 lb ) j − ( 2 lb ) k

where

r = ( 4.5 ft ) i − ( 9 ft ) j + ( 6 ft ) k Then

MO

i j k = 4.5 −9 6 lb ⋅ ft = (18 − 18 ) i + ( −9 + 9 ) j + (13.5 − 13.5 ) k  lb ⋅ ft −1.5 3 −2 or M O = 0

−1   r . This answer is expected since r and F are proportional  F = 3   Therefore, vector F has a line of action passing through the origin at O. MO = r × F

(c) Have

F = − (1.5 lb ) i − ( 3 lb ) j − ( 2 lb ) k

where

r = ( 4 ft ) i − (1 ft ) j + ( 7 ft ) k Then

MO

i j k = 4 −1 7 lb ⋅ ft = ( 2 − 21) i + ( −10.5 + 8 ) j + (12 − 1.5 ) k  lb ⋅ ft −1.5 3 −2 or M O = − (19 lb ⋅ ft ) i − ( 2.5 lb ⋅ ft ) j + (10.5 lb ⋅ ft ) k

PROBLEM 3.21 Before the trunk of a large tree is felled, cables AB and BC are attached as shown. Knowing that the tension in cables AB and BC are 777 N and 990 N, respectively, determine the moment about O of the resultant force exerted on the tree by the cables at B.

SOLUTION

M O = rB/O × FB

Have

rB/O = ( 8.4 m ) j

where

FB = TAB + TBC TAB = λ BATAB =

− ( 0.9 m ) i − ( 8.4 m ) j + ( 7.2 m ) k

TBC = λ BCTBC =

( 0.9 )2 + (8.4 )2 + ( 7.2 )2

m

( 777 N )

( 5.1 m ) i − (8.4 m ) j + (1.2 m ) k 990 N ( ) ( 5.1)2 + (8.4 )2 + (1.2 )2 m

PROBLEM 3.21 CONTINUED ∴ FB =  − ( 63.0 N ) i − ( 588 N ) j + ( 504 N ) k  + ( 510 N ) i − ( 840 N ) j + (120 N ) k 

= ( 447 N ) i − (1428 N ) j + ( 624 N ) k and

MO

i j k = 0 8.4 0 N ⋅ m = ( 5241.6 N ⋅ m ) i − ( 3754.8 N ⋅ m ) k 447 −1428 624 or M O = ( 5.24 kN ⋅ m ) i − ( 3.75 kN ⋅ m ) k

PROBLEM 3.22 Before a telephone cable is strung, rope BAC is tied to a stake at B and is passed over a pulley at A. Knowing that portion AC of the rope lies in a plane parallel to the xy plane and that the tension T in the rope is 124 N, determine the moment about O of the resultant force exerted on the pulley by the rope.

SOLUTION Have where

M O = rA/O × R rA/O = ( 0 m ) i + ( 9 m ) j + (1 m ) k R = T1 + T2 T1 = − (124 N ) cos10°  i − (124 N ) sin10° j

= − (122.116 N ) i − ( 21.532 N ) j  (1.5 m ) i − ( 9 m ) j + (1.8 m ) k T2 = λT2 =   2 2 2  (1.5 m ) + ( 9 m ) + (1.8 m )

  (124 N )  

= ( 20 N ) i − (120 N ) j + ( 24 N ) k ∴ R = − (102.116 N ) i − (141.532 N ) j + ( 24 N ) k

MO

i j k = 0 9 1 N⋅m −102.116 −141.532 24 = ( 357.523 N ⋅ m ) i − (102.116 N ⋅ m ) j + ( 919.044 N ⋅ m ) k or M O = ( 358 N ⋅ m ) i − (102.1 N ⋅ m ) j + ( 919 N ⋅ m ) k

PROBLEM 3.23 An 8-lb force is applied to a wrench to tighten a showerhead. Knowing that the centerline of the wrench is parallel to the x axis, determine the moment of the force about A.

SOLUTION Have

M A = rC/ A × F

where

rC/ A = ( 8.5 in.) i − ( 2.0 in.) j + ( 5.5 in.) k Fx = − ( 8cos 45° sin12° ) lb

Fy = − ( 8sin 45° ) lb Fz = − ( 8cos 45° cos12° ) lb ∴ F = − (1.17613 lb ) i − ( 5.6569 lb ) j − ( 5.5332 lb ) k

and

i j k MA = 8.5 −2.0 5.5 lb ⋅ in. −1.17613 −5.6569 −5.5332 = ( 42.179 lb ⋅ in.) i + ( 40.563 lb ⋅ in.) j − ( 50.436 lb ⋅ in.) k or M A = ( 42.2 lb ⋅ in.) i + ( 40.6 lb ⋅ in.) j − ( 50.4 lb ⋅ in.) k

PROBLEM 3.24 A wooden board AB, which is used as a temporary prop to support a small roof, exerts at point A of the roof a 228 N force directed along BA. Determine the moment about C of that force.

SOLUTION Have

M C = rA/C × FBA

where

rA/C = ( 0.96 m ) i − ( 0.12 m ) j + ( 0.72 m ) k

and

FBA = λ BA FBA   − ( 0.1 m ) i + (1.8 m ) j − ( 0.6 m ) k  = 228 N )  ( 2 2 2 ( 0.1) + (1.8) + ( 0.6 ) m  

= − (12.0 N ) i + ( 216 N ) j − ( 72 N ) k

∴ MC

i j k = 0.96 −0.12 0.72 N ⋅ m −12.0 216 −72 = − (146.88 N ⋅ m ) i + ( 60.480 N ⋅ m ) j + ( 205.92 N ⋅ m ) k or M C = − (146.9 N ⋅ m ) i + ( 60.5 N ⋅ m ) j + ( 206 N ⋅ m ) k

PROBLEM 3.25 The ramp ABCD is supported by cables at corners C and D. The tension in each of the cables is 360 lb. Determine the moment about A of the force exerted by (a) the cable at D, (b) the cable at C.

SOLUTION M A = rE/ A × TDE

(a) Have

rE/ A = ( 92 in.) j

where TDE = λ DETDE =

( 24 in.) i + (132 in.) j − (120 in.) k 360 lb ( ) ( 24 )2 + (132 )2 + (120 )2 in.

= ( 48 lb ) i + ( 264 lb ) j − ( 240 lb ) k i j k ∴ M A = 0 92 0 lb ⋅ in. = − ( 22, 080 lb ⋅ in.) i − ( 4416 lb ⋅ in ) k 48 264 −240 or M A = − (1840 lb ⋅ ft ) i − ( 368 lb ⋅ ft ) k M A = rG/ A × TCG

(b) Have

rG/ A = (108 in.) i + ( 92 in.) j

where TCG = λ CGTCG =

− ( 24 in.) i + (132 in.) j − (120 in.) k

( 24 )2 + (132 )2 + (120 )2 in.

( 360 lb )

= − ( 48 lb ) i + ( 264 lb ) j − ( 240 lb ) k i j k ∴ M A = 108 92 0 lb ⋅ in. −48 264 −240 = − ( 22, 080 lb ⋅ in.) i + ( 25,920 lb ⋅ in.) j + ( 32,928 lb ⋅ in.) k or M A = − (1840 lb ⋅ ft ) i + ( 2160 lb ⋅ ft ) j + ( 2740 lb ⋅ ft ) k

PROBLEM 3.26 The arms AB and BC of a desk lamp lie in a vertical plane that forms an o

angle of 30 with the xy plane. To reposition the light, a force of magnitude 8 N is applied at C as shown. Determine the moment of the force about O knowing that AB = 450 mm, BC = 325 mm, and line CD is parallel to the z axis.

SOLUTION Have

M O = rC/O × FC

where

( rC/O ) x = ( ABxz + BCxz ) cos 30° ABxz = ( 0.450 m ) sin 45° = 0.31820 m BC xz = ( 0.325 m ) sin 50° = 0.24896 m

( rC/O ) y = (OAy + ABy − BC y ) = 0.150 m + ( 0.450 m ) cos 45° − ( 0.325 m ) cos 50° = 0.25929 m

( rC/O ) z = ( ABxz + BCxz ) sin 30° = ( 0.31820 m + 0.24896 m ) sin 30° = 0.28358 m rC/O = ( 0.49118 m ) i + ( 0.25929 m ) j + ( 0.28358 m ) k

or

or

( FC ) x

= − ( 8 N ) cos 45° sin 20° = −1.93476 N

( FC ) y

= − ( 8 N ) sin 45° = −5.6569 N

( FC ) z

= ( 8 N ) cos 45° cos 20° = 5.3157 N

FC = − (1.93476 N ) i − ( 5.6569 N ) j + ( 5.3157 N ) k

∴ MO

i j k = 0.49118 0.25929 0.28358 N ⋅ m −1.93476 −5.6569 5.3157

= ( 2.9825 N ⋅ m ) i − ( 3.1596 N ⋅ m ) j − ( 2.2769 N ⋅ m ) k or M O = ( 2.98 N ⋅ m ) i − ( 3.16 N ⋅ m ) j − ( 2.28 N ⋅ m ) k

PROBLEM 3.27 In Problem 3.21, determine the perpendicular distance from point O to cable AB. Problem 3.21: Before the trunk of a large tree is felled, cables AB and BC are attached as shown. Knowing that the tension in cables AB and BC are 777 N and 990 N, respectively, determine the moment about O of the resultant force exerted on the tree by the cables at B.

SOLUTION

Have

| M O | = TBAd

where

d = perpendicular distance from O to line AB.

Now

Μ O = rB/O × TBA

and

rB/O = ( 8.4 m ) j TBA = λ BATAB =

− ( 0.9 m ) i − ( 8.4 m ) j + ( 7.2 m ) k

( 0.9 ) + (8.4 ) + ( 7.2 ) m 2

2

2

( 777 N )

= − ( 63.0 N ) i − ( 588 N ) j + ( 504 N ) k ∴ MO =

and

i j k 0 8.4 0 N ⋅ m = ( 4233.6 N ⋅ m ) i + ( 529.2 N ⋅ m ) k −63.0 −588 504

| MO | =

( 4233.6 )2 + ( 529.2 )2

= 4266.5 N ⋅ m

∴ 4266.5 N ⋅ m = ( 777 N ) d

or

d = 5.4911 m

or d = 5.49 m

PROBLEM 3.28 In Problem 3.21, determine the perpendicular distance from point O to cable BC. Problem 3.21: Before the trunk of a large tree is felled, cables AB and BC are attached as shown. Knowing that the tension in cables AB and BC are 777 N and 990 N, respectively, determine the moment about O of the resultant force exerted on the tree by the cables at B.

SOLUTION Have

| M O | = TBC d

where

d = perpendicular distance from O to line BC. M O = rB/O × TBC rB/O = 8.4 m j TBC = λ BCTBC =

( 5.1 m ) i − (8.4 m ) j + (1.2 m ) k 990 N ( ) ( 5.1)2 + (8.4 )2 + (1.2 )2 m

= ( 510 N ) i − ( 840 N ) j + (120 N ) k

∴ MO

and

i j k = 0 8.4 0 = (1008 N ⋅ m ) i − ( 4284 N ⋅ m ) k 510 −840 120

| MO | =

(1008)2 + ( 4284 )2

= 4401.0 N ⋅ m

∴ 4401.0 N ⋅ m = ( 990 N ) d d = 4.4454 m

or d = 4.45 m

PROBLEM 3.29 In Problem 3.24, determine the perpendicular distance from point D to a line drawn through points A and B. Problem 3.24: A wooden board AB, which is used as a temporary prop to support a small roof, exerts at point A of the roof a 228 N force directed along BA. Determine the moment about C of that force.

SOLUTION Have

| M D | = FBAd

where

d = perpendicular distance from D to line AB. M D = rA/D × FBA rA/D = − ( 0.12 m ) j + ( 0.72 m ) k

FBA = λ BA FBA =

( − ( 0.1 m ) i + (1.8 m ) j − ( 0.6 m ) k ) ( 228 N ) ( 0.1)2 + (1.8)2 + ( 0.6 )2

m

= − (12.0 N ) i + ( 216 N ) j − ( 72 N ) k ∴ MD

i j k = 0 −0.12 0.72 N ⋅ m −12.0 216 −72

= − (146.88 N ⋅ m ) i − ( 8.64 N ⋅ m ) j − (1.44 N ⋅ m ) k

and

|MD | =

(146.88)2 + (8.64 )2 + (1.44 )2

= 147.141 N ⋅ m

∴ 147.141 N ⋅ m = ( 228 N ) d d = 0.64536 m

or d = 0.645 m

PROBLEM 3.30 In Problem 3.24, determine the perpendicular distance from point C to a line drawn through points A and B. Problem 3.24: A wooden board AB, which is used as a temporary prop to support a small roof, exerts at point A of the roof a 228 N force directed along BA. Determine the moment about C of that force.

SOLUTION Have

| M C | = FBAd

where

d = perpendicular distance from C to line AB. M C = rA/C × FBA rA/C = ( 0.96 m ) i − ( 0.12 m ) j + ( 0.72 m ) k FBA = λ BA FBA =

( − ( 0.1 m ) i + (1.8 m ) j − ( 0.6 ) k ) ( 228 N ) ( 0.1)2 + (1.8)2 + ( 0.6 )2

m

= − (12.0 N ) i + ( 216 N ) j − ( 72 N ) k i j k ∴ M C = 0.96 −0.12 0.72 N ⋅ m −12.0 216 −72

= − (146.88 N ⋅ m ) i − ( 60.48 N ⋅ m ) j + ( 205.92 N ⋅ m ) k

and

| MC | =

(146.88)2 + ( 60.48)2 + ( 205.92 )2

= 260.07 N ⋅ m

∴ 260.07 N ⋅ m = ( 228 N ) d d = 1.14064 m

or d = 1.141 m

PROBLEM 3.31 In Problem 3.25, determine the perpendicular distance from point A to portion DE of cable DEF. Problem 3.25: The ramp ABCD is supported by cables at corners C and D. The tension in each of the cables is 360 lb. Determine the moment about A of the force exerted by (a) the cable at D, (b) the cable at C.

SOLUTION

Have

M A = TDE d

where

d = perpendicular distance from A to line DE. M A = rE/ A × TDE

rE/ A = ( 92 in.) j TDE = λ DETDE =

( 24 in.) i + (132 in.) j − (120 in.) k 360 lb ( ) ( 24 )2 + (132 )2 + (120 )2 in.

= ( 48 lb ) i + ( 264 lb ) j − ( 240 lb ) k i j k ∴ M A = 0 92 0 N⋅m 48 264 −240 = − ( 22, 080 lb ⋅ in.) i − ( 4416 lb ⋅ in.) k

PROBLEM 3.31 CONTINUED and

MA =

( 22, 080 )2 + ( 4416 )2

= 22,517 lb ⋅ in.

∴ 22,517 lb ⋅ in. = ( 360 lb ) d d = 62.548 in.

or d = 5.21 ft W

PROBLEM 3.32 In Problem 3.25, determine the perpendicular distance from point A to a line drawn through points C and G. Problem 3.25: The ramp ABCD is supported by cables at corners C and D. The tension in each of the cables is 360 lb. Determine the moment about A of the force exerted by (a) the cable at D, (b) the cable at C.

SOLUTION

Have

M A = TCG d

where

d = perpendicular distance from A to line CG.

M A = rG/ A × TCG rG/ A = (108 in.) i + ( 92 in.) j

TCG = λ CGTCG =

− ( 24 in.) i + (132 in.) j − (120 in.) k

( 24 )

2

+ (132 ) + (120 ) in. 2

2

( 360 lb )

= − ( 48 lb ) i + ( 264 lb ) j − ( 240 lb ) k i j k ∴ M A = 108 92 0 lb ⋅ in. − 48 264 −240 = − ( 22, 080 lb ⋅ in.) i + ( 25,920 lb ⋅ in.) j + ( 32,928 lb ⋅ in.) k and

MA =

( 22, 080 )2 + ( 25,920 )2 + ( 32,928)2

= 47,367 lb ⋅ in.

∴ 47,367 lb ⋅ in. = ( 360 lb ) d d = 131.575 in.

or d = 10.96 ft W

PROBLEM 3.33 In Problem 3.25, determine the perpendicular distance from point B to a line drawn through points D and E.

Problem 3.25: The ramp ABCD is supported by cables at corners C and D. The tension in each of the cables is 360 lb. Determine the moment about A of the force exerted by (a) the cable at D, (b) the cable at C.

SOLUTION

M B = TDE d

Have where

d = perpendicular distance from B to line DE. M B = rE/B × TDE rE/B = − (108 in.) i + ( 92 in.) j TDE = λ DETDE =

( 24 in.) i + (132 in.) j − (120 in.) k 360 lb ( ) 2 2 2 24 132 120 in. + + ( ) ( ) ( )

= ( 48 lb ) i + ( 264 lb ) j − ( 240 lb ) k i j k ∴ M B = −108 92 0 lb ⋅ in. 48 264 −240 = − ( 22, 080 lb ⋅ in.) i − ( 25,920 lb ⋅ in.) j − ( 32,928 lb ⋅ in.) k and

MB =

( 22, 080 )2 + ( 25,920 )2 + ( 32,928)2

= 47,367 lb ⋅ in.

∴ 47,367 lb ⋅ in. = ( 360 lb ) d d = 131.575 in. or d = 10.96 ft W

PROBLEM 3.34 Determine the value of a which minimizes the perpendicular distance from point C to a section of pipeline that passes through points A and B.

SOLUTION Assuming a force F acts along AB,

M C = rA/C × F = F ( d ) where

d = perpendicular distance from C to line AB F = λ AB F =

(8 m ) i + ( 7 m ) j − ( 9 m ) k F ( 8 )2 + ( 7 )2 + ( 9 )2 m

= F ( 0.57437 ) i + ( 0.50257 ) j − ( 0.64616 ) k rA/C = (1 m ) i − ( 2.8 m ) j − ( a − 3 m ) k i j k ∴ MC = 1 −2.8 3−a F 0.57437 0.50257 −0.64616 = ( 0.30154 + 0.50257a ) i + ( 2.3693 − 0.57437a ) j

+ 2.1108k ] F Since

MC =

rA/C × F 2

or

rA/C × F 2 = ( dF )

2

∴ ( 0.30154 + 0.50257a ) + ( 2.3693 − 0.57437a ) + ( 2.1108 ) = d 2 2

Setting

2

2

( )

d d 2 = 0 to find a to minimize d da 2 ( 0.50257 )( 0.30154 + 0.50257a ) + 2 ( −0.57437 )( 2.3693 − 0.57437a ) = 0

Solving

a = 2.0761 m or a = 2.08 m W

PROBLEM 3.35 Given the vectors P = 7i − 2j + 5k, Q = −3i − 4j + 6k, and S = 8i + j − 9k, compute the scalar products P ⋅ Q, P ⋅ S, and Q ⋅ S.

SOLUTION P ⋅ Q = ( 7i − 2 j + 5k ) ⋅ ( −3i − 4 j + 6k ) = ( 7 )( −3) + ( −2 )( −4 ) + ( 5 )( 6 )

= 17 or P ⋅ Q = 17 W

P ⋅ S = ( 7i − 2 j + 5k ) ⋅ ( 8i + j − 9k )

= ( 7 )( 8 ) + ( −2 )(1) + ( 5 )( −9 ) =9 or P ⋅ S = 9 W Q ⋅ S = ( −3i − 4 j + 6k ) ⋅ ( 8i + j − 9k ) = ( −3)( 8 ) + ( −4 )(1) + ( 6 )( −9 ) = −82 or Q ⋅ S = −82 W

PROBLEM 3.36 Form the scalar products B ⋅ C and B′ ⋅ C, where B = B′, and use the results obtained to prove the identity cos α cos β =

1 2

cos (α + β ) + 12 cos (α − β ) .

SOLUTION By definition B ⋅ C = BC cos (α − β ) where

B = B ( cos β ) i + ( sin β ) j C = C ( cos α ) i + ( sin α ) j

∴ ( B cos β )( C cos α ) + ( B sin β )( C sin α ) = BC cos (α − β ) cos β cos α + sin β sin α = cos (α − β )

or

(1)

By definition B′ ⋅ C = BC cos (α + β ) where

B′ = ( cos β ) i − ( sin β ) j

∴ ( B cos β )( C cos α ) + ( − B sin β )( C sin α ) = BC cos (α + β ) or

cos β cos α − sin β sin α = cos (α + β )

(2)

Adding Equations (1) and (2), 2 cos β cos α = cos (α − β ) + cos (α + β ) or cos α cos β =

1 1 cos (α + β ) + cos (α − β ) W 2 2

PROBLEM 3.37 Consider the volleyball net shown. Determine the angle formed by guy wires AB and AC.

SOLUTION First note AB = rB/ A =

( −1.95 m )2 + ( −2.4 m )2 + ( 0.6 m )2

= 3.15 m AC = rC/ A =

( 0 m )2 + ( −2.4 m )2 + (1.8 m )2

= 3.0 m and rB/ A = − (1.95 m ) i − ( 2.40 m ) j + ( 0.6 m ) k rC/ A = − ( 2.40 m ) j + (1.80 m ) k By definition rB/ A ⋅ rC/ A = rB/ A rC/ A cosθ or

( −1.95i − 2.40 j + 0.6k ) ⋅ ( −2.40 j + 1.80k ) = ( 3.15)( 3.0 ) cosθ ( −1.95)( 0 ) + ( −2.40 )( −2.40 ) + ( 0.6 )(1.8) = 9.45cosθ ∴ cosθ = 0.72381

and

θ = 43.630° or θ = 43.6° W

PROBLEM 3.38 Consider the volleyball net shown. Determine the angle formed by guy wires AC and AD.

SOLUTION First note AC = rC/ A = AD = rD/ A =

and

( −2.4 )2 + (1.8)2

m = 3m

(1.2 )2 + ( −2.4 )2 + ( 0.3)2

m = 2.7 m

rC/ A = − ( 2.4 m ) j + (1.8 m ) k rD/ A = (1.2 m ) i − ( 2.4 m ) j + ( 0.3 m ) k

By definition rC/ A ⋅ rD/ A = rC/ A rD/ A cosθ or

( −2.4 j + 1.8k ) ⋅ (1.2i − 2.4 j + 0.3k ) = ( 3)( 2.7 ) cosθ ( 0 )(1.2 ) + ( −2.4 )( −2.4 ) + (1.8)( 0.3) = 8.1cosθ

and

cosθ =

6.3 = 0.77778 8.1

θ = 38.942° or θ = 38.9° W

PROBLEM 3.39 Steel framing members AB, BC, and CD are joined at B and C and are braced using cables EF and EG. Knowing that E is at the midpoint of BC and that the tension in cable EF is 330 N, determine (a) the angle between EF and member BC, (b) the projection on BC of the force exerted by cable EF at point E.

SOLUTION

λ BC ⋅ λ EF = (1)(1) cosθ

(a) By definition where

λ BC =

λ EF =

∴

(16 m ) i − ( 4.5 m ) j − (12 m ) k (16 )2 + ( 4.5)2 + (12 )2 m − (7 m) i − (6 m) j + (6 m)k

( 7 ) 2 + ( 6 )2 + ( 6 ) 2 m

=

=

1 ( −7i − 6 j + 6k ) 11.0

(16i − 4.5j − 12k ) ⋅ ( −7i − 6 j + 6k ) 20.5

1 (16i − 4.5j − 12k ) 20.5

11.0

= cosθ

(16 )( −7 ) + ( −4.5)( −6 ) + ( −12 )( 6 ) = ( 20.5)(11.0 ) cosθ and

 −157 

θ = cos −1   = 134.125°  225.5  or θ = 134.1° W

(b) By definition

(TEF )BC

= TEF cosθ = ( 330 N ) cos134.125° = −229.26 N or (TEF ) BC = −230 N W

PROBLEM 3.40 Steel framing members AB, BC, and CD are joined at B and C and are braced using cables EF and EG. Knowing that E is at the midpoint of BC and that the tension in cable EG is 445 N, determine (a) the angle between EG and member BC, (b) the projection on BC of the force exerted by cable EG at point E.

SOLUTION

λ BC ⋅ λ EG = (1)(1) cosθ

(a) By definition where

λ BC =

(16 m ) i − ( 4.5 m ) j − (12 m ) k (16 m )2 + ( 4.5)2 + (12 )2 m

=

16i − 4.5j − 12k 20.5

= 0.78049i − 0.21951j − 0.58537k λ EG =

(8 m ) i − ( 6 m ) j + ( 4.875 m ) k (8)2 + ( 6 )2 + ( 4.875)2 m

=

8i − 6 j + 4.875k 11.125

= 0.71910i − 0.53933j + 0.43820k ∴ λ BC ⋅ λ EG = and

16 ( 8 ) + ( −4.5 )( −6 ) + ( −12 )( 4.875 ) = cosθ ( 20.5)(11.25)  96.5 

θ = cos −1   = 64.967°  228.06  or θ = 65.0° W

(b) By definition

(TEG )BC

= TEG cosθ = ( 445 N ) cos 64.967° = 188.295 N or (TEG ) BC = 188.3 N W

PROBLEM 3.41 Slider P can move along rod OA. An elastic cord PC is attached to the slider and to the vertical member BC. Knowing that the distance from O to P is 0.12 m and the tension in the cord is 30 N, determine (a) the angle between the elastic cord and the rod OA, (b) the projection on OA of the force exerted by cord PC at point P.

SOLUTION λ OA ⋅ λ PC = (1)(1) cosθ

(a) By definition λ OA =

where

=

( 0.24 m ) i + ( 0.24 m ) j − ( 0.12 m ) k ( 0.24 )2 + ( 0.24 )2 + ( 0.12 )2 m 2 2 1 i+ j− k 3 3 3

Knowing that | rA/O | = LOA = 0.36 m and that P is located 0.12 m from O, it follows that the coordinates of P are

1 the coordinates of A. 3 ∴ P ( 0.08 m, 0.08 m, − 0.040 m )

Then

λ PC =

( 0.10 m ) i + ( 0.22 m ) j + ( 0.28 m ) k ( 0.10 )2 + ( 0.22 )2 + ( 0.28)2 m

= 0.27037i + 0.59481j + 0.75703k 2 1  2 ∴  i + j − k  ⋅ ( 0.27037i + 0.59481j + 0.75703k ) = cosθ 3 3  3

θ = cos −1 ( 0.32445) = 71.068°

and

or θ = 71.1° W

(b)

(TPC )OA

= TPC cosθ = ( 30 N ) cos 71.068°

(TPC )OA

= 9.7334 N or (TPC )OA = 9.73 N W

PROBLEM 3.42 Slider P can move along rod OA. An elastic cord PC is attached to the slider and to the vertical member BC. Determine the distance from O to P for which cord PC and rod OA are perpendicular.

SOLUTION The requirement that member OA and the elastic cord PC be perpendicular implies that λ OA ⋅ λ PC = 0 where

λ OA =

=

or

λ OA ⋅ rC/P = 0

( 0.24 m ) i + ( 0.24 m ) j − ( 0.12 m ) k ( 0.24 )2 + ( 0.24 )2 + ( 0.12 )2 m 2 2 1 i+ j− k 3 3 3

Letting the coordinates of P be P ( x, y, z ) , we have rC/P = ( 0.18 − x ) i + ( 0.30 − y ) j + ( 0.24 − z ) k  m

2 1  2 ∴  i + j − k  ⋅ ( 0.18 − x ) i + ( 0.30 − y ) j + ( 0.24 − z ) k  = 0 3 3 3   Since Then

rP/O = λ OAdOP =

x=

2 dOP , 3

y =

(1)

dOP ( 2i + 2 j − k ) , 3

2 dOP, 3

z =

−1 dOP 3

(2)

Substituting the expressions for x, y, and z from Equation (2) into Equation (1),   1 2 2 1 ( 2i + 2 j − k ) ⋅  0.18 − dOP  i +  0.30 − dOP  j +  0.24 + dOP  k  = 0 3 3 3 3        or

3dOP = 0.36 + 0.60 − 0.24 = 0.72 ∴ dOP = 0.24 m or dOP = 240 mm W

PROBLEM 3.43 Determine the volume of the parallelepiped of Figure 3.25 when (a) P = −(7 in.)i − (1 in.)j + (2 in.)k, Q = (3 in.)i − (2 in.)j + (4 in.)k, and S = −(5 in.)i + (6 in.)j − (1 in.)k, (b) P = (1 in.)i + (2 in.)j − (1 in.)k, Q = −(8 in.)i − (1 in.)j + (9 in.)k, and S = (2 in.)i + (3 in.)j + (1 in.)k.

SOLUTION Volume of a parallelepiped is found using the mixed triple product. (a)

Vol = P ⋅ ( Q × S ) −7 −1 2 = 3 −2 4 in 3 = ( −14 + 168 + 20 − 3 + 36 − 20 ) in 3 −5 6 −1 = 187 in 3 or Volume = 187 in 3 W

(b)

Vol = P ⋅ ( Q × S ) 1 2 −1 = −8 −1 9 in 3 = ( −1 − 27 + 36 + 16 + 24 − 2 ) in 3 2 3 1 = 46 in 3 or Volume = 46 in 3 W

PROBLEM 3.44 Given the vectors P = 4i − 2j + Pzk, Q = i + 3j − 5k, and S = −6i + 2j − k, determine the value of Pz for which the three vectors are coplanar.

SOLUTION For the vectors to all be in the same plane, the mixed triple product is zero.

P ⋅ (Q × S ) = 0 4 −2 Pz ∴ O = 1 3 −5 = −12 + 40 − 60 − 2 + Pz ( 2 + 18 ) −6 2 −1 so that

Pz =

34 = 1.70 20 or Pz = 1.700 W

PROBLEM 3.45 The 0.732 × 1.2-m lid ABCD of a storage bin is hinged along side AB and is held open by looping cord DEC over a frictionless hook at E. If the tension in the cord is 54 N, determine the moment about each of the coordinate axes of the force exerted by the cord at D.

SOLUTION First note

z =

( 0.732 )2 − ( 0.132 )2

m

= 0.720 m Then

d DE =

( 0.360 )2 + ( 0.720 )2 + ( 0.720 )2

m

= 1.08 m and Have

rE/D = ( 0.360 m ) i + ( 0.720 m ) j − ( 0.720 m ) k TDE = =

(

TOE rE/D d DE

)

54 N ( 0.360i + 0.720 j − 0.720k ) 1.08

= (18.0 N ) i + ( 36.0 N ) j − ( 36.0 N ) k Now

M A = rD/ A × TDE

where

rD/ A = ( 0.132 m ) j + ( 0.720 m ) k

Then

i j k M A = 0 0.132 0.720 N ⋅ m 18.0 36.0 −36.0

PROBLEM 3.45 CONTINUED

{

∴ M A = ( 0.132 )( −36.0 ) − ( 0.720 )( 36.0 )  i + ( 0.720 )(18.0 ) − 0  j

}

+ 0 − ( 0.132 )(18.0 )  k N ⋅ m or

M A = − ( 30.7 N ⋅ m ) i + (12.96 N ⋅ m ) j − ( 2.38 N ⋅ m ) k ∴ M x = −30.7 N ⋅ m, M y = 12.96 N ⋅ m, M z = −2.38 N ⋅ m W

PROBLEM 3.46 The 0.732 × 1.2-m -m lid ABCD of a storage bin is hinged along side AB and is held open by looping cord DEC over a frictionless hook at E. If the tension in the cord is 54 N, determine the moment about each of the coordinate axes of the force exerted by the cord at C.

SOLUTION z =

First note

( 0.732 )2 − ( 0.132 )2

m

= 0.720 m Then

dCE =

( 0.840 )2 + ( 0.720 )2 + ( 0.720 )2

m

= 1.32 m and

TCE =

=

rE/C dCE

(TCE )

− ( 0.840 m ) i + ( 0.720 m ) j − ( 0.720 m ) k ( 54 N ) 1.32 m

= − ( 36.363 N ) i + ( 29.454 N ) j − ( 29.454 N ) k Now

M A = rE/ A × TCE

where

rE/ A = ( 0.360 m ) i + ( 0.852 m ) j

Then

i j k M A = 0.360 0.852 0 N⋅m −34.363 29.454 −29.454 = − ( 25.095 N ⋅ m ) i + (10.6034 N ⋅ m ) j + ( 39.881 N ⋅ m ) k ∴ M x = −25.1 N ⋅ m, M y = 10.60 N ⋅ m, M z = 39.9 N ⋅ m W

PROBLEM 3.47 A fence consists of wooden posts and a steel cable fastened to each post and anchored in the ground at A and D. Knowing that the sum of the moments about the z axis of the forces exerted by the cable on the posts at B and C is −66 N · m, determine the magnitude TCD when TBA = 56 N.

SOLUTION Based on

| M z | = k ⋅ ( rB ) y × TBA  + k ⋅ ( rC ) y × TCD     

where

M z = − ( 66 N ⋅ m ) k

( rB ) y

= ( rC ) y = (1 m ) j

TBA = λ BATBA =

(1.5 m ) i − (1 m ) j + ( 3 m ) k 3.5 m

( 56 N )

= ( 24 N ) i − (16 N ) j + ( 48 N ) k

TCD = λ CDTCD =

( 2 m ) i − (1 m ) j − ( 2 m ) k T

=

1 TCD ( 2i − j − 2k ) 3

3.0 m

CD

{

}

∴ − ( 66 N ⋅ m ) = k ⋅ (1 m ) j × ( 24 N ) i − (16 N ) j + ( 48 N ) k   1  + k ⋅ (1 m ) j ×  TCD ( 2i − j − 2k )   3   or

−66 = −24 − ∴ TCD =

2 TCD 3

3 ( 66 − 24 ) N 2 or TCD = 63.0 N W

PROBLEM 3.48 A fence consists of wooden posts and a steel cable fastened to each post and anchored in the ground at A and D. Knowing that the sum of the moments about the y axis of the forces exerted by the cable on the posts at B and C is 212 N · m, determine the magnitude of TBA when TCD = 33 N.

SOLUTION Based on

| M y | = j ⋅ ( rB ) z × TBA + ( rC ) z × TCD 

where

M y = ( 212 N ⋅ m ) j

( rB ) z

= (8 m ) k

( rC ) z

= (2 m) k

TBA = λ BATBA =

(1.5 m ) i − (1 m ) j − ( 3 m ) k T

=

TBA (1.5i − j + 3k ) 3.5

BA

3.5 m

TCD = λ CDTCD =

( 2 m ) i − (1 m ) j − ( 2 m ) k 3.0 m

( 33 N )

= ( 22i − 11j − 22k ) N  T  ∴ ( 212 N ⋅ m ) = j ⋅ ( 8 m ) k ×  BA (1.5i − j + 3k )    3.5   + j ⋅ ( 2 m ) k × ( 22 i − 11j − 22k ) N 

or

212 =

8 (1.5 ) TBA + 2 ( 22 ) 3.5

∴ TBA =

168 18.6667 or TBA = 49.0 N W

PROBLEM 3.49 To lift a heavy crate, a man uses a block and tackle attached to the bottom of an I-beam at hook B. Knowing that the moments about the y and z axes of the force exerted at B by portion AB of the rope are, respectively, 100 lb ⋅ ft and −400 lb ⋅ ft , determine the distance a.

SOLUTION M O = rA/O × TBA

Based on where

M O = M xi + M y j + M zk = M xi + (100 lb ⋅ ft ) j − ( 400 lb ⋅ ft ) k

rA/O = ( 6 ft ) i + ( 4 ft ) j TBA = λ BATBA =

( 6 ft ) i − (12 ft ) j − ( a ) k T

BA

d BA

i j k T ∴ M xi + 100 j − 400k = 6 4 0 BA d 6 −12 −a BA =

TBA  − ( 4a ) i + ( 6a ) j − ( 96 ) k  d BA 

100 d BA 6a

From j-coefficient:

100d AB = 6aTBA or TBA =

From k -coefficient:

−400d AB = −96TBA or TBA =

Equating Equations (1) and (2) yields

a=

400 d BA 96

(1) (2)

100 ( 96 ) 6 ( 400 ) or a = 4.00 ft W

PROBLEM 3.50 To lift a heavy crate, a man uses a block and tackle attached to the bottom of an I-beam at hook B. Knowing that the man applies a 200-lb force to end A of the rope and that the moment of that force about the y axis is 175 lb ⋅ ft , determine the distance a.

SOLUTION

(

| M y | = j ⋅ rA/O × TBA

Based on

)

rA/O = ( 6 ft ) i + ( 4 ft ) j

where

TBA = λ BATBA = =

=

rA/B

d BA

TBA

( 6 ft ) i − (12 ft ) j − ( a ) k d BA

( 200 lb )

200 ( 6i − 12 j − ak ) d BA

0 1 0 200 4 0 ∴ 175 lb ⋅ ft = 6 d 6 −12 −a BA 200 175 = 0 − 6 ( −a )  d BA

where

d BA =

( 6 )2 + (12 )2 + ( a )2

ft

= 180 + a 2 ft ∴ 175 180 + a 2 = 1200a

or

180 + a 2 = 6.8571a

Squaring each side 180 + a 2 = 47.020a 2 Solving

a = 1.97771 ft

or a = 1.978 ft W

PROBLEM 3.51 A force P is applied to the lever of an arbor press. Knowing that P lies in a plane parallel to the yz plane and that M x = 230 lb ⋅ in.,

M y = −200 lb ⋅ in., and M z = −35 lb ⋅ in., determine the magnitude of P and the values of φ and θ.

SOLUTION Based on

M x = ( P cos φ ) ( 9 in.) sin θ  − ( P sin φ ) ( 9 in.) cosθ 

M y = − ( P cos φ )( 5 in.)

(2)

M z = − ( P sin φ )( 5 in.)

(3)

− ( P sin φ ) (5) Equation (3) M z : = Equation (2) M y − ( P cos φ ) (5)

Then

or

(1)

tan φ =

−35 = 0.175 −200

φ = 9.9262° or φ = 9.93° W

Substituting φ into Equation (2) −200 lb ⋅ in. = − ( P cos 9.9262° ) (5 in.) P = 40.608 lb

or P = 40.6 lb W Then, from Equation (1) 230 lb ⋅ in. = ( 40.608 lb ) cos 9.9262° ( 9 in.) sin θ  − ( 40.608 lb ) sin 9.9262°  ( 9 in.) cosθ 

or

0.98503sin θ − 0.172380cosθ = 0.62932

Solving numerically,

θ = 48.9° W

PROBLEM 3.52 A force P is applied to the lever of an arbor press. Knowing that P lies in a plane parallel to the yz plane and that M y = −180 lb ⋅ in. and M z = −30 lb ⋅ in., determine the moment M x of P about the x axis when θ = 60°.

SOLUTION Based on

M x = ( P cos φ ) ( 9 in.) sin θ  − ( P sin φ ) ( 9 in.) cosθ 

(1)

M y = − ( P cos φ )( 5 in.)

(2)

M z = − ( P sin φ )( 5 in.)

(3)

− ( P sin φ )( 5 ) Equation (3) M z : = Equation (2) M y − ( P cos φ )( 5 )

Then

−30 = tan φ −180

or

∴ φ = 9.4623° From Equation (3), −30 lb ⋅ in. = − ( P sin 9.4623° )( 5 in.) ∴ P = 36.497 lb From Equation (1), M x = ( 36.497 lb )( 9 in.)( cos 9.4623° sin 60° − sin 9.4623° cos 60° )

= 253.60 lb ⋅ in. or M x = 254 lb ⋅ in. W

PROBLEM 3.53 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable AE at A is 220 lb, determine the moment of that force about the line joining points D and B.

SOLUTION

(

M DB = λ DB ⋅ rA/D × TAE

Have λ DB =

where

( 48 in.) i − (14 in.) j 50 in.

)

= 0.96i − 0.28 j

rA/D = − ( 4 in.) j + ( 8 in.) k

( 36 in.) i − ( 24 in.) j + ( 8 in.) k  TAE = λ AETAE =  ( 220 lb ) 44 in. = (180 lb ) i − (120 lb ) j + ( 40 lb ) k

∴ M DB

0.960 −0.280 0 8 lb ⋅ in. = 0 −4 180 −120 40 = ( 0.960 ) ( −4 )( 40 ) − ( 8 )( −120 )  + ( −0.280 ) 8 (180 ) − 0 

= 364.8 lb ⋅ in. or M DB = 365 lb ⋅ in. W

PROBLEM 3.54 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable CF at C is 132 lb, determine the moment of that force about the line joining points D and B.

SOLUTION

(

M DB = λ DB ⋅ rC/D × TCF

Have where

λ DB =

( 48 in.) i − (14 in.) j 50 in.

)

= 0.96i − 0.28 j

rC/D = ( 8 in.) j − (16 in.) k TCF = λ CFTCF =

( 24 in.) i − ( 36 in.) j − (8 in.) k 44 in.

(132 lb )

= ( 72 lb ) i − (108 lb ) j − ( 24 lb ) k

∴ M DB

0.96 −0.28 0 8 = 0 −16 lb ⋅ in. 72 −108 −24 = 0.96 ( 8 )( −24 ) − ( −16 )( −108 )  + ( −0.28 ) ( −16 )( 72 ) − 0 

= −1520.64 lb ⋅ in. or M DB = −1521 lb ⋅ in. W

PROBLEM 3.55 A mast is mounted on the roof of a house using bracket ABCD and is guyed by cables EF, EG, and EH. Knowing that the force exerted by cable EF at E is 66 N, determine the moment of that force about the line joining points D and I.

SOLUTION M DI = λ DI ⋅ rF /I × TEF 

Have where

λ DI =

(1.6 m ) i − ( 0.4 m ) j (1.6 )2 + ( 0.4 )2 m

=

1 ( 4i − j) 17

rF /I = ( 4.6 m + 0.8 m ) k = ( 5.4 m ) k

TEF = λ EF TEF =

(1.2 m ) i − ( 3.6 m ) j + ( 5.4 m ) k 6.6 m

( 66 N )

= (12 N ) i − ( 36 N ) j + ( 54 N ) k = 6 ( 2 N ) i − ( 6 N ) j + ( 9 N ) k 

−1 0 0 1 2 −6 9 4

∴ M DI =

( 6 N )( 5.4 m ) 0 17

= 7.8582 ( 0 + 24 ) + ( −2 − 0 ) 

= 172.879 N ⋅ m or M DI = 172.9 N ⋅ m W

PROBLEM 3.56 A mast is mounted on the roof of a house using bracket ABCD and is guyed by cables EF, EG, and EH. Knowing that the force exerted by cable EG at E is 61.5 N, determine the moment of that force about the line joining points D and I.

SOLUTION Have

M DI = λ DI ⋅ rG/I × TEG 

where

λ DI = =

(1.6 m ) i − ( 0.4 m ) j 0.4 17 m 1 ( 4i − j) 17

rG/I = − (10.9 m + 0.8 m ) k = − (11.7 m ) k

TEG = λ EGTEG =

(1.2 m ) i − ( 3.6 m ) j − (11.7 m ) k 12.3 m

( 61.5 N )

= 5 (1.2 N ) i − ( 3.6 N ) j − (11.7 N ) k 

∴ M DI =

5 N (11.7 m ) 17

4 −1 0 −1 0 0 1.2 −3.6 −11.7

{

}

= (14.1883 N ⋅ m )  0 − ( 4 )( −1)( −3.6 )  + ( −1)( −1)(1.2 ) − 0  = −187.286 N ⋅ m

or M DI = −187.3 N ⋅ m

PROBLEM 3.57 A rectangular tetrahedron has six edges of length a. A force P is directed as shown along edge BC. Determine the moment of P about edge OA.

SOLUTION

(

M OA = λ OA ⋅ rC/O × P

Have

)

where From triangle OBC a 2

( OA) x

=

( OA) z

= ( OA ) x tan 30° =

( OA)2

Since

a 1  a =   2 3  2 3

= ( OA) x + ( OA) y + ( OAz ) 2

2

2

 a  2 a a 2 =   + ( OA) y +   2   2 3

or

∴

( OA) y

a2 −

=

2

2

a2 a2 2 − =a 4 12 3

Then

rA/O =

a 2 a i +a j+ k 2 3 2 3

and

λ OA =

1 i+ 2

2 1 j+ k 3 2 3

P = λ BC P = =

( a sin 30° ) i − ( a cos30° ) k a

(

P i − 3k 2

)

rC/O = ai

( P)

PROBLEM 3.57 CONTINUED

∴ M OA

1 2 = 1 1

2 1 3 2 3 P ( a )   0 0 2 0 − 3

=

aP  2  −  (1) − 3 2  3 

=

aP 2

(

) M OA =

aP 2

PROBLEM 3.58 A rectangular tetrahedron has six edges of length a. (a) Show that two opposite edges, such as OA and BC, are perpendicular to each other. (b) Use this property and the result obtained in Problem 3.57 to determine the perpendicular distance between edges OA and BC.

SOLUTION (a) For edge OA to be perpendicular to edge BC, JJJG JJJG OA ⋅ BC = 0 where From triangle OBC

and

Then

or

so that

a 2

( OA) x

=

( OA) z

= ( OA ) x tan 30° =

a 1  a = 2  3  2 3

JJJG  a   a  ∴ OA =   i + ( OA)y j +  k 2 2 3 JJJG BC = ( a sin 30° ) i − ( a cos 30° ) k

=

a a 3 i− k 2 2

=

a i − 3k 2

(

)

a  a   a =0  i + ( OA ) y j +   k  ⋅ i − 3k 2 2 2 3    

(

)

a2 a2 + ( OA)y ( 0 ) − =0 4 4 JJJG JJJG ∴ OA ⋅ BC = 0 JJJG JJJG OA is perpendicular to BC.

PROBLEM 3.58 CONTINUED (b) Have M OA = Pd , with P acting along BC and d the JJJG JJJG perpendicular distance from OA to BC. From the results of Problem 3.57, M OA = ∴

or

Pa 2

Pa = Pd 2 d =

a 2

PROBLEM 3.59 The 8-ft-wide portion ABCD of an inclined, cantilevered walkway is partially supported by members EF and GH. Knowing that the compressive force exerted by member EF on the walkway at F is 5400 lb, determine the moment of that force about edge AD.

SOLUTION

(

M AD = λ AD ⋅ rE/ A × TEF

Having λ AD =

where

( 24 ft ) i + ( 3 ft ) j ( 24 )2 + ( 3)2 ft

)

1 (8i + j) 65

=

rE/ A = ( 7 ft ) i − ( 3 ft ) j TEF = λ EFTEF =

(8 ft − 7 ft ) i + 3 ft + ( 248 ) ( 3 ft ) j + (8 ft ) k (1)2 + ( 4 )2 + (8)2

ft

( 5400 lb )

= 600 (1 lb ) i + ( 4 lb ) j + ( 8 lb ) k 

∴ M AD =

8 1 0 600 600 7 −3 0 lb ⋅ ft = ( −192 − 56 ) lb ⋅ ft 65 65 1 4 8

= −18,456.4 lb ⋅ ft

or M AD = −18.46 kip ⋅ ft

PROBLEM 3.60 The 8-ft-wide portion ABCD of an inclined, cantilevered walkway is partially supported by members EF and GH. Knowing that the compressive force exerted by member GH on the walkway at H is 4800 lb, determine the moment of that force about edge AD.

SOLUTION

(

M AD = λ AD ⋅ rG/ A × TGH

Having λ AD =

where

( 24 ft ) i + ( 3 ft ) j ( 24 )2 + ( 3)2 ft

=

)

1 (8i + j) 65

rG/ A = ( 20 ft ) i − ( 6 ft ) j = 2 (10 ft ) i − ( 3 ft ) j TGH = λ GH TGH =

(16 ft − 20 ft ) i + 6 ft + ( 1624 ) ( 3 ft ) j + (8 ft ) k ( 4 )2 + ( 8 )2 + ( 8 )2

ft

( 4800 lb )

= 1600  − (1 lb ) i + ( 2 lb ) j + ( 2 lb ) k 

8

∴ M AD =

1 0 3200 lb ⋅ ft −3 0 = ( −48 − 20 ) 65 −1 2 2

(1600 lb )( 2 ft ) 10 65

= −26,989 lb ⋅ ft

or M AD = −27.0 kip ⋅ ft

PROBLEM 3.61 Two forces F1 and F2 in space have the same magnitude F. Prove that the moment of F1 about the line of action of F2 is equal to the moment of F2 about the line of action of F1.

SOLUTION

F1 = F1λ1

First note that

F2 = F2λ 2

and

Let M1 = moment of F2 about the line of action of M1 and M 2 = moment of F1 about the line of action of M 2 Now, by definition

(

)

(

)

(

)

(

)

M1 = λ1 ⋅ rB/ A × F2 = λ1 ⋅ rB/ A × λ 2 F2 M 2 = λ 2 ⋅ rA/B × F1 = λ 2 ⋅ rA/B × λ1 F1 F1 = F2 = F

Since

and

(

rA/B = −rB/ A

)

M1 = λ1 ⋅ rB/ A × λ 2 F

(

)

M 2 = λ 2 ⋅ −rB/ A × λ1 F Using Equation (3.39)

(

)

(

λ1 ⋅ rB/ A × λ 2 = λ 2 ⋅ −rB/ A × λ1 so that

(

) )

M 2 = λ1 ⋅ rB/ A × λ 2 F ∴ M12 = M 21

PROBLEM 3.62 In Problem 3.53, determine the perpendicular distance between cable AE and the line joining points D and B.

Problem 3.53: The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable AE at A is 220 lb, determine the moment of that force about the line joining points D and B.

SOLUTION

(

M DB = λ DB ⋅ rA/D × TAE

Have where

λ DB =

( 48 in.) i − (14 in.) j 50 in.

)

= 0.96i − 0.28 j

rA/D = − ( 4 in.) j + ( 8 in.) k TAE = λ AETAE =

( 36 in.) i − ( 24 in.) j + (8 in.) k 44 in.

( 220 lb )

= (180 lb ) i − (120 lb ) j + ( 40 lb ) k

∴ M DB

0.96 −0.28 0 8 lb ⋅ in. = 0 −4 180 −120 40 = 364.8 lb ⋅ in.

Only the perpendicular component of TAE contributes to the moment of

TAE about line DB. The parallel component of TAE will be used to find the perpendicular component.

PROBLEM 3.62 CONTINUED Have

(TAE )parallel

= λ DB ⋅ TAE = ( 0.96i − 0.28 j) ⋅ (180 lb ) i − (120 lb ) j + ( 40 lb ) k  = ( 0.96 )(180 ) + ( −0.28 )( −120 ) + ( 0 )( 40 )  lb

= (172.8 + 33.6 ) lb

= 206.4 lb Since TAE = ( TAE )perpendicular + ( TAE )parallel

∴

(TAE )perpendicular

=

(TAE )2 − (TAE )2parallel

=

( 220 )2 − ( 206.41)2

= 76.151 lb Then

M DB = (TAE )perpendicular ( d )

364.8 lb ⋅ in. = ( 76.151 lb ) d d = 4.7905 in. or d = 4.79 in.

PROBLEM 3.63 In Problem 3.54, determine the perpendicular distance between cable CF and the line joining points D and B. Problem 3.54: The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable CF at C is 132 lb, determine the moment of that force about the line joining points D and B.

SOLUTION

( M DB ) = λ DB ⋅ ( rC/D × TCF )

Have

λ DB =

where

( 48 in.) i − (14 in.) j 50 in.

= 0.96i − 0.28j rC/D = ( 8 in.) j − (16 in.) k TCF = λ CF TCF

=

( 24 in.) i − ( 36 in.) j − (8 in.) k 44 in.

(132 lb )

= ( 72 lb ) i − (108 lb ) j − ( 24 lb ) k

∴ M DB

0.96 −0.28 0 = 0 −16 lb ⋅ in 8 72 −108 −24 = −1520.64 lb ⋅ in.

Only the perpendicular component of TCF contributes to the moment of TCF about line DB. The parallel component of TCF will be used to obtain the perpendicular component.

PROBLEM 3.63 CONTINUED Have

(TCF )parallel

= λ DB ⋅ TCF = ( 0.96i − 0.28 j) ⋅ ( 72 lb ) i − (108 lb ) j − ( 24 lb ) k  = ( 0.96 )( 72 ) + ( −0.28 )( −108 ) + ( 0 )( −24 )  lb

= 99.36 lb Since TCF = ( TCF )perp. + ( TCF )parallel ∴

(TCF )perp.

=

(TCF )2 − (TCF )2parallel

=

(132 )2 − ( 99.36 )2

= 86.900 lb Then

M DB = (TCF )perp. ( d ) −1520.64 lb ⋅ in. = ( 86.900 lb ) d d = 17.4988 in. or d = 17.50 in.

PROBLEM 3.64 In Problem 3.55, determine the perpendicular distance between cable EF and the line joining points D and I. Problem 3.55: A mast is mounted on the roof of a house using bracket ABCD and is guyed by cables EF, EG, and EH. Knowing that the force exerted by cable EF at E is 66 N, determine the moment of that force about the line joining points D and I.

SOLUTION

(

M DI = λ DI ⋅ rF /I × TEF

Have λ DI =

where

(1.6 m ) i − ( 0.4 m ) j 0.4 17 m

=

)

1 ( 4i − j) 17

rF /I = ( 5.4 m ) k TEF = λ EFTEF =

(1.2 m ) i − ( 3.6 m ) j + ( 5.4 m ) k 6.6 m

( 66 N )

= 6 ( 2 N ) i − ( 6 N ) j + ( 9 N ) k 

∴ M DI =

4 −1 0 0 1 = 172.879 N ⋅ m 2 −6 9

( 6 N )( 5.4 m ) 0 17

Only the perpendicular component of TEF contributes to the moment of TEF about line DI. The parallel component of TEF will be used to find the perpendicular component.

Have

(TEF )parallel

= λ DI ⋅ TEF =

1 ( 4i − j) ⋅ (12 N ) i − ( 36 N ) j + ( 54 N ) k  17

=

1 ( 48 + 36 ) N 17

=

84 N 17

PROBLEM 3.64 CONTINUED Since TEF = ( TEF )perp. + ( TEF )parallel ∴

(TEF )perp.

=

(TEF )2 − (TEF )2parallel

=

( 66 )2 − 

 84    17 

2

= 62.777 N Then

M DI = (TEF )perp. ( d ) 172.879 N ⋅ m = ( 62.777 N )( d ) d = 2.7539 m or d = 2.75 m

PROBLEM 3.65 In Problem 3.56, determine the perpendicular distance between cable EG and the line joining points D and I. Problem 3.56: A mast is mounted on the roof of a house using bracket ABCD and is guyed by cables EF, EG, and EH. Knowing that the force exerted by cable EG at E is 61.5 N, determine the moment of that force about the line joining points D and I.

SOLUTION M DI = λ DI ⋅ rG/I × TEG 

Have λ DI =

where

(1.6 m ) i − ( 0.4 m ) j 0.4 17 m

=

1 ( 4i − j) 17

rG/I = − (10.9 m + 0.8 m ) k = − (11.7 m ) k TEG = λ EGTEG =

(1.2 m ) i − ( 3.6 m ) j − (11.7 m ) k 12.3 m

( 61.5 N )

= 5 (1.2 N ) i − ( 3.6 N ) j − (11.7 N ) k 

∴ M DI =

( 5 N )(11.7 m ) 17

4 −1 0 −1 0 0 1.2 −3.6 −11.7

= −187.286 N ⋅ m Only the perpendicular component of TEG contributes to the moment of TEG about line DI. The parallel component of TEG will be used to find the perpendicular component.

Have TEG ( parallel ) = λ DI ⋅ TEG

=

1 ( 4i − j) ⋅ 5 (1.2 N ) i − ( 3.6 N ) j − (11.7 N ) k  17

=

5 ( 4.8 + 3.6 ) N 17

=

42 N 17

PROBLEM 3.65 CONTINUED Since TEF = ( TEG )perp. + ( TEG )parallel ∴

(TEG )perp.

=

2 (TEG )2 − (TEG )parallel

=

( 61.5)

2

 42  −   17 

2

= 60.651 N Then

M DI = (TEG )perp. ( d ) 187.286 N ⋅ m = ( 60.651 N )( d ) d = 3.0880 m or d = 3.09 m

PROBLEM 3.66 In Problem 3.41, determine the perpendicular distance between post BC and the line connecting points O and A. Problem 3.41: Slider P can move along rod OA. An elastic cord PC is attached to the slider and to the vertical member BC. Knowing that the distance from O to P is 0.12 m and the tension in the cord is 30 N, determine (a) the angle between the elastic cord and the rod OA, (b) the projection on OA of the force exerted by cord PC at point P.

SOLUTION Assume post BC is represented by a force of magnitude FBC where

FBC = FBC j

Have

M OA = λ OA ⋅ rB/O × FBC

where

(

λ OA =

)

( 0.24 m ) i + ( 0.24 m ) j − ( 0.12 m ) k 0.36 m

2 2 1 i+ j− k 3 3 3

=

rB/O = ( 0.18 m ) i + ( 0.24 m ) k ∴ M OA

2 2 −1 1 F = FBC 0.18 0 0.24 = BC ( −0.48 − 0.18 ) = −0.22 FBC 3 3 0 1 0

Only the perpendicular component of FBC contributes to the moment of FBC about line OA. The parallel component will be found first so that the perpendicular component of FBC can be determined.

2 1  2 FBC ( parallel ) = λ OA ⋅ FBC =  i + j − k  ⋅ FBC j 3 3  3

=

2 FBC 3 FBC = ( FBC )parallel + ( FBC )perp.

Since

( FBC )perp.

=

( FBC )2 − ( FBC )2parallel

=

( FBC )2 − 

2 FBC    3 

2

= 0.74536 FBC

Then

M OA = ( FBC )perp. ( d )

0.22 FBC = ( 0.74536 FBC ) d

d = 0.29516 m or d = 295 mm

PROBLEM 3.67 In Problem 3.45, determine the perpendicular distance between cord DE and the y axis. Problem 3.45: The 0.732 × 1.2 -m lid ABCD of a storage bin is hinged along side AB and is held open by looping cord DEC over a frictionless hook at E. If the tension in the cord is 54 N, determine the moment about each of the coordinate axes of the force exerted by the cord at D.

SOLUTION First note z =

( 0.732 )2 − ( 0.132 )2

m

= 0.720 m

(

)

Have

M y = j ⋅ rD/ A × TDE

where

rD/ A = ( 0.132 j + 0.720k ) m TDE = λ DETDE

=

( 0.360 m ) i + ( 0.732 m ) j − ( 0.720 m ) k 1.08 m

( 54 N )

= (18 N ) i + ( 36 N ) j − ( 36 N ) k 0 1 0 ∴ M y = 0 0.132 0.720 = 12.96 N ⋅ m −36 18 36 Only the perpendicular component of TDE contributes to the moment of TDE about the y-axis. The parallel component will be found first so that

the perpendicular component of TDE can be determined. TDE ( parallel ) = j ⋅ TDE = 36 N

PROBLEM 3.67 CONTINUED

( TDE ) = ( TDE )parallel + ( TDE )perp.

Since

(TDE )perp.

Then

=

2 (TDE )2 − (TDE )parallel

=

( 54 )2 − ( 36 )2

= 40.249 N

M y = (TDE )perp. (d ) 12.96 N ⋅ m = ( 40.249 N )( d ) d = 0.32199 m or d = 322 mm

PROBLEM 3.68 A plate in the shape of a parallelogram is acted upon by two couples. Determine (a) the moment of the couple formed by the two 21-N forces, (b) the perpendicular distance between the 12-N forces if the resultant of the two couples is zero, (c) the value of α if the resultant couple is 1.8 N ⋅ m clockwise and d is 1.05 m.

SOLUTION (a) Have where

M1 = d1F1 d1 = 0.4 m F1 = 21 N ∴ M1 = ( 0.4 m )( 21 N ) = 8.4 N ⋅ m or M1 = 8.40 N ⋅ m

(b) Have or

M1 + M 2 = 0

8.40 N ⋅ m − d 2 (12 N ) = 0 ∴ d 2 = 0.700 m

(c) Have or

M total = M1 + M 2

1.8 N ⋅ m = 8.40 N ⋅ m − (1.05 m )( sin α )(12 N ) ∴ sin α = 0.52381

and

α = 31.588° or α = 31.6°

PROBLEM 3.69 A couple M of magnitude 10 lb ⋅ ft is applied to the handle of a screwdriver to tighten a screw into a block of wood. Determine the magnitudes of the two smallest horizontal forces that are equivalent to M if they are applied (a) at corners A and D, (b) at corners B and C, (c) anywhere on the block.

SOLUTION M = Pd

(a) Have

 1 ft  10 lb ⋅ ft = P (10 in.)    12 in. 

or

∴ P = 12 lb d BC =

(b)

=

or Pmin = 12.00 lb

( BE )2 + ( EC )2 (10 in.)2 + ( 6 in.)2

= 11.6619 in.

M = Pd

Have

 1 ft  10 lb ⋅ ft = P (11.6619 in.)    12 in.  P = 10.2899 lb d AC =

(c)

=

or P = 10.29 lb

( AD )2 + ( DC )2 (10 in.)2 + (16 in.)2

= 2 89 in.

M = Pd AC

Have

(

)

 1 ft  10 lb ⋅ ft = P 2 89 in.    12 in.  P = 6.3600 lb

or P = 6.36 lb

PROBLEM 3.70 Two 60-mm-diameter pegs are mounted on a steel plate at A and C, and two rods are attached to the plate at B and D. A cord is passed around the pegs and pulled as shown, while the rods exert on the plate 10-N forces as indicated. (a) Determine the resulting couple acting on the plate when T = 36 N. (b) If only the cord is used, in what direction should it be pulled to create the same couple with the minimum tension in the cord? (c) Determine the value of that minimum tension.

SOLUTION M = Σ ( Fd )

(a) Have

= ( 36 N )( 0.345 m ) − (10 N )( 0.380 m ) = 8.62 N ⋅ m M = 8.62 N ⋅ m

(b)

M = Td = 8.62 N ⋅ m

Have For T to be minimum, d must be maximum.

∴ Tmin must be perpendicular to line AC tan θ =

0.380 m = 1.33333 0.285 m

θ = 53.130°

and

or θ = 53.1° M = Tmin d max

(c) Have

M = 8.62 N ⋅ m

where  d max =  

( 0.380 )2 + ( 0.285)2

 + 2 ( 0.030 )  m = 0.535 m 

∴ 8.62 N ⋅ m = Tmin ( 0.535 m ) Tmin = 16.1121 N or Tmin = 16.11 N

PROBLEM 3.71 The steel plate shown will support six 50-mm-diameter idler rollers mounted on the plate as shown. Two flat belts pass around the rollers, and rollers A and D will be adjusted so that the tension in each belt is 45 N. Determine (a) the resultant couple acting on the plate if a = 0.2 m, (b) the value of a so that the resultant couple acting on the plate is 54 N ⋅ m clockwise.

SOLUTION

(a) Note when a = 0.2 m, rC/F is perpendicular to the inclined 45 N forces. Have M = Σ ( Fd ) = − ( 45 N )  a + 0.2 m + 2 ( 0.025 m ) 

− ( 45 N )  2a 2 + 2 ( 0.025 m )   

For a = 0.2 m, M = − ( 45 N )( 0.450 m + 0.61569 m ) = −47.956 N ⋅ m or M = 48.0 N ⋅ m M = 54.0 N ⋅ m

(b)

M = Moment of couple due to horizontal forces at A and D + Moment of force-couple systems at C and F about C. −54.0 N ⋅ m = −45 N  a + 0.2 m + 2 ( 0.025 m ) 

+  M C + M F + Fx ( a + 0.2 m ) + Fy ( 2a )  where

M C = − ( 45 N )( 0.025 m ) = −1.125 N ⋅ m M F = M C = −1.125 N ⋅ m

PROBLEM 3.71 CONTINUED

∴

Fx =

−45 N 2

Fy =

−45 N 2

− 54.0 N ⋅ m = −45 N ( a + 0.25 m ) − 1.125 N ⋅ m − 1.125 N ⋅ m −45 N 45 N ( a + 0.2 m ) − ( 2a ) 2 2 1.20 = a + 0.25 + 0.025 + 0.025 +

a 0.20 2a + + 2 2 2

3.1213a = 0.75858 a = 0.24303 m or a = 243 mm

PROBLEM 3.72 The shafts of an angle drive are acted upon by the two couples shown. Replace the two couples with a single equivalent couple, specifying its magnitude and the direction of its axis.

SOLUTION M = M1 + M 2

Based on where

M1 = − ( 8 N ⋅ m ) j M2 = − (6 N⋅m)k

∴ M = − (8 N ⋅ m ) j − ( 6 N ⋅ m ) k M =

and

( 8 )2 + ( 6 ) 2

= 10 N ⋅ m or M = 10.00 N ⋅ m

λ = or

− (8 N ⋅ m ) j − ( 6 N ⋅ m ) k M = = −0.8j − 0.6k 10 N ⋅ m M M = M λ = (10 N ⋅ m )( −0.8j − 0.6k )

cosθ x = 0

∴ θ x = 90°

cosθ y = −0.8

∴ θ y = 143.130°

cosθ z = −0.6

∴ θ z = 126.870°

or θ x = 90.0°, θ y = 143.1°, θ z = 126.9°

PROBLEM 3.73 Knowing that P = 0, replace the two remaining couples with a single equivalent couple, specifying its magnitude and the direction of its axis.

SOLUTION Have

M = M1 + M 2

where

M1 = rC/B × P1C rC/B = ( 0.96 m ) i − ( 0.40 m ) j P1C = − (100 N ) k i j k ∴ M1 = 0.96 −0.40 0 = ( 40 N ⋅ m ) i + ( 96 N ⋅ m ) j 0 0 −100 M 2 = rD/ A × P2 E

Also,

rD/ A = ( 0.20 m ) j − ( 0.55 m ) k P2 E = λ ED P2 E

=

− ( 0.48 m ) i + ( 0.55 m ) k

( 0.48)2 + ( 0.55)2

m

(146 N )

= − ( 96 N ) i + (110 N ) k i j k ∴ M 2 = 0 0.20 −0.55 N ⋅ m −96 0 110

= ( 22.0 N ⋅ m ) i + ( 52.8 N ⋅ m ) j + (19.2 N ⋅ m ) k

PROBLEM 3.73 CONTINUED M = ( 40 N ⋅ m ) i + ( 96 N ⋅ m ) j + ( 22.0 N ⋅ m ) i

and

+ ( 52.8 N ⋅ m ) j + (19.2 N ⋅ m ) k 

= ( 62.0 N ⋅ m ) i + (148.8 N ⋅ m ) j + (19.2 N ⋅ m ) k M =

M x2 + M y2 + M z2 =

( 62.0 )2 + (148.8)2 + (19.2 )2

= 162.339 N ⋅ m or M = 162.3 N ⋅ m λ =

M 62.0i + 148.8j + 19.2k = M 162.339

= 0.38192i + 0.91660 j + 0.118271k cosθ x = 0.38192

∴ θ x = 67.547° or θ x = 67.5°

cosθ y = 0.91660

∴ θ y = 23.566° or θ y = 23.6°

cosθ z = 0.118271

∴ θ z = 83.208° or θ z = 83.2°

PROBLEM 3.74 Knowing that P = 0, replace the two remaining couples with a single equivalent couple, specifying its magnitude and the direction of its axis.

SOLUTION Have

M = M4 + M7

where

M 4 = rG/C × F4G rG/C = − (10 in.) i F4G = ( 4 lb ) k

∴ M 4 = − (10 in.) i × ( 4 lb ) k = ( 40 lb ⋅ in.) j M 7 = rD/F × F7 D

Also,

rD/F = − ( 5 in.) i + ( 3 in.) j F7 D = λ ED F7 D

=

=

− ( 5 in.) i + ( 3 in.) j + ( 7 in.) k

( 5 )2 + ( 3)2 + ( 7 )2

in.

( 7 lb )

7 lb ( −5i + 3j + 7k ) 83

i 7 lb ⋅ in. ∴ M7 = −5 83 −5

j k 7 lb ⋅ in. 3 0 = ( 21i + 35j + 0k ) 83 3 7

= 0.76835 ( 21i + 35 j) lb ⋅ in.

PROBLEM 3.74 CONTINUED M = ( 40 lb ⋅ in.) j +  0.76835 ( 21i + 35j) lb ⋅ in.

and

= (16.1353 lb ⋅ in.) i + ( 66.892 lb ⋅ in.) j M =

( M x )2 + ( M y )

2

=

(16.1353)2 + ( 66.892 )2

= 68.811 lb ⋅ in. or M = 68.8 lb ⋅ in. λ =

(16.1353 lb ⋅in.) i + ( 66.892 lb ⋅in.) j M = M 68.811 lb ⋅ in. = 0.23449i + 0.97212 j cosθ x = 0.23449

∴ θ x = 76.438° or θ x = 76.4°

cosθ y = 0.97212

∴ θ y = 13.5615° or θ y = 13.56°

cosθ z = 0.0

∴ θ z = 90° or θ z = 90.0°

PROBLEM 3.75 Knowing that P = 5 lb, replace the three couples with a single equivalent couple, specifying its magnitude and the direction of its axis.

SOLUTION M = M 4 + M 7 + M5

Have where M 4 = rG/C × F4G

M 7 = rD/F × F7 D

i j k = −10 0 0 lb ⋅ in. = ( 40 lb ⋅ in.) j 0 0 4

i j k  7  = −5 3 0   lb ⋅ in. = 0.76835 ( 21i + 35 j) lb ⋅ in.  83  −5 3 7

(See Solution to Problem 3.74.) M 5 = rC/ A × F5C

i j k = 10 −6 7 lb ⋅ in. = − ( 35 lb ⋅ in.) i + ( 50 lb ⋅ in.) k 0 5 0

∴ M = (16.1353 − 35 ) i + ( 40 + 26.892 ) j + ( 50 ) k  lb ⋅ in.

= − (18.8647 lb ⋅ in.) i + ( 66.892 lb ⋅ in.) j + ( 50 lb ⋅ in.) k M =

M x2 + M y2 + M z2 =

(18.8647 )2 + ( 66.892 )2 + ( 50 )2

= 85.618 lb ⋅ in.

or M = 85.6 lb ⋅ in. λ =

M −18.8647i + 66.892 j + 50k = = −0.22034i + 0.78129 j + 0.58399k 85.618 M

cosθ x = −0.22034

∴ θ x = 102.729°

or θ x = 102.7°

cosθ y = 0.78129

∴ θ y = 38.621°

or θ y = 38.6°

cosθ z = 0.58399

∴ θ z = 54.268°

or θ z = 54.3°

PROBLEM 3.76 Knowing that P = 210 N, replace the three couples with a single equivalent couple, specifying its magnitude and the direction of its axis.

SOLUTION M = M1 + M 2 + M P

Have

M1 = rC/B × P1C

where

M 2 = rD/ A × P2 E

i j k = 0.96 −0.40 0 = ( 40 N ⋅ m ) i + ( 96 N ⋅ m ) j 0 0 −100

i j k = 0 0.20 −0.55 = ( 22.0 N ⋅ m ) i + ( 52.8 N ⋅ m ) j + (19.2 N ⋅ m ) k −96 0 110 (See Solution to Problem 3.73.)

M P = rE/ A

i j k × PE = 0.48 0.20 −1.10 = ( 231 N ⋅ m ) i + (100.8 N ⋅ m ) k 0 210 0

∴ M = ( 40 + 22 + 231) i + ( 96 + 52.8 ) j + (19.2 + 100.8 ) k  N ⋅ m

= ( 293 N ⋅ m ) i + (148.8 N ⋅ m ) j + (120 N ⋅ m ) k M =

M x2 + M y2 + M z2 =

( 293)2 + (148.8)2 + (120 )2

= 349.84 N ⋅ m

or M = 350 N ⋅ m λ =

293i + 148.8 j + 120k M = = 0.83752i + 0.42533j + 0.34301k 349.84 M

cosθ x = 0.83752

∴ θ x = 33.121°

or θ x = 33.1°

cosθ y = 0.42533

∴ θ y = 64.828°

or θ y = 64.8°

cosθ z = 0.34301

∴ θ z = 69.940°

or θ z = 69.9°

PROBLEM 3.77 In a manufacturing operation, three holes are drilled simultaneously in a workpiece. Knowing that the holes are perpendicular to the surfaces of the workpiece, replace the couples applied to the drills with a single equivalent couple, specifying its magnitude and the direction of its axis.

SOLUTION M = M1 + M 2 + M 3

Have

M1 = − (1.1 lb ⋅ ft )( cos 25° j + sin 25°k )

where

M 2 = − (1.1 lb ⋅ ft ) j M 3 = − (1.3 lb ⋅ ft )( cos 20° j − sin 20°k )

∴ M = ( −0.99694 − 1.1 − 1.22160 ) j + ( −0.46488 + 0.44463) k = − ( 3.3185 lb ⋅ ft ) j − ( 0.020254 lb ⋅ ft ) k

and

M =

M x2 + M y2 + M z2 =

( 0 )2 + ( 3.3185)2 + ( 0.020254 )2

= 3.3186 lb ⋅ ft

or M = 3.32 lb ⋅ ft λ =

( 0 ) i − 3.3185j − 0.020254k M = M 3.3186 = −0.99997 j − 0.0061032k

cosθ x = 0

∴ θ x = 90°

cosθ y = −0.99997

∴ θ y = 179.555°

cosθ z = −0.0061032

∴ θ z = 90.349°

or θ x = 90.0° or θ y = 179.6° or θ z = 90.3°

PROBLEM 3.78 The tension in the cable attached to the end C of an adjustable boom ABC is 1000 N. Replace the force exerted by the cable at C with an equivalent force-couple system (a) at A, (b) at B.

SOLUTION (a) Based on

ΣF : FA = T = 1000 N

or FA = 1000 N

20°

ΣM A : M A = (T sin 50° )( dA ) = (1000 N ) sin 50° ( 2.25 m ) = 1723.60 N ⋅ m or M A = 1724 N ⋅ m (b) Based on

ΣF : FB = T = 1000 N or FB = 1000 N

20°

ΣMB : M B = (T sin 50° )( d B ) = (1000 N ) sin 50° (1.25 m ) = 957.56 N ⋅ m or M B = 958 N ⋅ m

PROBLEM 3.79 The 20-lb horizontal force P acts on a bell crank as shown. (a) Replace P with an equivalent force-couple system at B. (b) Find the two vertical forces at C and D which are equivalent to the couple found in part a.

SOLUTION (a) Based on

ΣF : PB = P = 20 lb or PB = 20 lb ΣM : MB = Pd B = 20 lb ( 5 in.) = 100 lb ⋅ in. or M B = 100 lb ⋅ in.

(b) If the two vertical forces are to be equivalent to MB , they must be a couple. Further, the sense of the moment of this couple must be counterclockwise. Then, with PC and PD acting as shown, ΣM : M D = PC d 100 lb ⋅ in. = PC ( 4 in.) ∴ PC = 25 lb or PC = 25 lb ΣFy : 0 = PD − PC ∴ PD = 25 lb or PD = 25 lb

PROBLEM 3.80 A 700-N force P is applied at point A of a structural member. Replace P with (a) an equivalent force-couple system at C, (b) an equivalent system consisting of a vertical force at B and a second force at D.

SOLUTION ΣF : PC = P = 700 N

(a) Based on

or PC = 700 N

60°

ΣM C : M C = − Px dCy + Py dCx Px = ( 700 N ) cos60° = 350 N

where

Py = ( 700 N ) sin 60° = 606.22 N dCx = 1.6 m dCy = 1.1 m ∴ M C = − ( 350 N )(1.1 m ) + ( 606.22 N )(1.6 m ) = −385 N ⋅ m + 969.95 N ⋅ m = 584.95 N ⋅ m or M C = 585 N ⋅ m ΣFx : PDx = P cos 60°

(b) Based on

= ( 700 N ) cos 60° = 350 N ΣM D :

( P cos 60° )( d DA ) =

PB ( d DB )

( 700 N ) cos 60° ( 0.6 m ) = PB ( 2.4 m )

PB = 87.5 N or PB = 87.5 N

PROBLEM 3.80 CONTINUED ΣFy : P sin 60° = PB + PDy

( 700 N ) sin 60° = 87.5 N + PDy PDy = 518.72 N PD = =

( PDx )2 + ( PDy )

2

( 350 )2 + ( 518.72 )2

= 625.76 N

 PDy  −1  518.72   = tan   = 55.991°  350   PDx 

θ = tan −1 

or PD = 626 N

56.0°

PROBLEM 3.81 A landscaper tries to plumb a tree by applying a 240-N force as shown. Two helpers then attempt to plumb the same tree, with one pulling at B and the other pushing with a parallel force at C. Determine these two forces so that they are equivalent to the single 240-N force shown in the figure.

SOLUTION Based on ΣFx : − ( 240 N ) cos30° = − FB cosα − FC cosα − ( FB + FC ) cos α = − ( 240 N ) cos 30°

or

ΣFy :

( 240 N ) sin 30° =

( FB + FC ) sin α

or

(1)

FB sin α + FC sin α

= ( 240 N ) sin 30°

(2)

From

Equation (2) : tan α = tan 30° Equation (1) ∴ α = 30° Based on ΣM C : ( 240 N ) cos ( 30° − 20° )  ( 0.25 m ) = ( FB cos10° )( 0.60 m )

∴ FB = 100 N or FB = 100.0 N From Equation (1),

30°

− (100 N + FC ) cos30° = −240cos30° FC = 140 N

or FC = 140.0 N

30°

PROBLEM 3.82 A landscaper tries to plumb a tree by applying a 240-N force as shown. (a) Replace that force with an equivalent force-couple system at C. (b) Two helpers attempt to plumb the same tree, with one applying a horizontal force at C and the other pulling at B. Determine these two forces if they are to be equivalent to the single force of part a.

SOLUTION

ΣFx : − ( 240 N ) cos30° = − FC cos30°

(a) Based on

∴ FC = 240 N or FC = 240 N ΣM C : ( 240 N ) cos10° ( d A ) = M C

30°

d A = 0.25 m

∴ M C = 59.088 N ⋅ m or M C = 59.1 N ⋅ m ΣFy :

(b) Based on

( 240 N ) sin 30° =

FB sin α

FB sin α = 120

or

(1)

ΣM B : 59.088 N ⋅ m − ( 240 N ) cos10° ( dC ) = − FC ( dC cos 20° ) 59.088 N ⋅ m − ( 240 N ) cos10°  ( 0.60 m ) = − FC ( 0.60 m ) cos 20°

0.56382 FC = 82.724 FC = 146.722 N

or FC = 146.7 N and

ΣFx : − ( 240 N ) cos30° = −146.722 N − FB cosα FB cosα = 61.124

(2)

From Equation (1) : Equation (2)

tan α =

α = 63.007° From Equation (1),

FB =

120 = 1.96323 61.124 or α = 63.0° 120 = 134.670 N sin 63.007° or FB = 134.7 N

63.0°

PROBLEM 3.83 A dirigible is tethered by a cable attached to its cabin at B. If the tension in the cable is 250 lb, replace the force exerted by the cable at B with an equivalent system formed by two parallel forces applied at A and C.

SOLUTION Require the equivalent forces acting at A and C be parallel and at an angle of α with the vertical. Then for equivalence, ΣFx :

( 250 lb ) sin 30° =

FA sin α + FB sin α

ΣFy : − ( 250 lb ) cos 30° = − FA cos α − FB cos α

(1) (2)

Dividing Equation (1) by Equation (2),

( 250 lb ) sin 30° − ( 250 lb ) cos 30°

=

( FA + FB ) sin α − ( FA + FB ) cos α

Simplifying yields α = 30° Based on ΣM C : ( 250 lb ) cos 30° (12 ft ) = ( FA cos 30° )( 32 ft )

∴ FA = 93.75 lb or FA = 93.8 lb

60°

Based on ΣM A : − ( 250 lb ) cos 30° ( 20 ft ) = ( FC cos 30° ) ( 32 ft )

∴ FC = 156.25 lb or FC = 156.3 lb

60°

PROBLEM 3.84 Three workers trying to move a 3 × 3 × 4-ft crate apply to the crate the three horizontal forces shown. (a) If P = 60 lb, replace the three forces with an equivalent force-couple system at A. (b) Replace the force-couple system of part a with a single force, and determine where it should be applied to side AB. (c) Determine the magnitude of P so that the three forces can be replaced with a single equivalent force applied at B.

SOLUTION (a) Based on ΣFz : − 50 lb + 50 lb + 60 lb = FA FA = 60 lb or FA = ( 60.0 lb ) k Based on ΣM A :

( 50 lb )( 2 ft ) − ( 50 lb )( 0.6 ft ) = M A M A = 70 lb ⋅ ft

(a)

or M A = ( 70.0 lb ⋅ ft ) j (b) Based on ΣFz : − 50 lb + 50 lb + 60 lb = F

F = 60 lb or F = ( 60.0 lb ) k Based on ΣM A : 70 lb ⋅ ft = 60 lb (x )

(b)

x = 1.16667 ft or x = 1.167 ft from A along AB (c) Based on ΣMB : − ( 50 lb ) (1 ft ) + ( 50 lb ) (2.4 ft ) − P (3 ft ) = R (0 )

P=

70 = 23.333 lb 3 or P = 23.3 lb

(c)

PROBLEM 3.85 A force and a couple are applied to a beam. (a) Replace this system with a single force F applied at point G, and determine the distance d. (b) Solve part a assuming that the directions of the two 600-N forces are reversed.

SOLUTION (a)

ΣFy : FC + FD + FE = F

Have

F = −800 N + 600 N − 600 N F = −800 N Have

or F = 800 N

ΣM G : FC ( d − 1.5 m ) − FD ( 2 m ) = 0

(800 N )( d

− 1.5 m ) − ( 600 N )( 2 m ) = 0

d =

1200 + 1200 800

d = 3m

or d = 3.00 m

(b)

Changing directions of the two 600 N forces only changes sign of the couple. ∴ F = −800 N and

or F = 800 N

ΣM G : FC ( d − 1.5 m ) + FD ( 2 m ) = 0

(800 N )( d d =

− 1.5 m ) + ( 600 N )( 2 m ) 1200 − 1200 =0 800 or d = 0

PROBLEM 3.86 Three cables attached to a disk exert on it the forces shown. (a) Replace the three forces with an equivalent force-couple system at A. (b) Determine the single force which is equivalent to the force-couple system obtained in part a, and specify its point of application on a line drawn through points A and D.

SOLUTION

(a) Have

ΣF : FB + FC + FD = FA FB = −FD

Since

∴ FA = FC = 110 N

20° or FA = 110.0 N

Have

20.0°

ΣM A : − FBT ( r ) − FCT ( r ) + FDT ( r ) = M A − (140 N ) sin15° ( 0.2 m ) − (110 N ) sin 25° ( 0.2 m ) + (140 N ) sin 45° ( 0.2 m ) = M A

M A = 3.2545 N ⋅ m

or M A = 3.25 N ⋅ m (b) Have

ΣF : FA = FE

or FE = 110.0 N

20.0°

ΣM : M A = [ FE cos 20°] ( a ) ∴ 3.2545 N ⋅ m = (110 N ) cos 20° ( a )

a = 0.031485 m or a = 31.5 mm below A

PROBLEM 3.87 While tapping a hole, a machinist applies the horizontal forces shown to the handle of the tap wrench. Show that these forces are equivalent to a single force, and specify, if possible, the point of application of the single force on the handle.

SOLUTION Since the forces at A and B are parallel, the force at B can be replaced with the sum of two forces with one of the forces equal in magnitude to the force at A except with an opposite sense, resulting in a force-couple. Have FB = 26.5 N + 2.5 N, where the 26.5 N force be part of the couple. Combining the two parallel forces, M couple = ( 26.5 N ) ( 0.080 m + 0.070 m ) cos 25°

= 3.60 N ⋅ m

and, M couple = 3.60 N ⋅ m

A single equivalent force will be located in the negative z-direction. Based on

ΣMB : − 3.60 N ⋅ m = (2.5 N )cos 25°  ( a )

a = −1.590 m F′ = ( 2.5 N )( cos 25°i + sin 25° j)

and is applied on an extension of handle BD at a distance of 1.590 m to the right of B

PROBLEM 3.88 A rectangular plate is acted upon by the force and couple shown. This system is to be replaced with a single equivalent force. (a) For α = 40°, specify the magnitude and the line of action of the equivalent force. (b) Specify the value of α if the line of action of the equivalent force is to intersect line CD 12 in. to the right of D.

SOLUTION

ΣFx : − ( 3 lb ) sin 40° + ( 3 lb ) sin 40° = Fx

(a) Have

∴ Fx = 0

ΣFy : − ( 3 lb ) cos 40° − 10 lb + ( 3 lb ) cos 40° = Fy

Have

∴ Fy = −10 lb or F = 10.00 lb Note: The two 3-lb forces form a couple ΣM A : rC/ A × PC + rB/ A × PB = rX / A × F

and

i j k i j k i j k 3 16 −10 0 + 160 1 0 0 = 10 d 0 0 sin 40° cos 40° 0 0 −1 0 0 −1 0 k : 3 (16 ) cos 40° − ( −10 ) 3sin 40° − 160 = −10d 36.770 + 19.2836 − 160 = −10d ∴ d = 10.3946 in. or F = 10.00 lb (b) From part (a), Have

at 10.39 in. right of A or at 5.61 in. left of B

F = 10.00 lb ΣM A : rC/ A × PC + rB/ A × PB = (12 in.) i × F

i j k i j k i j k 3 16 −10 0 + 160 1 0 0 = 120 1 0 0 sin α cos α 0 0 −1 0 0 −1 0 k : 48cos α + 30sin α − 160 = −120 24cosα = 20 − 15sin α

PROBLEM 3.88 CONTINUED Squaring both sides of the equation, and using the identity cos 2 α = 1 − sin 2 α , results in sin 2 α − 0.74906sin α − 0.21973 = 0 Using quadratic formula sin α = 0.97453

sinα = − 0.22547

so that

α = 77.0°

and

α = −13.03°

PROBLEM 3.89 A hexagonal plate is acted upon by the force P and the couple shown. Determine the magnitude and the direction of the smallest force P for which this system can be replaced with a single force at E.

SOLUTION Since the minimum value of P acting at B is realized when Pmin is perpendicular to a line connecting B and E, α = 30° Then, ΣM E : rB/E × Pmin + rD/ A × PD = 0 where rB/E = − ( 0.30 m ) i +  2 ( 0.30 m ) cos 30° j

= − ( 0.30 m ) i + ( 0.51962 m ) j rD/ A = 0.30 m + 2 ( 0.3 m ) sin 30°  i

= ( 0.60 m ) i

PD = ( 450 N ) j Pmin = Pmin ( cos 30° ) i + ( sin 30° ) j

∴ Pmin

i j k i j k −0.30 0.51962 0 + 0.60 0 0 N ⋅ m = 0 0.86603 0.50 0 0 450 0

Pmin ( −0.15 m − 0.45 m ) k + ( 270 N ⋅ m ) k = 0 ∴ Pmin = 450 N or Pmin = 450 N

30°

PROBLEM 3.90 An eccentric, compressive 270-lb force P is applied to the end of a cantilever beam. Replace P with an equivalent force-couple system at G.

SOLUTION Have ΣF : − ( 270 lb ) i = F ∴ F = − ( 270 lb ) i Also, have ΣM G : rA/G × P = M

i j k 270 0 −4 −2.4 lb ⋅ in. = M −1 0 0 ∴ M = ( 270 lb ⋅ in.) ( −2.4 )( −1) j − ( −4 )( −1) k 

or M = ( 648 lb ⋅ in.) j − (1080 lb ⋅ in.) k

PROBLEM 3.91 Two workers use blocks and tackles attached to the bottom of an I-beam to lift a large cylindrical tank. Knowing that the tension in rope AB is 324 N, replace the force exerted at A by rope AB with an equivalent force-couple system at E.

SOLUTION ΣF : TAB = F

Have where

TAB = λ ABTAB =

( 0.75 m ) i − ( 6.0 m ) j + ( 3.0 m ) k 6.75 m

( 324 N )

∴ TAB = 36 N ( i − 8j + 4k ) so that

F = ( 36.0 N ) i − ( 288 N ) j + (144.0 N ) k

Have ΣM E : rA/E × TAB = M

or

i j k ( 7.5 m )( 36 N ) 0 1 0 = M 1 −8 4 ∴ M = ( 270 N ⋅ m )( 4i − k ) or M = (1080 N ⋅ m ) i − ( 270 N ⋅ m ) k

PROBLEM 3.92 Two workers use blocks and tackles attached to the bottom of an I-beam to lift a large cylindrical tank. Knowing that the tension in rope CD is 366 N, replace the force exerted at C by rope CD with an equivalent force-couple system at O.

SOLUTION ΣF : TCD = F

Have where

TCD = λ CDTCD =

− ( 0.3 m ) i − ( 5.6 m ) j + ( 2.4 m ) k ( 366 N ) 6.1 m

∴ TCD = ( 6.0 N )( −3i − 56 j + 24k ) so that

F = − (18.00 N ) i − ( 336 N ) j + (144.0 N ) k

Have ΣM O : rC/O × TCD = M

or

i j k ( 7.5 m )( 6 N ) 0 1 0 = M −3 −56 24 ∴ M = ( 45 N ⋅ m )( 24i + 3k ) or M = (1080 N ⋅ m ) i + (135.0 N ⋅ m ) k

PROBLEM 3.93 To keep a door closed, a wooden stick is wedged between the floor and the doorknob. The stick exerts at B a 45-lb force directed along line AB. Replace that force with an equivalent force-couple system at C.

SOLUTION Have ΣF : PAB = FC where

PAB = λ AB PAB =

( 2.0 in.) i + ( 38 in.) j − ( 24 in.) k 44.989 in.

( 45 lb )

or FC = ( 2.00 lb ) i + ( 38.0 lb ) j − ( 24.0 lb ) k Have ΣM C : rB/C × PAB = M C

MC

i j k = 2 29.5 −33 0 lb ⋅ in. 1 19 −12 = ( 2 lb ⋅ in.) {( −33)( −12 ) i − ( 29.5 )( −12 ) j

+ ( 29.5 )(19 ) − ( −33)(1)  k

}

or M C = ( 792 lb ⋅ in.) i + ( 708 lb ⋅ in.) j + (1187 lb ⋅ in.) k

PROBLEM 3.94 A 25-lb force acting in a vertical plane parallel to the yz plane is applied to the 8-in.-long horizontal handle AB of a socket wrench. Replace the force with an equivalent force-couple system at the origin O of the coordinate system.

SOLUTION Have ΣF : PB = F where PB = 25 lb  − ( sin 20° ) j + ( cos 20° ) k 

= − ( 8.5505 lb ) j + ( 23.492 lb ) k or F = − ( 8.55 lb ) j + ( 23.5 lb ) k Have ΣM O : rB/O × PB = M O where rB/O = ( 8cos 30° ) i + (15 ) j − ( 8sin 30° ) k  in.

= ( 6.9282 in.) i + (15 in.) j − ( 4 in.) k

∴

i j k 6.9282 15 −4 lb ⋅ in. = M O 0 −8.5505 23.492

M O = ( 318.18 ) i − (162.757 ) j − ( 59.240 ) k  lb ⋅ in.

or M O = ( 318 lb ⋅ in.) i − (162.8 lb ⋅ in.) j − ( 59.2 lb ⋅ in.) k

PROBLEM 3.95 A 315-N force F and 70-N · m couple M are applied to corner A of the block shown. Replace the given force-couple system with an equivalent force-couple system at corner D.

SOLUTION Have ΣF : F = FD = λ AI F =

( 0.360 m ) i − ( 0.120 m ) j + ( 0.180 m ) k 0.420 m

( 315 N )

= ( 750 N )( 0.360i − 0.120 j + 0.180k ) or FD = ( 270 N ) i − ( 90.0 N ) j + (135.0 N ) k Have ΣM D : M + rI /D × F = M D where M = λ AC M =

( 0.240 m ) i − ( 0.180 m ) k 0.300 m

( 70.0 N ⋅ m )

= ( 70.0 N ⋅ m )( 0.800i − 0.600k ) rI /D = ( 0.360 m ) k ∴ MD

i j k = ( 70.0 N ⋅ m )( 0.8i − 0.6k ) + 0 0 0.36 ( 750 N ⋅ m ) 0.36 −0.12 0.18 = ( 56.0 N ⋅ m ) i − ( 42.0 N ⋅ m ) k + ( 32.4 N ⋅ m ) i + ( 97.2 N ⋅ m ) j or M D = ( 88.4 N ⋅ m ) i + ( 97.2 N ⋅ m ) j − ( 42.0 N ⋅ m ) k

PROBLEM 3.96 The handpiece of a miniature industrial grinder weighs 2.4 N, and its center of gravity is located on the y axis. The head of the handpiece is offset in the xz plane in such a way that line BC forms an angle of 25° with the x direction. Show that the weight of the handpiece and the two couples M1 and M 2 can be replaced with a single equivalent force.

Further assuming that M1 = 0.068 N ⋅ m and M 2 = 0.065 N ⋅ m, determine (a) the magnitude and the direction of the equivalent force, (b) the point where its line of action intersects the xz plane.

SOLUTION First assume that the given force W and couples M1 and M 2 act at the origin. Now

W = −Wj

and

M = M1 + M 2 = − ( M 2 cos 25° ) i + ( M1 − M 2 sin 25° ) k

Note that since W and M are perpendicular, it follows that they can be replaced with a single equivalent force. F =W

(a) Have

or

F = −Wj = − ( 2.4 N ) j or F = − ( 2.40 N ) j

(b) Assume that the line of action of F passes through point P (x, 0, z). Then for equivalence M = rP/O × F where ∴

rP/O = xi + zk − ( M 2 cos 25° ) i + ( M1 − M 2 sin 25° ) k i j k = x 0 z = (Wz ) i − (Wx ) k 0 −W 0

PROBLEM 3.96 CONTINUED Equating the i and k coefficients,

z = (b) For

−M z cos 25° W

and

 M − M 2 sin 25°  x = − 1  W  

W = 2.4 N, M1 = 0.068 N ⋅ m, M 2 = 0.065 N ⋅ m

x=

0.068 − 0.065sin 25° = −0.0168874 m −2.4 or x = −16.89 mm

z =

−0.065cos 25° = −0.024546 m 2.4 or z = −24.5 mm

PROBLEM 3.97 A 20-lb force F1 and a 40- lb ⋅ ft couple M1 are applied to corner E of the bent plate shown. If F1 and M1 are to be replaced with an equivalent force-couple system ( F2 , M 2 ) at corner B and if ( M 2 ) z = 0, determine (a) the distance d, (b) F2 and M 2.

SOLUTION

ΣM Bz : M 2 z = 0

(a) Have

(

)

k ⋅ rH /B × F1 + M1z = 0

where

(1)

rH /B = ( 31 in.) i − ( 2 in.) j F1 = λ EH F1

= =

( 6 in.) i + ( 6 in.) j − ( 7 in.) k 11.0 in.

( 20 lb )

20 lb ( 6i + 6 j − 7k ) 11.0

M1z = k ⋅ M1 M1 = λ EJ M1 =

−di + ( 3 in.) j − ( 7 in.) k d 2 + 58 in.

Then from Equation (1), 0 0 1 20 lb ⋅ in. ( −7 )( 480 lb ⋅ in.) 31 −2 0 + =0 11.0 d 2 + 58 6 6 −7

( 480 lb ⋅in.)

PROBLEM 3.97 CONTINUED Solving for d, Equation (1) reduces to 20 lb ⋅ in. 3360 lb ⋅ in. =0 (186 + 12 ) − 2 11.0 d + 58

d = 5.3955 in.

From which

or d = 5.40 in. (b)

F2 = F1 =

20 lb ( 6i + 6 j − 7k ) 11.0

= (10.9091i + 10.9091j − 12.7273k ) lb or F2 = (10.91 lb ) i + (10.91 lb ) j − (12.73 lb ) k M 2 = rH /B × F1 + M1 i j k 20 lb ⋅ in. ( −5.3955 ) i + 3j − 7k = 31 −2 0 + ( 480 lb ⋅ in.) 11.0 9.3333 6 6 −7 = ( 25.455i + 394.55 j + 360k ) lb ⋅ in. + ( −277.48i + 154.285 j − 360k ) lb ⋅ in. M 2 = − ( 252.03 lb ⋅ in.) i + ( 548.84 lb ⋅ in.) j or M 2 = − ( 21.0 lb ⋅ ft ) i + ( 45.7 lb ⋅ ft ) j

PROBLEM 3.98 A 4-ft-long beam is subjected to a variety of loadings. (a) Replace each loading with an equivalent force-couple system at end A of the beam. (b) Which of the loadings are equivalent?

SOLUTION (a)

ΣFy : − 200 lb − 100 lb = Ra

(a) Have

or R a = 300 lb and

ΣM A : 900 lb ⋅ ft − (100 lb )( 4 ft ) = M a or M a = 500 lb ⋅ ft ΣFy : − 300 lb = Rb

(b) Have

or R b = 300 lb ΣM A : − 450 lb ⋅ ft = M b

and

or M b = 450 lb ⋅ ft ΣFy : 150 lb − 450 lb = Rc

(c) Have

or R c = 300 lb and

ΣM A : 2250 lb ⋅ ft − ( 450 lb )( 4 ft ) = M c or M c = 450 lb ⋅ ft ΣFy : − 200 lb + 400 lb = Rd

(d) Have

or R d = 200 lb and

ΣM A :

( 400 lb )( 4 ft ) − 1150 lb ⋅ ft

= Md

or M d = 450 lb ⋅ ft (e) Have

ΣFy : − 200 lb − 100 lb = Re or R e = 300 lb

and

ΣM A : 100 lb ⋅ ft + 200 lb ⋅ ft − (100 lb )( 4 ft ) = M e or M e = 100 lb ⋅ ft

PROBLEM 3.98 CONTINUED (f) Have

ΣFy : − 400 lb + 100 lb = R f or R f = 300 lb

and

ΣM A : −150 lb ⋅ ft + 150 lb ⋅ ft + (100 lb )( 4 ft ) = M f or M f = 400 lb ⋅ ft

(g) Have

ΣFy : −100 lb − 400 lb = Rg or R g = 500 lb

and

ΣM A : 100 lb ⋅ ft + 2000 lb ⋅ ft − ( 400 lb )( 4 ft ) = M g or M g = 500 lb ⋅ ft

(h) Have

ΣFy : −150 lb − 150 lb = Rh or R h = 300 lb

and

ΣM A : 1200 lb ⋅ ft − 150 lb ⋅ ft − (150 lb )(4 ft ) = M h or M h = 450 lb ⋅ ft

(b)

Therefore, loadings (c) and (h) are equivalent

PROBLEM 3.99 A 4-ft-long beam is loaded as shown. Determine the loading of Problem 3.98 which is equivalent to this loading.

SOLUTION ΣFy : −100 lb − 200 lb = R

Have

or R = 300 lb and

ΣM A : − 200 lb ⋅ ft + 1400 lb ⋅ ft − ( 200 lb )( 4 ft ) = M or M = 400 lb ⋅ ft Equivalent to case (f) of Problem 3.98

Problem 3.98 Equivalent force-couples at A

case

R

(a)

300 lb

500 lb ⋅ ft

(b)

300 lb

450 lb ⋅ ft

(c)

300 lb

450 lb ⋅ ft

(d )

200 lb

450 lb ⋅ ft

(e)

300 lb

100 lb ⋅ ft

(f )

300 lb

400 lb ⋅ ft

(g )

500 lb

500 lb ⋅ ft

(h)

300 lb

450 lb ⋅ ft

M

PROBLEM 3.100 Determine the single equivalent force and the distance from point A to its line of action for the beam and loading of (a) Problem 3.98b, (b) Problem 3.98d, (c) Problem 3.98e. Problem 3.98: A 4-ft-long beam is subjected to a variety of loadings. (a) Replace each loading with an equivalent force-couple system at end A of the beam. (b) Which of the loadings are equivalent?

SOLUTION (a)

For equivalent single force at distance d from A ΣFy : − 300 lb = R

Have

or R = 300 lb ΣM C :

and

( 300 lb )( d ) − 450 lb ⋅ ft

=0

or d = 1.500 ft (b)

ΣFy : − 200 lb + 400 lb = R

Have

or R = 200 lb and

ΣM C :

( 200 lb )( d ) + ( 400 lb )( 4 − d ) − 1150 lb ⋅ ft

=0

or d = 2.25 ft (c)

Have

ΣFy : − 200 lb − 100 lb = R

or R = 300 lb and

ΣM C : 100 lb ⋅ ft + ( 200 lb )( d ) − (100 lb )( 4 − d ) + 200 lb ⋅ ft = 0

or d = 0.333 ft

PROBLEM 3.101 Five separate force-couple systems act at the corners of a metal block, which has been machined into the shape shown. Determine which of these systems is equivalent to a force F = (10 N ) j and a couple of moment M = ( 6 N ⋅ m ) i + ( 4 N ⋅ m ) k located at point A.

SOLUTION The equivalent force-couple system at A for each of the five force-couple systems will be determined. Each will then be compared to the given force-couple system to determine if they are equivalent. Force-couple system at B ΣF :

Have

F = (10 N ) j

or and

(10 N ) j = F

(

)

ΣM A : ΣM B + rB/ A × F = M

( 4 N ⋅ m ) i + ( 2 N ⋅ m ) k + ( 0.2 m ) i × (10 N ) j = M M = ( 4 N⋅m) i + ( 4 N⋅m)k

Comparing to given force-couple system at A, Is Not Equivalent Force-couple system at C

Have or and

ΣF :

(10 N ) j = F

F = (10 N ) j

(

)

ΣM A : M C + rC/ A × F = M

(8.5 N ⋅ m ) i + ( 0.2 m ) i + ( 0.25 m ) k  × (10 N ) j = M M = ( 6 N ⋅ m ) i + ( 2.0 N ⋅ m ) k

Comparing to given force-couple system at A, Is Not Equivalent

PROBLEM 3.101 CONTINUED Force-couple system at E ΣF :

Have

(10 N ) j = F

F = (10 N ) j

or

(

)

ΣM A : M E + rE/ A × F = M

and

( 6 N ⋅ m ) i + ( 0.4 m ) i − ( 0.08 m ) j × (10 N ) j = M M = (6 N⋅m) i + ( 4 N⋅m)k

Comparing to given force-couple system at A, Is Equivalent Force-couple system at G ΣF :

Have

(10 N ) i + (10 N ) j = F

F = (10 N ) i + (10 N ) j

or F has two force components

∴ force-couple system at G Is Not Equivalent Force-couple system at I

(10 N ) j = F

Have

ΣF :

or

F = (10 N ) j

and

(

)

ΣM A : ΣM I + rI / A × F = M

(10 N ⋅ m ) i − ( 2 N ⋅ m ) k + ( 0.4 m ) i − ( 0.2 m ) j + ( 0.4 m ) k  × (10 N ) j = M

or

M = (6 N ⋅ m) i + (2 N ⋅m)k

Comparing to given force-couple system at A, Is Not Equivalent

PROBLEM 3.102 The masses of two children sitting at ends A and B of a seesaw are 38 kg and 29 kg, respectively. Where should a third child sit so that the resultant of the weights of the three children will pass through C if she has a mass of (a) 27 kg, (b) 24 kg.

SOLUTION First

WA = mA g = ( 38 kg ) g WB = mB g = ( 29 kg ) g

(a)

WC = mC g = ( 27 kg ) g

For resultant weight to act at C,

ΣM C = 0

Then ( 38 kg ) g  ( 2 m ) − ( 27 kg ) g  ( d ) − ( 29 kg ) g  ( 2 m ) = 0 ∴ d =

76 − 58 = 0.66667 m 27 or d = 0.667 m

(b)

WC = mC g = ( 24 kg ) g

For resultant weight to act at C,

ΣM C = 0

Then ( 38 kg ) g  ( 2 m ) − ( 24 kg ) g  ( d ) − ( 29 kg ) g  ( 2 m ) = 0 ∴ d =

76 − 58 = 0.75 m 24 or d = 0.750 m

PROBLEM 3.103 Three stage lights are mounted on a pipe as shown. The mass of each light is mA = mB = 1.8 kg and mC = 1.6 kg . (a) If d = 0.75 m, determine the distance from D to the line of action of the resultant of the weights of the three lights. (b) Determine the value of d so that the resultant of the weights passes through the midpoint of the pipe.

SOLUTION WA = WB = m A g = (1.8 kg ) g

First

WC = mC g = (1.6 kg ) g d = 0.75 m

(a) Have

R = WA + WB + WC R = (1.8 + 1.8 + 1.6 ) kg  g

R = ( 5.2 g ) N

or Have

ΣM D : −1.8g ( 0.3 m ) − 1.8g (1.3 m ) − 1.6 g ( 2.05 m ) = −5.2 g ( D ) ∴ D = 1.18462 m

or D = 1.185 m D=

(b)

L = 1.25 m 2

Have ΣM D : − (1.8 g )( 0.3 m ) − (1.8g )(1.3 m ) − (1.6 g )(1.3 m + d ) = − ( 5.2 g )(1.25 m ) ∴ d = 0.9625 m

or d = 0.963 m

PROBLEM 3.104 Three hikers are shown crossing a footbridge. Knowing that the weights of the hikers at points C, D, and E are 800 N, 700 N, and 540 N, respectively, determine (a) the horizontal distance from A to the line of action of the resultant of the three weights when a = 1.1 m, (b) the value of a so that the loads on the bridge supports at A and B are equal.

SOLUTION a = 1.1 m

(a)

ΣF : −WC − WD − WE = R

Have

∴ R = −800 N − 700 N − 540 N R = 2040 N

(a)

R = 2040 N

or Have

ΣM A : − (800 N )(1.5 m ) − (700 N )(2.6 m ) − (540 N )(4.25 m ) = −R ( d ) ∴

− 5315 N ⋅ m = − ( 2040 N ) d d = 2.6054 m

and

or d = 2.61 m to the right of A (b) For equal reaction forces at A and B, the resultant, R, must act at the center of the span. (b)

L ΣM A = − R   2

From ∴

− ( 800 N )(1.5 m ) − ( 700 N )(1.5 m + a ) − ( 540 N )(1.5 m + 2.5a ) = − ( 2040 N )( 3 m )

3060 + 2050a = 6120 and

a = 1.49268 m

or a = 1.493 m

PROBLEM 3.105 Gear C is rigidly attached to arm AB. If the forces and couple shown can be reduced to a single equivalent force at A, determine the equivalent force and the magnitude of the couple M.

SOLUTION

For equivalence ΣFx : − ( 90 N ) sin 30° + (125 N ) cos 40° = Rx

or Rx = 50.756 N ΣFy : − ( 90 N ) cos30° − 200 N − (125 N ) sin 40° = Ry

or Ry = −358.29 N

and

( 50.756 )2 + ( −358.29 )2

R=

Then tan θ =

Ry Rx

=

−358.29 = −7.0591 50.756

= 361.87 N ∴ θ = −81.937°

or R = 362 N

81.9°

Also ΣM A : M − ( 90 N ) sin 35° ( 0.6 m ) − ( 200 N ) cos 25°  ( 0.85 m ) − (125 N ) sin 65°  (1.25 m ) = 0

∴ M = 326.66 N ⋅ m

or M = 327 N ⋅ m

PROBLEM 3.106 To test the strength of a 25 × 20-in. suitcase, forces are applied as shown. If P = 18 lb, (a) determine the resultant of the applied forces, (b) locate the two points where the line of action of the resultant intersects the edge of the suitcase.

SOLUTION

(a) P = 18 lb Have

ΣF : − ( 20 lb ) i +

42 lb ( −3i + 2 j) + (18 lb ) j + ( 36 lb ) i = Rxi + Ry j 13

∴ − (18.9461 lb ) i + ( 41.297 lb ) j = Rxi + Ry j R = − (18.95 lb ) i + ( 41.3 lb ) j

or R=

Rx2 + Ry2 =

(18.9461)2 + ( 41.297 )2

= 45.436 lb

 Ry  −1  41.297   = tan   = −65.355°  −18.9461   Rx 

θ x = tan −1 

or R = 45.4 lb (b) Have

65.4°

ΣM B = M B

 42 lb  M B = ( 4 in.) j × ( −20 lb ) i + ( 21 in.) i ×  ( −3i + 2 j) + (12 in.) j × ( 36 lb ) i + ( 3 in.) i × (18 lb ) j 13  

∴ M B = (191.246 lb ⋅ in.) k

PROBLEM 3.106 CONTINUED M B = rB × R

Since

i j k ∴ (191.246 lb ⋅ in.) k = x y 0 = ( 41.297 x + 18.9461y ) k −18.9461 41.297 0 For

y = 0,

x=

191.246 = 4.6310 in. 41.297

or x = 4.63 in.

For

x = 0,

y =

191.246 = 10.0942 in. 18.9461

or y = 10.09 in.

PROBLEM 3.107 Solve Problem 3.106 assuming that P = 28 lb. Problem 3.106: To test the strength of a 25 × 20-in. suitcase, forces are applied as shown. If P = 18 lb, (a) determine the resultant of the applied forces, (b) locate the two points where the line of action of the resultant intersects the edge of the suitcase.

SOLUTION

(a) P = 28 lb Have

ΣF : − ( 20 lb ) i +

42 ( −3i + 2 j) + ( 28 lb ) j + ( 36 lb ) i = Rxi + Ry j 13

∴ − (18.9461 lb ) i + ( 51.297 lb ) j = Rxi + Ry j R = − (18.95 lb ) i + ( 51.3 lb ) j

or R=

Rx2 + Ry2 =

(18.9461)2 + ( 51.297 )2

= 54.684 lb

 Ry  −1  51.297   = tan   = −69.729°  −18.9461   Rx 

θ x = tan −1 

or R = 54.7 lb (b) Have

69.7°

ΣM B = M B

 42 lb  M B = ( 4 in.) j × ( −20 lb ) i + ( 21 in.) i ×  ( −3i + 2j) + (12 in.) j × ( 36 lb ) i + ( 3 in.) i × ( 28 lb ) j 13  

∴ M B = ( 221.246 lb ⋅ in.) k

PROBLEM 3.107 CONTINUED M B = rB × R

Since

i j k ∴ ( 221.246 lb ⋅ in.) k = x y 0 = ( 51.297 x + 18.9461y ) k −18.9461 51.297 0

For

y = 0,

x=

221.246 = 4.3130 in. 51.297

or x = 4.31 in.

For

x = 0,

y =

221.246 = 11.6776 in. 18.9461

or y = 11.68 in.

PROBLEM 3.108 As four holes are punched simultaneously in a piece of aluminum sheet metal, the punches exert on the piece the forces shown. Knowing that the forces are perpendicular to the surfaces of the piece, determine (a) the resultant of the applied forces when α = 45° and the point of intersection of the line of action of that resultant with a line drawn through points A and B, (b) the value of α so that the line of action of the resultant passes through fold EF.

SOLUTION Position the origin for the coordinate system along the centerline of the sheet metal at the intersection with line EF. ΣF = R

(a) Have

R = −  2.6 j − 5.25 j − 10.5 ( cos 45°i + sin 45° j) − 3.2i  kN

∴ R = − (10.6246 kN ) i − (15.2746 kN ) j R=

Rx2 + Ry2 =

(10.6246 )2 + (15.2746 )2

= 18.6064 kN  Ry  −1  −15.2746   = tan   = 55.179°  −10.6246   Rx 

θ = tan −1 

or R = 18.61 kN

55.2°

M EF = ΣM EF

Have where

MEF = (2.6 kN )(90 mm ) + (5.25 kN )(40 mm ) − (10.5 kN )(20 mm ) − ( 3.2 kN ) ( 40 mm ) sin 45° + 40 mm 

∴ MEF = 15.4903 N ⋅ m

To obtain distance d left of EF, M EF = dRy = d ( −15.2746 kN )

Have ∴d =

15.4903 N ⋅ m = −1.01412 × 10−3 m −3 −15.2746 × 10 N or d = 1.014 mm left of EF

PROBLEM 3.108 CONTINUED (b) Have

M EF = 0 M EF = 0 = ( 2.6 kN )( 90 mm ) + ( 5.25 kN )( 40 mm ) − (10.5 kN )( 20 mm ) − ( 3.2 kN ) ( 40 mm ) sin α + 40 mm 

∴ (128 N ⋅ m ) sin α = 106 N ⋅ m

sin α = 0.828125

α = 55.907° or α = 55.9°

PROBLEM 3.109 As four holes are punched simultaneously in a piece of aluminum sheet metal, the punches exert on the piece the forces shown. Knowing that the forces are perpendicular to the surfaces of the piece, determine (a) the value of α so that the resultant of the applied forces is parallel to the 10.5 N force, (b) the corresponding resultant of the applied forces and the point of intersection of its line of action with a line drawn through points A and B.

SOLUTION (a) For the resultant force, R, to be parallel to the 10.5 kN force,

α =φ ∴ tan α = tan φ =

Ry Rx

where Rx = −3.2 kN − (10.5 kN ) sin α Ry = −2.6 kN − 5.25 kN − (10.5 kN ) cos α ∴ tan α =

3.2 + 10.5 sinα 7.85 + 10.5cos α

tan α =

and

3.2 = 0.40764 7.85

α = 22.178°

or α = 22.2°

α = 22.178°

(b) From

Rx = −3.2 kN − (10.5 kN ) sin 22.178° = −7.1636 kN Ry = −7.85 kN − (10.5 kN ) cos 22.178° = −17.5732 kN R=

Rx2 + Ry2 =

( 7.1636 )2 + (17.5732 )2

= 18.9770 kN

R = 18.98 kN

or

67.8°

Then M EF = ΣM EF where M EF = ( 2.6 kN )( 90 mm ) + ( 5.25 kN )( 40 mm ) − (10.5 kN )( 20 mm ) − ( 3.2 kN ) ( 40 mm ) sin 22.178° + 40 mm 

= 57.682 N ⋅ m

PROBLEM 3.109 CONTINUED To obtain distance d left of EF, M EF = dRy = d ( −17.5732 )

Have ∴d =

57.682 N ⋅ m = −3.2824 × 10−3 m −17.5732 × 103 N or d = 3.28 mm left of EF

PROBLEM 3.110 A truss supports the loading shown. Determine the equivalent force acting on the truss and the point of intersection of its line of action with a line through points A and G.

SOLUTION

R = ΣF

Have

R = ( 240 N )( cos 70°i − sin 70° j) − (160 N ) j

+ ( 300 N )( − cos 40°i − sin 40° j) − (180 N ) j ∴ R = − (147.728 N ) i − ( 758.36 N ) j R=

Rx2 + Ry2 =

(147.728)2 + ( 758.36 )2

= 772.62 N  Ry  −1  −758.36   = tan   = 78.977°  −147.728   Rx 

θ = tan −1 

or R = 773 N

79.0°

ΣM A = dRy

Have where

ΣM A = − [ 240 N cos 70°] ( 6 m ) − [ 240 N sin 70°] ( 4 m ) − (160 N )(12 m ) +

[300 N cos 40°] ( 6 m )

− [300 N sin 40°] ( 20 m ) − (180 N )( 8 m ) = −7232.5 N ⋅ m ∴d =

−7232.5 N ⋅ m = 9.5370 m −758.36 N

or d = 9.54 m to the right of A

PROBLEM 3.111 Three forces and a couple act on crank ABC. For P = 5 lb and α = 40°, (a) determine the resultant of the given system of forces, (b) locate the point where the line of action of the resultant intersects a line drawn through points B and C, (c) locate the point where the line of action of the resultant intersects a line drawn through points A and B.

SOLUTION P = 5 lb,

(a)

α = 40°

R = ΣF

Have

= ( 5 lb )( cos 40°i + sin 40° j) − ( 3 lb ) i − ( 2 lb ) j ∴ R = ( 0.83022 lb ) i + (1.21394 lb ) j R=

Rx2 + Ry2 =

( 0.83022 )2 + (1.21394 )2

= 1.47069 lb  Ry  −1  1.21394   = tan   = 55.632°  0.83022   Rx 

θ = tan −1 

or R = 1.471 lb

55.6°

MB = ΣMB = dRy

(b) From where

MB = − (5 lb )cos 40°  (15 in. )sin 50°  − (5 lb )sin 40°  × (15 in. )sin 50°  + ( 3 lb ) ( 6 in.) sin 50°

− ( 2 lb )( 6 in.) + 50 lb ⋅ in.

∴ M B = −23.211 lb ⋅ in. and

d =

MB −23.211 lb ⋅ in. = = −19.1205 in. Ry 1.21394 lb

or d = 19.12 in. to the left of B

PROBLEM 3.111 CONTINUED M B = rD/B × R

(c) From

− ( 23.211 lb ⋅ in.) k = ( −d1 cos 50°i + d1 sin 50° j) × ( −0.83022 lb ) i + (1.21394 lb ) j

− ( 23.211 lb ⋅ in.) k = ( −0.78028d1 − 0.63599d1 ) k ∴ d1 = or

23.211 = 16.3889 in. 1.41627

d1 = 16.39 in. from B along line AB or 1.389 in. above and to the left of A

PROBLEM 3.112 Three forces and a couple act on crank ABC. Determine the value of d so that the given system of forces is equivalent to zero at (a) point B, (b) point D.

SOLUTION ΣFx = 0

Based on

P cos α − 3 lb = 0 ∴ P cos α = 3 lb

(1)

ΣFy = 0

and

P sin α − 2 lb = 0 ∴ P sin α = 2 lb

(2)

Dividing Equation (2) by Equation (1), tan α =

2 3

∴ α = 33.690° Substituting into Equation (1), P=

3 lb = 3.6056 lb cos 33.690°

P = 3.61 lb

or (a) Based on

33.7°

ΣM B = 0

− ( 3.6056 lb ) cos 33.690° ( d + 6 in.) sin 50° − ( 3.6056 lb ) sin 33.690° ( d + 6 in.) cos 50° + ( 3 lb ) ( 6 in.) sin 50° − ( 2 lb )( 6 in.) + 50 lb ⋅ in. = 0

−3.5838d = −30.286 ∴ d = 8.4509 in. or d = 8.45 in.

PROBLEM 3.112 CONTINUED (b) Based on

ΣM D = 0

− ( 3.6056 lb ) cos 33.690°  ( d + 6 in.) sin 50°  − ( 3.6056 lb ) sin 33.690°  ( d + 6 in.) cos 50° + 6 in. + ( 3 lb ) ( 6 in.) sin 50° + 50 lb ⋅ in. = 0

−3.5838d = −30.286 ∴ d = 8.4509 in. or d = 8.45 in. This result is expected, since R = 0 and M RB = 0 for d = 8.45 in. implies that R = 0 and M = 0 at any other point for the value of d found in part a.

PROBLEM 3.113 Pulleys A and B are mounted on bracket CDEF. The tension on each side of the two belts is as shown. Replace the four forces with a single equivalent force, and determine where its line of action intersects the bottom edge of the bracket.

SOLUTION Equivalent force-couple at A due to belts on pulley A ΣF : −120 N − 160 N = RA

Have

∴ R A = 280 N ΣM A : − 40 N ( 0.02 m ) = M A

Have

∴ M A = 0.8 N ⋅ m Equivalent force-couple at B due to belts on pulley B ΣF :

Have

( 210 N + 150 N )

∴ R B = 360 N

25° = R B

25°

ΣM B : − 60 N ( 0.015 m ) = M B

Have

∴ M B = 0.9 N ⋅ m Equivalent force-couple at F Have

ΣF : R F = ( −280 N ) j + ( 360 N )( cos 25°i + sin 25° j) = ( 326.27 N ) i − (127.857 N ) j

R = RF =

2 2 RFx + RFy =

( 326.27 )2 + (127.857 )2

= 350.43 N

 RFy  −1  −127.857   = tan   = −21.399°  326.27   RFx 

θ = tan −1 

or R F = R = 350 N

21.4°

PROBLEM 3.113 CONTINUED Have ΣM F : M F = − ( 280 N )( 0.06 m ) − 0.80 N ⋅ m − ( 360 N ) cos 25° ( 0.010 m ) + ( 360 N ) sin 25° ( 0.120 m ) − 0.90 N ⋅ m

M F = − ( 3.5056 N ⋅ m ) k To determine where a single resultant force will intersect line FE, M F = dR y ∴ d =

MF −3.5056 N ⋅ m = = 0.027418 m = 27.418 mm Ry −127.857 N

or d = 27.4 mm

PROBLEM 3.114 As follower AB rolls along the surface of member C, it exerts a constant force F perpendicular to the surface. (a) Replace F with an equivalent force-couple system at the point D obtained by drawing the perpendicular from the point of contact to the x axis (b) For a = 1 m and b = 2 m, determine the value of x for which the moment of the equivalent forcecouple system at D is maximum.

SOLUTION (a) The slope of any tangent to the surface of member C is dy d   x 2   −2b = b 1 − 2   = 2 x dx dx   a   a Since the force F is perpendicular to the surface,

 dy  tan α = −    dx 

−1

=

a2  1    2b  x 

For equivalence ΣF : F = R ΣM D :

( F cosα )( y A ) = M D

where cos α =

2bx

(a ) 2

2

+ ( 2bx )

∴ MD =

2

,

 x2  y A = b 1 − 2  a  

 x3  2 Fb 2  x − 2  a   a 4 + 4b 2 x 2

Therefore, the equivalent force-couple system at D is

R = F

 a2  tan −1    2bx 

 x3  2Fb 2  x − 2  a   M = a 4 + 4b 2 x 2

PROBLEM 3.114 CONTINUED dM =0 dx

(b) To maximize M, the value of x must satisfy where, for a = 1 m, b = 2 m M =

∴

dM = 8F dx

)

1 + 16 x 2

1 1 + 16 x 2 1 − 3x 2 − x − x3  ( 32 x ) 1 + 16 x 2 2 

(

) ( ) ( (1 + 16x ) (1 + 16x )(1 − 3x ) − 16x ( x − x ) = 0 2

2

2

)

−

1 2

 = 0

3

32 x 4 + 3x 2 − 1 = 0

or x2 =

(

8F x − x 3

−3 ± 9 − 4 ( 32 )( −1) 2 ( 32 )

= 0.136011 m 2 and − 0.22976 m 2

Using the positive value of x 2 , x = 0.36880 m or x = 369 mm

PROBLEM 3.115 As plastic bushings are inserted into a 3-in.-diameter cylindrical sheet metal container, the insertion tool exerts the forces shown on the enclosure. Each of the forces is parallel to one of the coordinate axes. Replace these forces with an equivalent force-couple system at C.

SOLUTION For equivalence Σ F: FA + FB + FC + FD = R C

R C = − ( 5 lb ) j − ( 3 lb ) j − ( 4 lb ) k − ( 7 lb ) i ∴ R C = ( −7 lb ) i − ( 8 lb ) j − ( 4 lb ) k Also for equivalence ΣM C : rA′/C × FA + rB′/C × FB + rD′/C × FD = M C or

MC

i j k i j k i j k = 0 0 −1.5 in. + 1 in. 0 −1.5 in. + 0 1.5 in. 1.5 in. 0 5 lb 0 0 −3 lb 0 −7 lb 0 0 = ( −7.50 lb ⋅ in. − 0 ) i  + ( 0 − 4.50 lb ⋅ in.) i + ( −3.0 lb ⋅ in. − 0 ) k  + (10.5 lb ⋅ in. − 0 ) j + ( 0 + 10.5 lb ⋅ in.) k 

or M C = − (12.0 lb ⋅ in.) i + (10.5 lb ⋅ in.) j + ( 7.5 lb ⋅ in.) k

PROBLEM 3.116 Two 300-mm-diameter pulleys are mounted on line shaft AD. The belts B and C lie in vertical planes parallel to the yz plane. Replace the belt forces shown with an equivalent force-couple system at A.

SOLUTION Equivalent force-couple at each pulley Pulley B R B = ( 290 N )( − cos 20° j + sin 20°k ) − 430 Nj = − ( 702.51 N ) j + ( 99.186 N ) k M B = − ( 430 N − 290 N )( 0.15 m ) i = − ( 21 N ⋅ m ) i Pulley C R C = ( 310 N + 480 N )( − sin10° j − cos10°k ) = − (137.182 N ) j − ( 778.00 N ) k M C = ( 480 N − 310 N )( 0.15 m ) i = ( 25.5 N ⋅ m ) i Then

R = R B + R C = − ( 839.69 N ) j − ( 678.81 N ) k or R = − ( 840 N ) j − ( 679 N ) k

M A = M B + M C + rB/ A × R B + rC/ A × R C

i j k = − ( 21 N ⋅ m ) i + ( 25.5 N ⋅ m ) i + 0.45 0 0 N⋅m 0 −702.51 99.186 i j k + 0.90 0 0 N⋅m 0 −137.182 −778.00 = ( 4.5 N ⋅ m ) i + ( 655.57 N ⋅ m ) j − ( 439.59 N ⋅ m ) k or M A = ( 4.50 N ⋅ m ) i + ( 656 N ⋅ m ) j − ( 440 N ⋅ m ) k

PROBLEM 3.117 A mechanic uses a crowfoot wrench to loosen a bolt at C. The mechanic holds the socket wrench handle at points A and B and applies forces at these points. Knowing that these forces are equivalent to a force-couple system at C consisting of the force C = − ( 40 N ) i + ( 20 N ) k and the couple M C = ( 40 N ⋅ m ) i , determine the forces applied at A and B when

Az = 10 N.

SOLUTION ΣF : A + B = C

Have

Fx : Ax + Bx = −40 N

or

∴ Bx = − ( Ax + 40 N )

(1)

ΣFy : Ay + By = 0 Ay = − By

or

(2)

ΣFz : 10 N + Bz = 20 N Bz = 10 N

or

ΣM C : rB/C × B + rA/C × A = M C

Have

∴

or

(3)

i j k i j k 0.2 0 −0.05 + 0.2 0 0.2 N ⋅ m = ( 40 N ⋅ m ) i Bx By 10 Ax Ay 10

( 0.05By − 0.2 Ax ) i + ( −0.05Bx − 2 + 0.2 Ax − 2) j (

)

+ 0.2By + 0.2 Ay k = ( 40 N ⋅ m ) i From

i - coefficient

0.05By − 0.2 Ay = 40 N ⋅ m

(4)

j - coefficient

− 0.05Bx + 0.2 Ax = 4 N ⋅ m

(5)

k - coefficient

0.2 By + 0.2 Ay = 0

(6)

PROBLEM 3.117 CONTINUED From Equations (2) and (4):

(

)

0.05By − 0.2 − By = 40

By = 160 N, Ay = −160 N From Equations (1) and (5):

−0.05 ( − Ax − 40 ) + 0.2 Ax = 4 Ax = 8 N

From Equation (1):

Bx = − ( 8 + 40 ) = −48 N ∴ A = ( 8 N ) i − (160 N ) j + (10 N ) k B = − ( 48 N ) i + (160 N ) j + (10 N ) k

PROBLEM 3.118 While using a pencil sharpener, a student applies the forces and couple shown. (a) Determine the forces exerted at B and C knowing that these forces and the couple are equivalent to a force-couple system at A consisting of the force R = ( 3.9 lb ) i + Ry j − (1.1 lb ) k and the couple M RA = M xi + (1.5 lb ⋅ ft ) j − (1.1 lb ⋅ ft ) k. . (b) Find the corresponding values of Ry and M x .

SOLUTION ΣF : B + C = R

Have

ΣFx : Bx + Cx = 3.9 lb

or

Bx = 3.9 lb − Cx

(1)

ΣFy : C y = Ry

(2)

ΣFz : C z = −1.1 lb

(3)

ΣM A : rB/ A × B + rC/ A × C + M B = M RA

Have

i j k i j k 1 1 4 0 2.0 + ( 2 lb ⋅ ft ) i = M xi + (1.5 lb ⋅ ft ) j − (1.1 lb ⋅ ft ) k ∴ x 0 4.5 + 12 12 Bx 0 0 C x C y −1.1

( 2 − 0.166667C y ) i + ( 0.375Bx + 0.166667Cx + 0.36667 ) j + ( 0.33333C y ) k = M xi + (1.5 ) j − (1.1) k From

i - coefficient

2 − 0.166667C y = M x

(4)

j - coefficient

0.375Bx + 0.166667Cx + 0.36667 = 1.5

(5)

k - coefficient

0.33333C y = −1.1

(6)

or

C y = −3.3 lb

(a) From Equations (1) and (5): 0.375 ( 3.9 − Cx ) + 0.166667Cx = 1.13333 Cx = From Equation (1):

0.32917 = 1.58000 lb 0.20833 Bx = 3.9 − 1.58000 = 2.32 lb ∴ B = ( 2.32 lb ) i

C = (1.580 lb ) i − ( 3.30 lb ) j − (1.1 lb ) k (b) From Equation (2): From Equation (4):

Ry = C y = −3.30 lb

or R y = − ( 3.30 lb ) j

M x = −0.166667 ( −3.30 ) + 2.0 = 2.5500 lb ⋅ ft or M x = ( 2.55 lb ⋅ ft ) i

PROBLEM 3.119 A portion of the flue for a furnace is attached to the ceiling at A. While supporting the free end of the flue at F, a worker pushes in at E and pulls out at F to align end E with the furnace. Knowing that the 10-lb force at F lies in a plane parallel to the yz plane, determine (a) the angle α the force at F should form with the horizontal if duct AB is not to tend to rotate about the vertical, (b) the force-couple system at B equivalent to the given force system when this condition is satisfied.

SOLUTION (a) Duct AB will not have a tendency to rotate about the vertical or y-axis if:

(

)

R M By = j ⋅ ΣM RB = j ⋅ rF /B × FF + rE/B × FE = 0

where

rF /B = ( 45 in.) i − ( 23 in.) j + ( 28 in.) k rE/B = ( 54 in.) i − ( 34 in.) j + ( 28 in.) k FF = 10 lb ( sin α ) j + ( cos α ) k 

FE = − ( 5 lb ) k

∴

ΣM RB

i j k i j k = (10 lb ) 45 in. −23 in. 28 in. + ( 5 lb )( 2 in.) 27 −17 14 0 sin α cos α 0 0 −1 = ( −230 cos α − 280sin α + 170 ) i − ( 450 cos α − 270 ) j + ( 450sin α ) k  lb ⋅ in.

Thus,

R M By = −450 cos α + 270 = 0

cos α = 0.60

α = 53.130° or α = 53.1°

PROBLEM 3.119 CONTINUED (b) R = FE + FF where FE = − ( 5 lb ) k FF = (10 lb )( sin 53.130° j + cos 53.130°k ) = ( 8 lb ) j + ( 6 lb ) k

∴ R = ( 8 lb ) j + (1 lb ) k and

M = ΣM RB = −  230 ( 0.6 ) + 280 ( 0.8 ) − 170  i −  450 ( 0.6 ) − 270  j +  450 ( 0.8 ) k

= − (192 lb ⋅ in.) i − ( 0 ) j + ( 360 lb ⋅ in.) k or M = − (192 lb ⋅ in.) i + ( 360 lb ⋅ in.) k

PROBLEM 3.120 A portion of the flue for a furnace is attached to the ceiling at A. While supporting the free end of the flue at F, a worker pushes in at E and pulls out at F to align end E with the furnace. Knowing that the 10-lb force at F lies in a plane parallel to the yz plane and that α = 60°, (a) replace the given force system with an equivalent force-couple system at C, (b) determine whether duct CD will tend to rotate clockwise or counterclockwise relative to elbow C, as viewed from D to C.

SOLUTION R = ΣF = FF + FE

(a) Have

FF = 10 lb ( sin 60° ) j + ( cos 60° ) k  = ( 8.6603 lb ) j + ( 5.0 lb ) k

where

FE = − ( 5 lb ) k

∴ R = ( 8.6603 lb ) j

or R = ( 8.66 lb ) j

Have

M CR = Σ ( r × F ) = rF /C × FF + rE/C × FE

where

rF /C = ( 9 in.) i − ( 2 in.) j rE/C = (18 in.) i − (13 in.) j

∴

M CR

i j k i j k = 9 −2 0 lb ⋅ in. + 18 −13 0 lb ⋅ in. 0 8.6603 5.0 0 0 −5

= ( 55 lb ⋅ in.) i + ( 45 lb ⋅ in.) j + ( 77.942 lb ⋅ in.) k or M CR = ( 55.0 lb ⋅ in.) i + ( 45.0 lb ⋅ in.) j + ( 77.9 lb ⋅ in.) k (b) To determine which direction duct section CD has a tendency to turn, have R M CD = λ DC ⋅ M CR

where λ DC = Then

− (18 in.) i + ( 4 in.) j 2 85 in.

R M CD =

=

1 ( −9i + 2 j) 85

1 ( −9i + 2 j) ⋅ ( 55i + 45j + 77.942k ) lb ⋅ in. 85

= ( −53.690 + 9.7619 ) lb ⋅ in. = −43.928 lb ⋅ in. Since λ DC ⋅ M CR < 0, duct DC tends to rotate clockwise relative to elbow C as viewed from D to C.

PROBLEM 3.121 The head-and-motor assembly of a radial drill press was originally positioned with arm AB parallel to the z axis and the axis of the chuck and bit parallel to the y axis. The assembly was then rotated 25o about the y axis and 20o about the centerline of the horizontal arm AB, bringing it into the position shown. The drilling process was started by switching on the motor and rotating the handle to bring the bit into contact with the workpiece. Replace the force and couple exerted by the drill press with an equivalent force-couple system at the center O of the base of the vertical column.

SOLUTION

R =F

Have

= ( 44 N ) ( sin 20° cos 25° ) i − ( cos 20° ) j − ( sin 20° sin 25° ) k 

= (13.6389 N ) i − ( 41.346 N ) j − ( 6.3599 N ) k

or R = (13.64 N ) i − ( 41.3 N ) j − ( 6.36 N ) k M O = rB/O × F + M C

Have where

rB/O = ( 0.280 m ) sin 25°  i + ( 0.300 m ) j + ( 0.280 m ) cos 25°  k

= ( 0.118333 m ) i + ( 0.300 m ) j + ( 0.25377 m ) k M C = ( 7.2 N ⋅ m ) ( sin 20° cos 25° ) i − ( cos 20° ) j − ( sin 20° sin 25° ) k 

= ( 2.2318 N ⋅ m ) i − ( 6.7658 N ⋅ m ) j − (1.04072 N ⋅ m ) k

∴ MO

i j k = 0.118333 0.300 0.25377 N ⋅ m 13.6389 −41.346 −6.3599 + ( 2.2318i − 6.7658 j − 1.04072k ) N ⋅ m = (10.8162 N ⋅ m ) i − ( 2.5521 N ⋅ m ) j − (10.0250 N ⋅ m ) k or M O = (10.82 N ⋅ m ) i − ( 2.55 N ⋅ m ) j − (10.03 N ⋅ m ) k

PROBLEM 3.122 While a sagging porch is leveled and repaired, a screw jack is used to support the front of the porch. As the jack is expanded, it exerts on the porch the force-couple system shown, where R = 300 N and M = 37.5 N ⋅ m. Replace this force-couple system with an equivalent force-couple system at C.

SOLUTION  − ( 0.2 m ) i + (1.4 m ) j − ( 0.5 m ) k  R C = R = ( 300 N ) λ AB = 300 N   1.50 m  

From

R C = − ( 40.0 N ) i + ( 280 N ) j − (100 N ) k M C = rA/C × R + M

From where

rA/C = ( 2.6 m ) i + ( 0.5 m ) k  ( 0.2 m ) i − (1.4 m ) j + ( 0.5 m ) k  M = ( 37.5 N ⋅ m ) λ BA = ( 37.5 N ⋅ m )   1.50 m  

= ( 5.0 N ⋅ m ) i − ( 35.0 N ⋅ m ) j + (12.5 N ⋅ m ) k

∴ MC

i j k = (10 N ⋅ m ) 2.6 0 0.5 + ( 5.0 N ⋅ m ) i − ( 35.0 N ⋅ m ) j + (12.5 N ⋅ m ) k −4 28 −10 = ( −140 + 5 ) N ⋅ m  i + ( −20 + 260 − 35 ) N ⋅ m  j + ( 728 + 12.5 ) N ⋅ m  k

or M C = − (135.0 N ⋅ m ) i + ( 205 N ⋅ m ) j + ( 741 N ⋅ m ) k

PROBLEM 3.123 Three children are standing on a 15 × 15-ft raft. If the weights of the children at points A, B, and C are 85 lb, 60 lb, and 90 lb, respectively, determine the magnitude and the point of application of the resultant of the three weights.

SOLUTION

Have

ΣF : FA + FB + FC = R − ( 85 lb ) j − ( 60 lb ) j − ( 90 lb ) j = R − ( 235 lb ) j = R

Have

or R = 235 lb

ΣM x : FA ( z A ) + FB ( z B ) + FC ( zC ) = R ( z D )

(85 lb)( 9 ft ) + ( 60 lb )(1.5 ft ) + ( 90 lb )(14.25 ft ) = ( 235 lb )( zD ) ∴ z D = 9.0957 ft Have

or z D = 9.10 ft

ΣM z : FA ( x A ) + FB ( xB ) + FC ( xC ) = R ( xD )

(85 lb )( 3 ft ) + ( 60 lb )( 4.5 ft ) + ( 90 lb )(14.25 ft ) = ( 235 lb )( xD ) ∴ xD = 7.6915 ft

or xD = 7.69 ft

PROBLEM 3.124 Three children are standing on a 15 × 15-ft raft. The weights of the children at points A, B, and C are 85 lb, 60 lb, and 90 lb, respectively. If a fourth child of weight 95 lb climbs onto the raft, determine where she should stand if the other children remain in the positions shown and the line of action of the resultant of the four weights is to pass through the center of the raft.

SOLUTION

Have

ΣF : FA + FB + FC + FD = R − ( 85 lb ) j − ( 60 lb ) j − ( 90 lb ) j − ( 95 lb ) j = R ∴ R = − ( 330 lb ) j

Have

ΣM x : FA ( z A ) + FB ( z B ) + FC ( zC ) + FD ( z D ) = R ( z H )

(85 lb )( 9 ft ) + ( 60 lb )(1.5 ft ) + ( 90 lb )(14.25 ft ) + ( 95 lb )( zD ) = ( 330 lb )( 7.5 ft ) ∴ z D = 3.5523 ft Have

or z D = 3.55 ft

ΣM z : FA ( x A ) + FB ( xB ) + FC ( xC ) + FD ( xD ) = R ( xH )

(85 lb )( 3 ft ) + ( 60 lb )( 4.5 ft ) + ( 90 lb )(14.25 ft ) + ( 95 lb )( xD ) = ( 330 lb )( 7.5 ft ) ∴ xD = 7.0263 ft

or xD = 7.03 ft

PROBLEM 3.125 The forces shown are the resultant downward loads on sections of the flat roof of a building because of accumulated snow. Determine the magnitude and the point of application of the resultant of these four loads.

SOLUTION

Have

ΣF : FA + FB + FC + FD = R − ( 580 kN ) j − ( 2350 kN ) j − ( 330 kN ) j − (140 kN ) j = R ∴ R = − ( 3400 kN ) j

Have

R = 3400 kN

ΣM x : FA ( z A ) + FB ( z B ) + FC ( zC ) + FD ( z D ) = R ( z E )

( 580 kN )(8 m ) + ( 2350 kN )(16 m ) + ( 330 kN )( 6 m ) + (140 kN )( 33.5 m ) = ( 3400 kN )( zE ) ∴ z E = 14.3853 m Have

or z E = 14.39 m

ΣM z : FA ( x A ) + FB ( xB ) + FC ( xC ) + FD ( xD ) = R ( xE )

( 580 kN )(10 m ) + ( 2350 kN )( 32 m ) + ( 330 kN )( 54 m ) + (140 kN )( 32 m ) = ( 3400 kN )( xE ) ∴ xE = 30.382 m

or xE = 30.4 m

PROBLEM 3.126 The forces shown are the resultant downward loads on sections of the flat roof of a building because of accumulated snow. If the snow represented by the 580-kN force is shoveled so that the this load acts at E, determine a and b knowing that the point of application of the resultant of the four loads is then at B.

SOLUTION

Have

ΣF : FB + FC + FD + FE = R − ( 2350 kN ) j − ( 330 kN ) j − (140 kN ) j − ( 580 kN ) j = R ∴ R = − ( 3400 kN ) j

Have

ΣM x : FB ( z B ) + FC ( zC ) + FD ( z D ) + FE ( z E ) = R ( z B )

( 2350 kN )(16 m ) + ( 330 kN )( 6 m ) + (140 kN )( 33.5 m ) + ( 580 kN )( b ) = ( 3400 kN )(16 m ) ∴ b = 17.4655 m Have

or b = 17.47 m

ΣM z : FB ( xB ) + FC ( xC ) + FD ( xD ) + FE ( xE ) = R ( xB )

( 2350 kN )( 32 m ) + ( 330 kN )( 54 m ) + (140 kN )( 32 m ) + ( 580 kN )( a ) = ( 3400 kN )( 32 m ) ∴ a = 19.4828 m

or a = 19.48 m

PROBLEM 3.127 A group of students loads a 2 × 4-m flatbed trailer with two 0.6 × 0.6 × 0.6-m boxes and one 0.6 × 0.6 × 1.2-m box. Each of the boxes at the rear of the trailer is positioned so that it is aligned with both the back and a side of the trailer. Determine the smallest load the students should place in a second 0.6 × 0.6 × 1.2-m box and where on the trailer they should secure it, without any part of the box overhanging the sides of the trailer, if each box is uniformly loaded and the line of action of the resultant of the weights of the four boxes is to pass through the point of intersection of the centerlines of the trailer and the axle. (Hint: Keep in mind that the box may be placed either on its side or on its end.)

SOLUTION

For the smallest weight on the trailer so that the resultant force of the four weights acts over the axle at the intersection with the center line of the trailer, the added 0.6 × 0.6 × 1.2-m box should be placed adjacent to one of the edges of the trailer with the 0.6 × 0.6-m side on the bottom. The edges to be considered are based on the location of the resultant for the three given weights.

ΣF : − ( 200 N ) j − ( 400 N ) j − (180 N ) j = R

Have

∴ R = − ( 780 N ) j Have

ΣM z :

( 200 N )( 0.3 m ) + ( 400 N )(1.7 m ) + (180 N )(1.7 m ) = ( 780 N )( x ) ∴ x = 1.34103 m

Have

ΣM x :

( 200 N )( 0.3 m ) + ( 400 N )( 0.6 m ) + (180 N )( 2.4 m ) = ( 780 N )( z ) ∴ z = 0.93846 m

From the statement of the problem, it is known that the resultant of R from the original loading and the lightest load W passes through G, the point of intersection of the two center lines. Thus, ΣM G = 0. Further, since the lightest load W is to be as small as possible, the fourth box should be placed as far from G as possible without the box overhanging the trailer. These two requirements imply

( 0.3 m ≤

x ≤ 1 m ) (1.8 m ≤ z ≤ 3.7 m )

PROBLEM 3.127 CONTINUED Let x = 0.3 m,

ΣM Gz :

( 200 N )( 0.7 m ) − ( 400 N )( 0.7 m ) − (180 N )( 0.7 m ) + W ( 0.7 m ) = 0 ∴ W = 380 N

ΣM Gx : − ( 200 N )(1.5 m ) − ( 400 N )(1.2 m ) + (180 N )( 0.6 m ) + ( 380 N )( z − 1.8 m ) = 0 ∴ z = 3.5684 m < 3.7 m Let z = 3.7 m,

∴ acceptable

ΣM Gx : − ( 200 N )(1.5 m ) − ( 400 N )(1.2 m ) + (180 N )( 0.6 m ) + W (1.7 m ) = 0 ∴ W = 395.29 N > 380 N

Since the weight W found for x = 0.3 m is less than W found for z = 3.7 m, x = 0.3 m results in the smallest weight W. or W = 380 N at

( 0.3 m, 0, 3.57 m )

PROBLEM 3.128 Solve Problem 3.127 if the students want to place as much weight as possible in the fourth box and that at least one side of the box must coincide with a side of the trailer.

Problem 3.127: A group of students loads a 2 × 4-m flatbed trailer with two 0.6 × 0.6 × 0.6-m boxes and one 0.6 × 0.6 × 1.2-m box. Each of the boxes at the rear of the trailer is positioned so that it is aligned with both the back and a side of the trailer. Determine the smallest load the students should place in a second 0.6 × 0.6 × 1.2-m box and where on the trailer they should secure it, without any part of the box overhanging the sides of the trailer, if each box is uniformly loaded and the line of action of the resultant of the weights of the four boxes is to pass through the point of intersection of the centerlines of the trailer and the axle. (Hint: Keep in mind that the box may be placed either on its side or on its end.)

SOLUTION

For the largest additional weight on the trailer with the box having at least one side coinsiding with the side of the trailer, the box must be as close as possible to point G. For x = 0.6 m, with a small side of the box touching the z-axis, satisfies this condition. Let x = 0.6 m,

ΣM Gz :

( 200 N )( 0.7 m ) − ( 400 N )( 0.7 m ) − (180 N )( 0.7 m ) + W ( 0.4 m ) = 0 ∴ W = 665 N

and

ΣM GX : − ( 200 N )(1.5 m ) − ( 400 N )(1.2 m ) + (180 N )( 0.6 m ) + ( 665 N )( z − 1.8 m ) = 0 ∴ z = 2.8105 m

(2 m <

z < 4 m)

∴ acceptable or W = 665 N at

( 0.6 m, 0, 2.81 m )

PROBLEM 3.129 A block of wood is acted upon by three forces of the same magnitude P and having the directions shown. Replace the three forces with an equivalent wrench and determine (a) the magnitude and direction of the resultant R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xy plane.

SOLUTION

First, reduce the given force system to a force-couple at the origin. Have

ΣF : Pi − Pi − Pk = R ∴ R = − Pk

Have

ΣM O : − P ( 3a ) k − P ( 3a ) j + P ( −ai + 3aj) = M OR ∴ M OR = Pa ( −i − 3k )

Then let vectors ( R, M1 ) represent the components of the wrench, where their directions are the same.

R = − Pk

(a)

or Magnitude of R = P Direction of R : θ x = 90°, θ y = 90°, θ z = −180°

(b) Have

M1 = λ R ⋅ M OR = −k ⋅  Pa ( −i − 3k ) 

= 3Pa and pitch

p=

M1 3Pa = = 3a R P

or p = 3a

PROBLEM 3.129 CONTINUED (c) Have

M OR = M1 + M 2 ∴ M 2 = M OR − M1 = Pa ( −i − 3k ) − ( −3Pak ) = − Pai

Require

M 2 = rQ/O × R − Pai = ( xi + yj) × ( − P ) k = Pxj − Pyi

From

i : − Pa = − Py

or

y =a

j: x = 0

∴ The axis of the wrench is parallel to the z-axis and intersects the xy plane at x = 0, y = a

PROBLEM 3.130 A piece of sheet metal is bent into the shape shown and is acted upon by three forces. Replace the three forces with an equivalent wrench and determine (a) the magnitude and direction of the resultant R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the yz plane.

SOLUTION

First, reduce the given force system to a force-couple system at the origin.

( 2P ) i − ( P ) j + ( P ) j = R

ΣF :

Have

∴ R = ( 2P ) i ΣM O : Σ ( rO × F ) = M OR

Have

M OR

i j k i j k = Pa 2 2 2.5 + 0 0 4 = Pa ( −1.5i + 5j − 6k ) 2 −1 0 0 1 0 R = 2 Pi

(a)

or Magnitude of R = 2 P Direction of R : θ x = 0°, θ y = −90°, θ z = 90°

(b) Have

M1 = λ R ⋅ M OR

λR =

R R

= i ⋅ ( −1.5Pai + 5Paj − 6 Pak ) = −1.5Pa and pitch

p=

M1 −1.5Pa = = −0.75a R 2P

or p = −0.75a

PROBLEM 3.130 CONTINUED M OR = M1 + M 2

(c) Have

∴ M 2 = M OR − M1 = ( 5Pa ) j − ( 6Pa ) k Require

M 2 = rQ/O × R

( 5Pa ) j − ( 6Pa ) k = ( yj + zk ) × ( 2Pi ) = − ( 2Py ) k + ( 2Pz ) j From

i : 5Pa = 2 Pz

∴ z = 2.5a From

k : − 6 Pa = −2 Py

∴ y = 3a ∴ The axis of the wrench is parallel to the x-axis and intersects the yz-plane at y = 3a, z = 2.5a

PROBLEM 3.131 The forces and couples shown are applied to two screws as a piece of sheet metal is fastened to a block of wood. Reduce the forces and the couples to an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz plane.

SOLUTION

First, reduce the given force system to a force-couple at the origin. ΣF : − (10 N ) j − (11 N ) j = R

Have

∴ R = − ( 21 N ) j

ΣM O : Σ ( rO × F ) + ΣM C = M OR

Have

M OR

i j k i j k = 0 0 0.5 N ⋅ m + 0 0 −0.375 N ⋅ m − (12 N ⋅ m ) j 0 −10 0 0 −11 0 = ( 0.875 N ⋅ m ) i − (12 N ⋅ m ) j R = − ( 21 N ) j

(a) (b) Have

M1 = λ R ⋅ M OR

λR =

or R = − ( 21 N ) j

R R

= ( − j) ⋅ ( 0.875 N ⋅ m ) i − (12 N ⋅ m ) j

= 12 N ⋅ m and pitch

p=

and

M1 = − (12 N ⋅ m ) j

M1 12 N ⋅ m = = 0.57143 m R 21 N

or p = 0.571 m

PROBLEM 3.131 CONTINUED M OR = M1 + M 2

(c) Have

∴ M 2 = M OR − M1 = ( 0.875 N ⋅ m ) i M 2 = rQ/O × R

Require ∴

( 0.875 N ⋅ m ) i = ( xi + zk ) × − ( 21 N ) j 0.875i = − ( 21x ) k + ( 21z ) i

From i:

0.875 = 21z ∴ z = 0.041667 m

From k:

0 = −21x ∴ z =0

∴ The axis of the wrench is parallel to the y-axis and intersects the xz-plane at x = 0, z = 41.7 mm

PROBLEM 3.132 The forces and couples shown are applied to two screws as a piece of sheet metal is fastened to a block of wood. Reduce the forces and the couples to an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz plane.

SOLUTION First, reduce the given force system to a force-couple system. Have Have

ΣF : − ( 6 lb ) i − ( 4.5 lb ) j = R

R = 7.5 lb

ΣM O : ∑ ( rO × F ) + ∑ M C = M OR M OR = −6 lb ( 8 in.) j − (160 lb ⋅ in.) i − ( 72 lb ⋅ in.) j

= − (160 lb ⋅ in.) i − (120 lb ⋅ in.) j M OR = 200 lb ⋅ in. R = − ( 6 lb ) i − ( 4.5 lb ) j

(a) (b) Have

M1 = λ R ⋅ M OR

λ =

R R

= ( −0.8i − 0.6 j) ⋅  − (160 lb ⋅ in.) i − (120 lb ⋅ in.) j

= 200 lb ⋅ in. M1 = 200 lb ⋅ in. ( −0.8i − 0.6j)

and Pitch

p=

M1 200 lb ⋅ in. = = 26.667 in. R 7.50 lb or p = 26.7 in.

(c) From above note that M1 = M OR

Therefore, the axis of the wrench goes through the origin. The line of action of the wrench lies in the xy plane with a slope of dy 3 = dx 4

PROBLEM 3.133 Two bolts A and B are tightened by applying the forces and couple shown. Replace the two wrenches with a single equivalent wrench and determine (a) the resultant R, (b) the pitch of the single equivalent wrench, (c) the point where the axis of the wrench intersects the xz plane.

SOLUTION

First, reduce the given force system to a force-couple at the origin. Have

ΣF : − ( 20 lb ) k − ( 21 lb ) j = − ( 21 lb ) j − ( 20 lb ) k = R

R = 29 lb

ΣM O : ∑ ( rO × F ) + ∑ M C = M OR

and

i j k i j k 20 lb ( 4 in.) 4 3 0 + 21 lb ( 4 in.) 6 0 1 + ( −300 j − 320k ) lb ⋅ in. = M OR 0 0 −1 0 −1 0

∴ M OR = − (156 lb ⋅ in.) i + ( 20 lb ⋅ in.) j − ( 824 lb ⋅ in.) k R = − ( 21 lb ) j − ( 20 lb ) k

(a) (b) Have

M1 = λ R ⋅ M OR

=−

λR =

R R

−21j − 20k ⋅  − (156 lb ⋅ in.) i + ( 20 lb ⋅ in.) j − ( 824 lb ⋅ in.) k  29

= 553.80 lb ⋅ in.

PROBLEM 3.133 CONTINUED M1 = M1λ R = − ( 401.03 lb ⋅ in.) j − ( 381.93 lb ⋅ in.) k

and Then pitch

p=

M1 553.80 lb ⋅ in. = = 19.0964 in. R 29 lb

or p = 19.10 in.

M OR = M1 + M 2

(c) Have

∴ M 2 = M OR − M1 = ( −156i + 20 j − 824k ) − ( −401.03j − 381.93k )  lb ⋅ in.

= − (156.0 lb ⋅ in.) i + ( 421.03 lb ⋅ in.) j − ( 442.07 lb ⋅ in.) k M 2 = rQ/O × R

Require

( −156i + 421.03j − 442.07k ) = ( xi + zk ) × ( −21j − 20k ) = ( 21z ) i + ( 20 x ) j − ( 21x ) k From i:

−156 = 21z ∴ z = −7.4286 in.

or From k:

z = −7.43 in.

−442.07 = −21x ∴ x = 21.051 in.

or

x = 21.1 in.

∴ The axis of the wrench intersects the xz-plane at x = 21.1 in., z = −7.43 in.

PROBLEM 3.134 Two bolts A and B are tightened by applying the forces and couple shown. Replace the two wrenches with a single equivalent wrench and determine (a) the resultant R, (b) the pitch of the single equivalent wrench, (c) the point where the axis of the wrench intersects the xz plane.

SOLUTION

First reduce the given force system to a force-couple at the origin at B. 15   8 ΣF : − ( 79.2 lb ) k − ( 51 lb )  i + j = R 17   17

(a) Have

∴ R = − ( 24.0 lb ) i − ( 45.0 lb ) j − ( 79.2 lb ) k

R = 94.2 lb

and

ΣM B : rA/B × FA + M A + M B = M RB

Have

M RB

i j k 15   8 0 − 660k − 714  i + = 0 −20 j  = 1584i − 660k − 42 ( 8i + 15 j) 17   17 0 0 −79.2 ∴ M RB = (1248 lb ⋅ in.) i − ( 630 lb ⋅ in.) j − ( 660 lb ⋅ in.) k

(b) Have

M1 = λ R ⋅ M OR =

λR =

R R

−24.0i − 45.0 j − 79.2k ⋅ (1248 lb ⋅ in.) i − ( 630 lb ⋅ in.) j − ( 660 lb ⋅ in.) k  94.2

= 537.89 lb ⋅ in.

PROBLEM 3.134 CONTINUED M1 = M1λ R

and

= − (137.044 lb ⋅ in.) i − ( 256.96 lb ⋅ in.) j − ( 452.24 lb ⋅ in.) k Then pitch

p=

M1 537.89 lb ⋅ in. = = 5.7101 in. R 94.2 lb

or p = 5.71 in.

M RB = M1 + M 2

(c) Have

∴ M 2 = M RB − M1 = (1248i − 630 j − 660k ) − ( −137.044i − 256.96 j − 452.24k ) = (1385.04 lb ⋅ in.) i − ( 373.04 lb ⋅ in.) j − ( 207.76 lb ⋅ in.) k M 2 = rQ/B × R

Require

i j k 1385.04i − 373.04 j − 207.76k = x 0 z −24 −45 −79.2 = ( 45 z ) i − ( 24 z ) j + ( 79.2 x ) j − ( 45 x ) k From i: From k:

1385.04 = 45 z

−207.76 = −45x

∴ z = 30.779 in. ∴ x = 4.6169 in. ∴ The axis of the wrench intersects the xz-plane at x = 4.62 in., z = 30.8 in.

PROBLEM 3.135 A flagpole is guyed by three cables. If the tensions in the cables have the same magnitude P, replace the forces exerted on the pole with an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz plane.

SOLUTION

(a) First reduce the given force system to a force-couple at the origin.

ΣF : Pλ BA + Pλ DC + Pλ DE = R

Have

 4 3  3 4   −9 4 12   R = P  j − k  +  i − j  +  i − j + k 5 5 5 5 25 5 25        ∴ R =

R=

3P 25

( 2 )2 + ( 20 )2 + (1)2

=

3P ( 2i − 20 j − k ) 25

27 5 P 25

ΣM : Σ ( rO × P ) = M OR

Have

−4 P 3P  4P  4P 12 P   3P  −9 P j− k  + ( 20a ) j ×  i− j  + ( 20a ) j ×  i− j+ k  = M OR 5  5  5 25   5  5  25

( 24a ) j × 

∴ M OR =

M1 = λ R ⋅ M OR

(b) Have where

24 Pa ( −i − k ) 5

λR =

3P 25 1 R = = ( 2i − 20 j − k ) ( 2i − 20 j − k ) R 25 27 5 P 9 5

PROBLEM 3.135 CONTINUED Then

M1 =

and pitch

p=

M1 = M 1λ R =

(c)

Then

1 9 5

( 2i − 20 j − k ) ⋅

24 Pa −8Pa ( −i − k ) = 5 15 5

M1 −8Pa  25  −8a =  = R 81 15 5  27 5 P 

or p = −0.0988a

−8Pa  1  8Pa ( −2i + 20 j + k )   ( 2i − 20 j − k ) = 675 15 5  9 5 

M 2 = M OR − M1 =

24Pa 8Pa 8Pa ( −i − k ) − ( −2i + 20 j + k ) = ( −403i − 20 j − 406k ) 5 675 675 M 2 = rQ/O × R

Require

 8Pa   3P    ( −403i − 20 j − 406k ) = ( xi + zk ) ×   ( 2i − 20 j − k ) 675    25   3P  =   20 zi + ( x + 2 z ) j − 20 xk   25  From i:

8 ( −403)

Pa  3P  = 20 z   675  25 

∴ z = −1.99012a

From k:

8 ( −406 )

Pa  3P  = −20 x   675  25 

∴ x = 2.0049a

∴ The axis of the wrench intersects the xz-plane at x = 2.00a, z = −1.990a

PROBLEM 3.136 Determine whether the force-and-couple system shown can be reduced to a single equivalent force R. If it can, determine R and the point where the line of action of R intersects the yz plane. If it cannot be so reduced, replace the given system with an equivalent wrench and determine its resultant, its pitch, and the point where its axis intersects the yz plane.

SOLUTION

First, reduce the given force system to a force-couple at D. Have

ΣF : FDA + FED = FDAλ DA + FEDλ ED = R

where

 − ( 0.300 m ) i + ( 0.225 m ) j + ( 0.200 m ) k  FDA = 136 N   0.425 m  

= − ( 96 N ) i + ( 72 N ) j + ( 64 N ) k  − ( 0.150 m ) i − ( 0.200 m ) k  FED = 120 N   = − ( 72 N ) i − ( 96 N ) k 0.250 m  

∴ R = − (168 N ) i + ( 72 N ) j − ( 32 N ) k

Have or

R ΣM D : M A = M D

 − ( 0.150 m ) i − ( 0.150 m ) j + ( 0.450 m ) k  16 N ⋅ m M RD = (16 N ⋅ m )  ( −i − j + 3k )  = 0.150 11 m 11  

PROBLEM 3.136 CONTINUED The force-couple at D can be replaced by a single force if R is perpendicular to M RD . To be perpendicular, R ⋅ M RD = 0. Have

R ⋅ M RD = ( −168i + 72 j − 32k ) ⋅ =

16 ( −i − j + 3k ) 11

128 ( 21 − 9 − 12 ) 11

=0

∴ Force-couple can be reduced to a single equivalent force. To determine the coordinates where the equivalent single force intersects the yz-plane, M RD = rQ/D × R where

rQ/D = ( 0 − 0.300 ) m  i + ( y − 0.075 ) m  j + ( z − 0 ) m  k

i 16 N ⋅ m ∴ ( −i − j + 3k ) = (8 N ) −0.3 11 −21

j k ( y − 0.075) z m 9 −4

or 16 N ⋅ m ( −i − j + 3k ) = (8 N ) −4 ( y − 0.075) − 9 z  i + ( −21z − 1.2 ) j + −2.7 + 21 ( y − 0.075) k m 11

{

From j:

From k:

−16 = 8 ( −21z − 1.2 ) 11 48 = 8  −2.7 + 21( y − 0.075 )  11

}

∴ z = −0.028427 m = −28.4 mm ∴ y = 0.28972 m = 290 mm

∴ line of action of R intersects the yz-plane at

y = 290 mm, z = −28.4 mm

PROBLEM 3.137 Determine whether the force-and-couple system shown can be reduced to a single equivalent force R. If it can, determine R and the point where the line of action of R intersects the yz plane. If it cannot be so reduced, replace the given system with an equivalent wrench and determine its resultant, its pitch, and the point where its axis intersects the yz plane.

SOLUTION

First, reduce the given force system to a force-couple at the origin. ΣF : FA + FG = R

Have

 ( 4 in.) i + ( 6 in.) j − (12 in.) k  ∴ R = (10 lb ) k + 14 lb   = ( 4 lb ) i + ( 6 lb ) j − ( 2 lb ) k 14 in.   R=

and Have

56 lb

ΣM O : ∑ ( rO × F ) + ∑ M C = M OR

{

}

M OR = (12 in.) j × (10 lb ) k  + (16 in.) i × ( 4 lb ) i + ( 6 lb ) j − (12 lb ) k 

 (16 in.) i − (12 in.) j   ( 4 in.) i − (12 in.) j + ( 6 in.) k  + ( 84 lb ⋅ in.)   + ( 120 lb ⋅ in.)   20 in. 14 in.    

∴ M 0R = ( 221.49 lb ⋅ in.) i + ( 38.743 lb ⋅ in.) j + (147.429 lb ⋅ in.) k = (18.4572 lb ⋅ ft ) i + ( 3.2286 lb ⋅ ft ) j + (12.2858 lb ⋅ ft ) k

PROBLEM 3.137 CONTINUED The force-couple at O can be replaced by a single force if the direction of R is perpendicular to M OR . To be perpendicular R ⋅ M OR = 0 Have

R ⋅ M OR = ( 4i + 6 j − 2k ) ⋅ (18.4572i + 3.2286 j + 12.2858k ) = 0? = 73.829 + 19.3716 − 24.572 ≠0

∴ System cannot be reduced to a single equivalent force. To reduce to an equivalent wrench, the moment component along the line of action of P is found. M1 = λ R ⋅ M OR

λR =

R R

 ( 4i + 6 j − 2k )  =  ⋅ (18.4572i + 3.2286 j + 12.2858k ) 56  

= 9.1709 lb ⋅ ft M1 = M1λ R = ( 9.1709 lb ⋅ ft )( 0.53452i + 0.80178 j − 0.26726k )

and And pitch

p=

M1 9.1709 lb ⋅ ft = = 1.22551 ft R 56 lb

or p = 1.226 ft Have M 2 = M OR − M1 = (18.4572i + 3.2286 j + 12.2858k ) − ( 9.1709 )( 0.53452i + 0.80178 j − 0.26726k ) = (13.5552 lb ⋅ ft ) i − ( 4.1244 lb ⋅ ft ) j + (14.7368 lb ⋅ ft ) k Require

M 2 = rQ/O × R

(13.5552i − 4.1244 j + 14.7368k ) = ( yj + zk ) × ( 4i + 6 j − 2k ) = − ( 2 y + 6z ) i + ( 4z ) j − ( 4 y ) k From j:

−4.1244 = 4z

From k:

14.7368 = −4 y

or or

z = −1.0311 ft y = −3.6842 ft

∴ line of action of the wrench intersects the yz plane at y = −3.68 ft, z = 1.031 ft

PROBLEM 3.138 Replace the wrench shown with an equivalent system consisting of two forces perpendicular to the y axis and applied respectively at A and B.

SOLUTION Express the forces at A and B as A = Axi + Az k B = Bxi + Bzk Then, for equivalence to the given force system ΣFx : Ax + Bx = 0

(1)

ΣFz : Az + Bz = R

(2)

ΣM x : Az ( a ) + Bz ( a + b ) = 0

(3)

ΣM z : − Ax ( a ) − Bx ( a + b ) = M

(4)

Bx = − Ax

From Equation (1), Substitute into Equation (4)

− Ax ( a ) + Ax ( a + b ) = M ∴ Ax = From Equation (2), and Equation (3),

M b

and

Bx = −

M b

Bz = R − Az Az a + ( R − Az )( a + b ) = 0 a  ∴ Az = R 1 +  b 

PROBLEM 3.138 CONTINUED and

a  Bz = R − R 1 +  b  a ∴ Bz = − R b

Then

M A=  b

a    i + R 1 +  k b  

M B = −  b

 a  i −  Rk  b 

PROBLEM 3.139 Show that, in general, a wrench can be replaced with two forces chosen in such a way that one force passes through a given point while the other force lies in a given plane.

SOLUTION

First, choose a coordinate system so that the xy plane coincides with the given plane. Also, position the coordinate system so that the line of action of the wrench passes through the origin as shown in Figure a. Since the orientation of the plane and the components (R, M) of the wrench are known, it follows that the scalar components of R and M are known relative to the shown coordinate system. A force system to be shown as equivalent is illustrated in Figure b. Let A be the force passing through the given point P and B be the force that lies in the given plane. Let b be the x-axis intercept of B. The known components of the wrench can be expressed as R = Rxi + Ry j + Rzk

M = M xi + M y j + M zk

and

while the unknown forces A and B can be expressed as A = Axi + Ay j + Azk

and

B = Bxi + Bzk

Since the position vector of point P is given, it follows that the scalar components (x, y, z) of the position vector rP are also known. Then, for equivalence of the two systems ΣFx : Rx = Ax + Bx

(1)

ΣFy : Ry = Ay

(2)

ΣFz : Rz = Az + Bz

(3)

ΣM x : M x = yAz − zAy

(4)

ΣM y : M y = zAx − xAz − bBz

(5)

ΣM z : M z = xAy − yAx

(6)

PROBLEM 3.139 CONTINUED

(

)

Based on the above six independent equations for the six unknowns Ax , Ay , Az , Bx , Bz , b , there exists a unique solution for A and B. Ay = Ry

From Equation (2) Equation (6)

1 Ax =   xRy − M z  y

)

Equation (1)

1 Bx = Rx −   xRy − M z  y

(

)

Equation (4)

1 Az =   M x + zRy  y

)

Equation (3)

1 Bz = Rz −   M x + zRy  y

)

Equation (5)

(

(

(

b=

( xM x + yM y + zM z ) ( M x − yRz + zRy )

PROBLEM 3.140 Show that a wrench can be replaced with two perpendicular forces, one of which is applied at a given point.

SOLUTION

First, observe that it is always possible to construct a line perpendicular to a given line so that the constructed line also passes through a given point. Thus, it is possible to align one of the coordinate axes of a rectangular coordinate system with the axis of the wrench while one of the other axes passes through the given point. See Figures a and b. Have

R = Rj

and

M = Mj

and are known.

The unknown forces A and B can be expressed as A = Axi + Ay j + Azk

and

B = Bxi + By j + Bzk

The distance a is known. It is assumed that force B intersects the xz plane at (x, 0, z). Then for equivalence ∑ Fx : 0 = Ax + Bx

(1)

∑ Fy : R = Ay + By

(2)

∑ Fz : 0 = Az + Bz

(3)

∑ M x : 0 = − zBy

(4)

∑ M y : M = −aAz − xBz + zBx

(5)

∑ M z : 0 = aAy + xBy

(6)

Since A and B are made perpendicular, A⋅B = 0 There are eight unknowns:

or

Ax Bx + Ay By + Az Bz = 0 Ax , Ay , Az , Bx , By , Bz , x, z

But only seven independent equations. Therefore, there exists an infinite number of solutions.

(7)

PROBLEM 3.140 CONTINUED 0 = − zBy

Next consider Equation (4): If By = 0, Equation (7) becomes

Ax Bx + Az Bz = 0 Ax2 + Az2 = 0

Using Equations (1) and (3) this equation becomes

Since the components of A must be real, a nontrivial solution is not possible. Thus, it is required that By ≠ 0, so that from Equation (4), z = 0. To obtain one possible solution, arbitrarily let Ax = 0. (Note: Setting Ay , Az , or Bz equal to zero results in unacceptable solutions.) The defining equations then become. 0 = Bx

Then

(1)′

R = Ay + By

(2)

0 = Az + Bz

(3)

M = −aAz − xBz

(5)′

0 = aAy + xBy

(6)

Ay By + Az Bz = 0

(7)′

Equation (2) can be written

Ay = R − By

Equation (3) can be written

Bz = − Az

Equation (6) can be written

x=−

aAy By

Substituting into Equation (5)′,  R − By  M = −aAz −  −a  ( − Az )  By  

or

Az = −

M By aR

Substituting into Equation (7)′, M  M  By  By  = 0 ( R − By ) By +  − aR  aR 

(8)

PROBLEM 3.140 CONTINUED By =

or

a 2 R3 a R2 + M 2 2

Then from Equations (2), (8), and (3) Ay = R − Az = −

Bz =

a 2 R3 RM 2 = a2R2 + M 2 a2R2 + M 2

 M  a 2 R3 aR 2 M = −  2 2  aR  a R + M 2  a2R2 + M 2

aR 2 M a2R2 + M 2

In summary A=

RM ( Mj − aRk ) a R2 + M 2

B=

aR 2 ( aRj + Mk ) a2R2 + M 2

2

Which shows that it is possible to replace a wrench with two perpendicular forces, one of which is applied at a given point. Lastly, if R > 0 and M > 0, it follows from the equations found for A and B that Ay > 0 and By > 0. From Equation (6), x < 0 (assuming a > 0). Then, as a consequence of letting Ax = 0, force A lies in a plane parallel to the yz plane and to the right of the origin, while force B lies in a plane parallel to the yz plane but to the left of the origin, as shown in the figure below.

PROBLEM 3.141 Show that a wrench can be replaced with two forces, one of which has a prescribed line of action.

SOLUTION

First, choose a rectangular coordinate system where one axis coincides with the axis of the wrench and another axis intersects the prescribed line of action ( AA′ ) . Note that it has been assumed that the line of action of force B intersects the xz plane at point P ( x, 0, z ) . Denoting the known direction of line AA′ by λ A = λxi + λ y j + λzk it follows that force A can be expressed as

(

A = Aλ A = A λxi + λ y j + λz k

)

Force B can be expressed as B = Bxi + By j + Bzk Next, observe that since the axis of the wrench and the prescribed line of action AA′ are known, it follows that the distance a can be determined. In the following solution, it is assumed that a is known. Then, for equivalence

(

ΣFx : 0 = Aλx + Bx

(1)

ΣFy : R = Aλ y + By

(2)

ΣFz : 0 = Aλz + Bz

(3)

ΣM x : 0 = − zBy

(4)

ΣM y : M = −aAλz + zBx − xBz

(5)

ΣM z : 0 = aAλ y + xBy

(6)

)

Since there are six unknowns A, Bx , By , Bz , x, z and six independent equations, it will be possible to obtain a solution.

PROBLEM 3.141 CONTINUED Case 1: Let z = 0 to satisfy Equation (4)

Aλ y = R − By

Now Equation (2)

Bz = − Aλz

Equation (3) x=−

Equation (6)

aAλ y By

 a  = −  R − By  By   

(

)

Substitution into Equation (5)   a   M = −aAλz −  −   R − By ( − Aλz )    By  

(

∴ A=−

1 M 

λz  aR 

)

By

Substitution into Equation (2) R=−

1 M  B λ + By λz  aR  y y

∴ By = Then

A=−

λz aR 2 λz aR − λ y M

MR R = aR λz aR − λ y M λy − λz M

Bx = − Aλx =

λx MR λz aR − λ y M

Bz = − Aλz =

λz MR λz aR − λ y M

In summary A=

B=

and

P λA aR λy − λz M

R ( λ Mi + λz aRj + λz Mk ) λz aR − λ y M x

   λz aR − λ y M R  x = a 1 −  = a 1 − R   By  λz aR 2   

    

or x = Note that for this case, the lines of action of both A and B intersect the x axis.

λy M λz R

PROBLEM 3.141 CONTINUED Case 2: Let By = 0 to satisfy Equation (4) A=

Now Equation (2)

R

λy

Equation (1)

λ Bx = − R  x  λy 

   

Equation (3)

λ Bz = − R  z  λy 

   

aAλ y = 0

Equation (6)

which requires a = 0

Substitution into Equation (5)  λ M = z −R  x   λ y

  λ    − x −R  z     λ y   

or

This last expression is the equation for the line of action of force B. In summary  R A=  λy   R B=  λy 

 λ A  

  ( −λ x i − λ z k )  

Assuming that λx , λ y , λz > 0, the equivalent force system is as shown below.

Note that the component of A in the xz plane is parallel to B.

M 

λz x − λx z =   λ y  R

PROBLEM 3.142 A worker tries to move a rock by applying a 360-N force to a steel bar as shown. (a) Replace that force with an equivalent force-couple system at D. (b) Two workers attempt to move the same rock by applying a vertical force at A and another force at D. Determine these two forces if they are to be equivalent to the single force of part a.

SOLUTION

(a) Have

ΣF : 360 N ( − sin 40°i − cos 40° j) = − ( 231.40 N ) i − ( 275.78 N ) j = F

or F = 360 N ΣM D : rB/D × R = M

Have where

50°

rB/D = − ( 0.65 m ) cos 30°  i + ( 0.65 m ) sin 30°  j

= − ( 0.56292 m ) i + ( 0.32500 m ) j i j k ∴ M = −0.56292 0.32500 0 N ⋅ m = (155.240 + 75.206 ) N ⋅ m  k −231.40 −275.78 0 = ( 230.45 N ⋅ m ) k (b) Have where

or M = 230 N ⋅ m ΣM D : M = rA/D × FA

rA/D = − (1.05 m ) cos 30° i + (1.05 m ) sin 30°  j

= − ( 0.90933 m ) i + ( 0.52500 m ) j

PROBLEM 3.142 CONTINUED i j k ∴ FA −0.90933 0.52500 0 N ⋅ m = [ 230.45 N ⋅ m ] k 0 −1 0

( 0.90933FA ) k

or

= 230.45k

∴ FA = 253.42 N

or FA = 253 N

ΣF : F = FA + FD

Have

− ( 231.40 N ) i − ( 275.78 N ) j = − ( 253.42 N ) j + FD ( − cosθ i − sin θ j) From

i : 231.40 N = FD cosθ

(1)

j: 22.36 N = FD sin θ

(2)

Equation (2) divided by Equation (1) tan θ = 0.096629 ∴ θ = 5.5193°

or

θ = 5.52°

Substitution into Equation (1)

FD =

231.40 = 232.48 N cos5.5193° or FD = 232 N

5.52°

PROBLEM 3.143 A worker tries to move a rock by applying a 360-N force to a steel bar as shown. If two workers attempt to move the same rock by applying a force at A and a parallel force at C, determine these two forces so that they will be equivalent to the single 360-N force shown in the figure.

SOLUTION

ΣF : R = FA + FC

Have

− ( 360 N ) sin 40° i − ( 360 N ) cos 40° j = − ( FA + FC ) sin θ  i − ( FA + FC ) cosθ  j

From

i:

( 360 N ) sin 40° = ( FA + FC ) sin θ

(1)

j:

( 360 N ) cos 40° = ( FA + FC ) cosθ

(2)

Dividing Equation (1) by Equation (2), tan 40° = tan θ ∴ θ = 40° Substituting θ = 40° into Equation (1), FA + FC = 360 N Have where

ΣM C : rB/C × R = rA/C × FA

rB/C = ( 0.35 m )( −cos30°i + sin 30° j) = − ( 0.30311 m ) i + ( 0.175 m ) j

(3)

PROBLEM 3.143 CONTINUED R = ( 360 N )( −sin40°i − cos 40° j) = − ( 231.40 N ) i − ( 275.78 N ) j rA/C = ( 0.75 m )( −cos30°i + sin 30 j) = − ( 0.64952 m ) i + ( 0.375 m ) j FA = FA ( − sin 40°i − cos 40° j) = FA ( −0.64279i − 0.76604 j) ∴

i j k i j k −0.30311 0.175 0 N ⋅ m = FA −0.64952 0.375 0 N ⋅ m −231.40 −275.78 0 −0.64279 −0.76604 0 83.592 + 40.495 = ( 0.49756 + 0.24105 ) FA ∴ FA = 168.002 N

or

FA = 168.0 N

Substituting into Equation (3), FC = 360 − 168.002 = 191.998 N

or

FC = 192.0 N

or FA = 168.0 N

50°

FC = 192.0 N

50°

PROBLEM 3.144 A force and a couple are applied as shown to the end of a cantilever beam. (a) Replace this system with a single force F applied at point C, and determine the distance d from C to a line drawn through points D and E. (b) Solve part a if the directions of the two 360-N forces are reversed.

SOLUTION (a)

(a) Have

ΣF : F = ( 360 N ) j − ( 360 N ) j − ( 600 N ) k or F = − ( 600 N ) k

and

ΣM D :

( 360 N )( 0.15 m ) = ( 600 N )( d )

∴ d = 0.09 m or d = 90.0 mm below ED F = − ( 600 N ) k

(b) Have from part a (b)

and

ΣM D : − ( 360 N )( 0.15 m ) = − ( 600 N )( d ) ∴ d = 0.09 m or d = 90.0 mm above ED

PROBLEM 3.145 A crate of mass 80 kg is held in the position shown. Determine (a) the moment produced by the weight W of the crate about E, (b) the smallest force applied at B which creates a moment of equal magnitude and opposite sense about E.

SOLUTION

(

)

W = mg = 80 kg 9.81 m/s 2 = 784.8 N

(a) By definition

ΣM E : M E = ( 784.8 N )( 0.25 m )

Have

∴ M E = 196.2 N ⋅ m (b) For the force at B to be the smallest, resulting in a moment ( M E ) about E, the line of action of force FB must be perpendicular to the line connecting E to B. The sense of FB must be such that the force produces a counterclockwise moment about E. Note: Have

d =

( 0.85 m )2 + ( 0.5 m )2

= 0.98615 m

ΣM E : 196.2 N ⋅ m = FB ( 0.98615 m ) ∴ FB = 198.954 N

and

 0.85 m 

θ = tan −1   = 59.534°  0.5 m  or FB = 199.0 N

59.5°

PROBLEM 3.146 A crate of mass 80 kg is held in the position shown. Determine (a) the moment produced by the weight W of the crate about E, (b) the smallest force applied at A which creates a moment of equal magnitude and opposite sense about E, (c) the magnitude, sense, and point of application on the bottom of the crate of the smallest vertical force which creates a moment of equal magnitude and opposite sense about E.

SOLUTION

(

)

W = mg = 80 kg 9.81 m/s 2 = 784.8 N

(a) By definition

ΣM E : M E = ( 784.8 N )( 0.25 m )

Have

∴ M E = 196.2 N ⋅ m (b) For the force at A to be the smallest, resulting in a moment about E, the line of action of force FA must be perpendicular to the line connecting E to A. The sense of FA must be such that the force produces a counterclockwise moment about E. Note: Have

d =

( 0.35 m )2 + ( 0.5 m )2

= 0.61033 m

ΣM E : 196.2 N ⋅ m = FA ( 0.61033 m ) ∴ FA = 321.47 N

and

 0.35 m 

θ = tan −1   = 34.992°  0.5 m  or FA = 321 N

35.0°

(c) The smallest force acting on the bottom of the crate resulting in a moment about E will be located at the point on the bottom of the crate farthest from E and acting perpendicular to line CED. The sense of the force will be such as to produce a counterclockwise moment about E. A force acting vertically upward at D satisfies these conditions.

PROBLEM 3.146 CONTINUED Have

ΣM E : M E = rD/E × FD

(196.2 N ⋅ m ) k = ( 0.85 m ) i × ( FD ) j (196.2 N ⋅ m ) k = ( 0.85FD ) k ∴ FD = 230.82 N or FD = 231 N

PROBLEM 3.147 A farmer uses cables and winch pullers B and E to plumb one side of a small barn. Knowing that the sum of the moments about the x axis of the forces exerted by the cables on the barn at points A and D is equal to 4728 lb ⋅ ft, determine the magnitude of TDE when TAB = 255 lb.

SOLUTION The moment about the x-axis due to the two cable forces can be found using the z-components of each force acting at their intersection with the xy-plane (A and D). The x-components of the forces are parallel to the xaxis, and the y-components of the forces intersect the x-axis. Therefore, neither the x or y components produce a moment about the x-axis. ΣM x :

Have

(TAB ) z

where

(TAB ) z ( y A ) + (TDE ) z ( yD ) = M x

= k ⋅ TAB = k ⋅ (TABλ AB )   −i − 12 j + 12k   = k ⋅  255 lb    = 180 lb 17   

(TDE ) z

= k ⋅ TDE = k ⋅ (TDE λ DE )   1.5i − 14 j + 12k   = k ⋅ TDE    = 0.64865TDE 18.5   

y A = 12 ft yD = 14 ft M x = 4728 lb ⋅ ft ∴ and

(180 lb )(12 ft ) + ( 0.64865TDE )(14 ft ) = 4728 lb ⋅ ft TDE = 282.79 lb or TDE = 283 lb

PROBLEM 3.148 Solve Problem 3.147 when the tension in cable AB is 306 lb. Problem 3.147: A farmer uses cables and winch pullers B and E to plumb one side of a small barn. Knowing that the sum of the moments about the x axis of the forces exerted by the cables on the barn at points A and D is equal to 4728 lb ⋅ ft, determine the magnitude of TDE when TAB = 255 lb.

SOLUTION The moment about the x-axis due to the two cable forces can be found using the z components of each force acting at the intersection with the xy plane (A and D). The x components of the forces are parallel to the x axis, and the y components of the forces intersect the x axis. Therefore, neither the x or y components produce a moment about the x axis. ΣM x :

Have

(TAB ) z

where

(TAB ) z ( y A ) + (TDE ) z ( yD ) = M x

= k ⋅ TAB = k ⋅ (TABλ AB )   −i − 12 j + 12k   = k ⋅ 306 lb    = 216 lb 17   

(TDE ) z

= k ⋅ TDE = k ⋅ (TDE λ DE )   1.5i − 14 j + 12k   = k ⋅ TDE    = 0.64865TDE 18.5   

y A = 12 ft yD = 14 ft M x = 4728 lb ⋅ ft ∴ and

( 216 lb )(12 ft ) + ( 0.64865TDE )(14 ft ) = 4728 lb ⋅ ft TDE = 235.21 lb or TDE = 235 lb

PROBLEM 3.149 As an adjustable brace BC is used to bring a wall into plumb, the forcecouple system shown is exerted on the wall. Replace this force-couple system with an equivalent force-couple system at A knowing that R = 21.2 lb and M = 13.25 lb ⋅ ft.

SOLUTION ΣF : R = R A = Rλ BC

Have λ BC =

where

∴ RA =

( 42 in.) i − ( 96 in.) j − (16 in.) k 106 in. 21.2 lb ( 42i − 96 j − 16k ) 106

or R A = ( 8.40 lb ) i − (19.20 lb ) j − ( 3.20 lb ) k ΣM A : rC/ A × R + M = M A

Have where

rC/ A = ( 42 in.) i + ( 48 in.) k =

1 ( 42i + 48k ) ft 12

= ( 3.5 ft ) i + ( 4.0 ft ) k R = ( 8.40 lb ) i − (19.20 lb ) j − ( 3.20 lb ) k M = −λ BC M =

−42i + 96 j + 16k (13.25 lb ⋅ ft ) 106

= − ( 5.25 lb ⋅ ft ) i + (12 lb ⋅ ft ) j + ( 2 lb ⋅ ft ) k

PROBLEM 3.149 CONTINUED Then

i j k 3.5 0 4.0 lb ⋅ ft + ( −5.25i + 12 j + 2k ) lb ⋅ ft = M A 8.40 −19.20 −3.20 ∴ M A = ( 71.55 lb ⋅ ft ) i + ( 56.80 lb ⋅ ft ) j − ( 65.20 lb ⋅ ft ) k or M A = ( 71.6 lb ⋅ ft ) i + ( 56.8 lb ⋅ ft ) j − ( 65.2 lb ⋅ ft ) k

PROBLEM 3.150 Two parallel 60-N forces are applied to a lever as shown. Determine the moment of the couple formed by the two forces (a) by resolving each force into horizontal and vertical components and adding the moments of the two resulting couples, (b) by using the perpendicular distance between the two forces, (c) by summing the moments of the two forces about point A.

SOLUTION

(a) Have where

ΣM B : − d1Cx + d 2C y = M

d1 = ( 0.360 m ) sin 55° = 0.29489 m d 2 = ( 0.360 m ) cos 55° = 0.20649 m Cx = ( 60 N ) cos 20° = 56.382 N C y = ( 60 N ) sin 20° = 20.521 N ∴ M = − ( 0.29489 m )( 56.382 N ) k + ( 0.20649 m )( 20.521 N ) k = − (12.3893 N ⋅ m ) k or M = 12.39 N ⋅ m

(b) Have

M = Fd ( −k ) = 60 N ( 0.360 m ) sin ( 55° − 20° )  ( −k )

= − (12.3893 N ⋅ m ) k or M = 12.39 N ⋅ m

PROBLEM 3.150 CONTINUED (c) Have

ΣM A : Σ ( rA × F ) = rB/ A × FB + rC/ A × FC = M

i j k i j k ∴ M = ( 0.520 m )( 60 N ) cos 55° sin 55° 0 + ( 0.880 m )( 60 N ) cos 55° sin 55° 0 − cos 20° − sin 20° 0 cos 20° sin 20° 0 = (17.8956 N ⋅ m − 30.285 N ⋅ m ) k = − (12.3892 N ⋅ m ) k or M = 12.39 N ⋅ m

PROBLEM 3.151 A 32-lb motor is mounted on the floor. Find the resultant of the weight and the forces exerted on the belt, and determine where the line of action of the resultant intersects the floor.

SOLUTION

ΣF :

Have

( 60 lb ) i − ( 32 lb ) j + (140 lb )( cos 30°i + sin 30°j) = R ∴ R = (181.244 lb ) i + ( 38.0 lb ) j or R = 185.2 lb

11.84°

ΣM O : ΣM O = xRy

Have ∴

− (140 lb ) cos 30° ( 4 + 2 cos 30° ) in. − (140 lb ) sin 30° ( 2 in.) sin 30° 

− ( 60 lb )( 2 in.) = x ( 38.0 lb )

x= and

1 ( −694.97 − 70.0 − 120 ) in. 38.0

x = −23.289 in.

Or, resultant intersects the base (x axis) 23.3 in. to the left of the vertical centerline (y axis) of the motor.

PROBLEM 3.152 To loosen a frozen valve, a force F of magnitude 70 lb is applied to the handle of the valve. Knowing that θ = 25°, M x = −61 lb ⋅ ft, and

M z = −43 lb ⋅ ft, determine θ and d.

SOLUTION Have

ΣM O : rA/O × F = M O

where

rA/O = − ( 4 in.) i + (11 in.) j − ( d ) k F = F ( cosθ cos φ i − sin θ j + cosθ sin φ k ) F = 70 lb, θ = 25°

For

F = ( 70 lb ) ( 0.90631cos φ ) i − 0.42262 j + ( 0.90631sin φ ) k 

∴ MO

i j k = ( 70 lb ) −4 11 −d in. −0.90631cos φ −0.42262 0.90631sin φ = ( 70 lb ) ( 9.9694sin φ − 0.42262d ) i + ( −0.90631d cos φ + 3.6252sin φ ) j + (1.69048 − 9.9694 cos φ ) k  in.

and

M x = ( 70 lb )( 9.9694sin φ − 0.42262d ) in. = − ( 61 lb ⋅ ft )(12 in./ft )

(1)

M y = ( 70 lb )( −0.90631d cos φ + 3.6252sin φ ) in.

(2)

M z = ( 70 lb )(1.69048 − 9.9694cos φ ) in. = −43 lb ⋅ ft (12 in./ft )

(3)

PROBLEM 3.152 CONTINUED From Equation (3)

 634.33 

φ = cos −1   = 24.636°  697.86  or φ = 24.6° From Equation (1)  1022.90  d =  = 34.577 in.  29.583  or d = 34.6 in.

PROBLEM 3.153 When a force F is applied to the handle of the valve shown, its moments about the x and z axes are, respectively, M x = −77 lb ⋅ ft and M z = −81 lb ⋅ ft. For d = 27 in., determine the moment M y of F about the y axis.

SOLUTION Have

ΣM O : rA/O × F = M O

where

rA/O = − ( 4 in.) i + (11 in.) j − ( 27 in.) k F = F ( cosθ cos φ i − sin θ j + cosθ sin φ k ) ∴ MO

i j k = F −4 11 −27 lb ⋅ in. cosθ cos φ − sin θ cosθ sin φ = F (11cosθ sin φ − 27sin θ ) i + ( −27 cosθ cos φ + 4cosθ sin φ ) j + ( 4sin θ − 11cosθ cos φ ) k  ( lb ⋅ in.)

and

M x = F (11cosθ sin φ − 27sin θ )( lb ⋅ in.)

(1)

M y = F ( −27 cosθ cos φ + 4cosθ sin φ )( lb ⋅ in.)

(2)

M z = F ( 4sin θ − 11cosθ cos φ )( lb ⋅ in.)

(3)

Now, Equation (1)

cosθ sin φ =

1  Mx  + 27sin θ   11  F 

(4)

and

cosθ cos φ =

1 Mz   4sin θ −  F  11 

(5)

Equation (3)

Substituting Equations (4) and (5) into Equation (2), 1 1 M M      M y = F −27   4sin θ − z   + 4   x + 27sin θ    F     11   11  F 

or

My =

1 ( 27M z + 4M x ) 11

PROBLEM 3.153 CONTINUED Noting that the ratios

27 4 and are the ratios of lengths, have 11 11 My =

27 4 ( −81 lb ⋅ ft ) + ( −77 lb ⋅ ft ) = 226.82 lb ⋅ ft 11 11 or M y = −227 lb ⋅ ft