Statics

TECHNOLOGICAL INSTITUTE OF THE PHILIPINES 938 Aurora Boulevard, Cubao, Quezon City

COMPILATION OF ASSIGNMENTS IN CE001 (Midterm Half)

Submitted by: Coronado, Patrick Nikolai L.

Submitted to: Engr. Jeniffer Camino

CPE31FA1

August 29, 2014

Homework No. 1 (pp. 151-152)

4.1-4.3 Each of the bodies shown is homogeneous and has a mass of 30 kg. Assume friction at all contact surfaces. Draw the fully dimensioned FBD for each body and determine the number of unknowns.

Ans: 4.1) 3 unknowns

4.2) 3 unknowns 4.3) 3 unknowns

a) 3 unknowns b) 3 unknowns c) 3 unknowns d) 4 unknowns 4.4 The homogeneous bar weighs 45 N. It is resting on friction surfaces at A and B. Draw the FBD of the bar and determine the number of unknowns.

Ans: 4 unknowns

4.5 The homogeneous beam AB weighs 2000 N. For each support condition shown in (a) through (d), draw the FBD of the beam and determine the number of unknowns.

Ans:

4.6 The homogeneous triangular plate has a mass of 12 Kg. Draw the FBD of the plate for each set of supports shown in (a)-(d) and determine the number of unknowns. Ans: a) 3 unknowns b) 4 unknowns c) 3 unknowns d) 3 unknowns

4.7 The Bracket of negligible weight is supported by a pin at A and a frictionless peg at B, which can slide in the slot in the bracket. Draw the FBD of the bracket if (a) θ =45 ° ; and (b)

θ

= 90 ° . What are the

unknowns?

4.8 To open the high-pressure water cock, a 60-N horizontal force must be applied to handle at A. Draw the FBD of the handle, neglecting its weight. Count the unknowns.

4.9 The high-pressure water cock is rigidly attached to the support at D. Neglecting the weight of the members, draw the FBD of the entire assembly and count the unknowns. Ans: 6 unknowns

Ans: 3 unknowns

4.10 Draw the FBD of the entire frame, assuming that friction and the weights of the members are negligible. How many unknowns appear on this FBD?

Ans: 4 unknowns

4.11 Draw a FBD of member CE of the frame described in the previous problem. How many unknowns appear on this FBD?

∑Fx= W sin33.7 - Na = 0, Na = 0.56 W

Homework No. 2 (pp. 161-165)

4.13 Calculate the force P that is required to hold the 600-N roller at rest on the rough incline.

4.12 The homogeneous cylinder of weight W rests in a frictionless rightangled corner. Determine the contact forces NA and NB in terms of W and θ .

∑My = 0 = - A (2m) + 600 N (2m) = 0, ∑Mx = - F (2m) + P (2m) = 0 F = P ∑Fx = - F + 600N (sin15) = 0 F = 155.29 P = 155.29 N

Ans: θ

∑Fy= - A + 600N (cos15) = 0

= 33.7 °

A = 579.56N

NA = 0.555 W NB = 0.832 W

4.14 Solve problem 4.13 if the force P pushes rather than pulls. ∑Fy = W cos θ

- Nb = 0,

Ans:179.3 N

∑Fx = Wsin0 – Na= 0 θ = 33.7 ° ∑Fy = W cos33.7 - Nb = 0, Nb = 0.832 W

∑My = 0 = - A (2m) + 600N (2m) = 0, ∑Mx =F(2m)+P(2m)=0 F=-P ∑Fx=-F+600N (sin15) =0 F=155.29

P=-155.29N ∑Fy=-A+600N(cos15)=0 A=579.56N

4.15 The 480-kg bent bar ABC of uniform cross section is supported by a pin at A and a vertical angle cable at C. Determine the pin reactions and the force in the cable. Ans:5754 N mm

ΣMA = 0 CA + 300 mm (4 sin 16°) (300 mm - (600 mm sin 16°)) (4 N) ∑Fx = Ax = 0 ∑Fy = Ay – B – 480 (9.81) + T = 0 ∑Mab = A (2m) – 160 (9.81) (2m) =

-

(720 mm - (600 sin 16°)) (10 N) = 0 CA = 5,753.88 N • mm

0, Ay = 1569.6 N θ

∑Mbc = 320 (9.81) (2m) – T (4m) = 0

4.17 At what

will the lamp in Prob.

T = 1569.6 N

4.16 in equilibrium without the couple CA?

B = 480 kG (9.81) - 1569.6 N - 1569.6 N =0 B = 1569.6 N

4.16 The table lamp consists of two uniform arms, each weighing 4 N, and a 10-N bulb fixture. If θ = 16 ° , calculate the couple CA that must be supplied by the friction in joint A.

∑Fx = Ax = 0 ∑Fy = - 10 N – 4 N + Ay - 0.8 N (sin θ )=0 ∑Ma = 0.8 N (0.3m) + 4 N (0.9m) + 10 N (1.32m) = 0

8.64 N m / 0.6 m = Ay

Fx=0

Ay = 14.4 N

3 B( 5 )-Cx=0

∑Fy = [0.4 N / 0.8 N = sin θ θ

] sin¯¹

= 30 °

4.18 The bent beam ABC is attached to a pin at C and rests against a roller support at B. Neglecting the weight of the beam, find the reactions at B and C caused by the 150-kg load.

5 Cx=(150kg)(9.81m/s²)(5m)( 12.5 3 5

)(

)

Cx=1766N Fy=0 4 B( 5 )-(150kg)(9.81m/s²)+Cy=0 Cy=1471.5N-(150kg)(9.81m/s²)(5m)( 5 4 ¿ )( 12.5 5 Cy=-883N

Ans: RB = 2940 N Cx = 1766N Cy = -883 N

Mc = 0 4 - (150kg) (9.8m/s²) (5m) + B ( 5 ) (2) 3 + B ( 5 )(1.5) = 0 8 4.5 B + B =7350 5 5 RB=2940N

4.19 Compute all reactions at the base A of the traffic light standard, given that the tension in the cable BC is (a) T = 2720 N; and (b) T = 0. The weight of the standard is negligible compared with the 1600-N weight of the traffic light.

4.20 The man is holding up the 35-kg ladder ABC by pushing perpendicular to the ladder. If the maximum force that the man can exert is 400 N, determine the smallest θ at which he can support the ladder.

a. T = 2720 N

Ans:39.0 °

Θ = tan-1 (5/7.5) Θ = 33.69°

ΣMA = 0

ΣFx= 0

35(9.81) N (cos Θ) (3m) = (400

Ax – 2720 N (sin 33.69°) = 0 Ax = 1,508.78 N ΣFy = 0

N) (2m) Θ = 39.04°

Ay – 2720 N (cos 33.69°) – 1600 N = 0 Ay = 3,863.18 N ΣMA = 0 -CA + (4m) (1600 N) – (7.5m) (2720 N sin 33.69°) = 0 CA = -4,915.86 N • m

b. T = 0 N Ax = 0 Ay = 1600 N ΣMA = 0 CA = (4m) (1600 N) CA = 6400 N • m

4.21 The machine part of negligible weight is supported by a pin at A and a roller at C. Determine the magnitudes of the forces acting on the part A and C.

Ans: T = 112.2 N Ax = -38.4 N Ay = 182.4 N

∑MA = 0 3kN (.75m) + 2.4kN. m + Nc sin50 (.144m) – Nc cos50 (.257m) = 0

Tb=Tc ∑Ma=400N(1.5m)+Tb(2.5m)+Tc(3m)

Nc = 84.72 kN

=0

∑MB = 0

(2.5m)Tb+(3m)Tb=-600Nm

AY (.75m) - 84.72 sin50 (0.606m) + 2.4kN.m = 0

T=109.09 counterclockwise

AY = 49.24kN

Ax=-(109.09cos70 ° ),

∑Mc= 0 49.24kN (.144m) + 2.4kN.m (0.606m) - AX(0.257m) = 0

∑Fx=Tbcos70 ° +Ax=0,

+

AX = 44kN

4.22 The uniform plank ABC weighs 400 N. It is supported by a pin at A and a cable that runs the pulley D. Determine the tension in the cable and the components of the pin reaction at A. Note the tension in the cable is constant (see Sample Problem 4.5).

Ax=-37.31N ∑Fy=Tbsin70 ° +Tc+Ay – 400=0, Ay=400 – Tbsin70 ° -Tc, Ay=188.40N

4.23 The center of the gravity of the 850-N man is at G. If the man pulls on the rope with a 380-N force, determine the horizontal distance b between the man’s feet and G.

4 T3 ( 5 ) - T1 = 0 4 T1 = 817.5 N ( 5 ) T1 = 654 N

∑Md = 388 N (1.2m) – 850 (b) = 0 388 N (1.2m) / 850 N = b, b = 0.55m

4.24 The homogeneous 100-kg sign is suspended from three wires. Find the tension in each wire.

4.25 When the truck is empty, it weighs 30000 N and its center of gravity is at G. Determine the total weight W of the logs, knowing that the load on the rear axle is twice the load on the front axle.

Mx = 0 Ans:

- NA (1.44m) +

T1 = 654 N

4 5

w (2.88m) +

30000 (5.64m) – NB (7.44m) = 0

T2 = 490.5 N

NA = 2NB

T3 = 817.5 N

NB (7.44m) + NB (2.88m) = MA = 0 3 - (100) (9.81) N (1m) - T3 ( 5

(2.88) + 30000 N (5.64m) ) (2m)

=0 T3 = 817.5 N

NB =

( 45 ) (2.88 )+ 30000 ( 5.64 ) 10.32

MA = 0

MB = 0

w 5

(1.44m) +

4 5

- (100) (9.81) N (1m) + T2 (2m) = 0

-

T2 = 490.5 N

30000 N (4.2m) – NB (6m) = 0

Fx = 0

4w 5

(1.44m) +

3w 5

(1.44m) + 30000 N (4.2m) -

([ 45w )+30000 N ( 5.64 m ) ] 10.32

108 125

w + 126000 Nm -

6=0 288 215

w -

98372.09 Nm =0 2556 w = 27627.91 5375 W = 58098.6 N

Homework No. 3 (pp. 229-231)

4.148 Show that all diagonal members of the truss carry the same force, and find the magnitude of this force. Ans:1.25 P (T)

∑ F v =0

4000=17880

4 DJ=P 5

BG =13420(C)

∑ M A =0

DJ=1.25 P(T )

4000 ( 4 ) + BG 4.149 Determine the forces members FG and AB in terms of P.

( 4.472 ) ( 8)=0

in

∑ M A =0

BG =4470

∑ F V =0

P (12 ) +FG ( 4 )=0 8000= AB

FG=3 P (C)

∑ M G =0

( 4.474 )+ BG ( 4.474 )

( 4.472 )

AB=17880(T )

\

∑ F h=0

9P=FS (4) 9 AB= (T ) 4

FG=17880

4.150 Determine the forces members BC, BG, and FG.

in

( 4.474 )

4.151 Find the forces in members BC and DE.

Ans: PBC = 13420 N (C) PBG = 4470 N (C) PFG = 16000 N (T)

∑ M E=0

∑ M E=0

6200 ( 9 )=T ( 3.5 )

A y ( 16 )=4000 (12 ) +6000 (8) +8000(4)

T =15942.86 N

A y =8000

∑ M B =0

∑ M A =0

DE

¿ ( ) E 16 =4000 ( 4 )+ 6000¿ 8) + 8000(4)

DE=14 694 (T )

4.5 ( 4.74 )( 2)−6200 ( 4.5)

E=10000 N

∑ F V =0

4.152 Compute the forces in members EF, NF, and NO.

4.153 Repeat Prob. 4.152 assuming that the 300-kN force is applied at O instead of L.

Ans: PEF = 240 kN (C) PNF = 82.0 kN (T) Ry= 75KN

PNO = 187.5 kN (T)

Jx= 0 MN (CLOCKWISE) = 0

Jy= 225KN

MN (CLOCKWISE) = - 75 kN (20m) -

ΣMN= 0

EF (6.25m)

ΣFX = 0

- 1500 kN.m = EF (6.25m)

300KN(5m)-EF(6.25)-Ry(20m) = 0

EF = 240kN(C)

NF(5/8)+EF+NO = 0

MF (CLOCKWISE) = 0

EF = 0 (c)

MF (CLOCKWISE) = NO (6m) -

NO = 180KN (c)

75kN (15m)

ΣFy = 0

NO (6m) = 1125kNm

75KN – NF(6.25/8) – 300= 0

NO = 186.5kN (T)

NF = -288(T) KN

ΣFx = EF – NO – NF (

5 √ 61 )

0 = 240kN - 187.5kN – NF ( 5 - 52.5 kN = - NF ( √ 61 ) NF = 82.0 kN NF = 82.0kN (T)

5 √61 ) 4.154 Determine the forces members BG, CI, and CD.

in

Ans: PBG = 0.250 P (T) PCI = 0.354 P (T) PCD = 0.750 P (C) 4.156 Determine the angle

∑ M E=0

that

maximizes the tensile force in member BC and calculate the maximum value of his force.

4 a )=P (a) Ay¿ A y=

θ

P 4

∑ F v =0 P =BG 4 BG =0.25 P

∑ F v =0 0.25 P=CI

Ans:

( √12 )

(PBC)max = 168.0 kN θ

CI =0.354 P

∑ M D=0 3 a ( 0.25 P )+ 0.354

= 45 °

θ=tan−1 1 =CD ( a ) √2

( )

CD=0.750 P

2 2

θ=45 °

∑ M G =0 42 ( 2 √ 2 ) ( 2 )=BC ( √2)

4.155 Assuming that P = 48000 N and that it may be applied at any joint on the line FJ, determine the location of P that would cause (a) maximum tension in member HI; (b) maximum compression in member CI; and (c) maximum tension in member CI. Also determine the magnitude of the indicated force in each case.

BC =168 N

4.157 Find the forces in members CD, DH, and HI.

4.158 Determine the members CD and DF.

forces

in

4.159 Computer the forces in members CD and JK, given that P = 3000 N and Q = 1000 N. (Hint: Use the section indicated by the dashed line.)

Ans: PCD = 12.75 kN (T) PDF = 7.39 kN (C)

∑ F v =0 5=−FB

( 45 )

ΣMK= 0

FB=6.25 ( C ¿

Dy(2m) – 1000(45) – 1000(3) –

∑ F h=0

1000(45) – 1000(2m)= 0

4=F sin33.69

Dy=7500 N

F=7.211 N

ΣMA= 0

SFY= 0

ΣFy= 0

3 + DF (4/9.85) = 0 DF= -7.3875 kN (C)

-JK(2)-(-4500)(2)= 0 -Dy-Ky+F= 0 JK= 4500KN (T)

∑ F h=0 BC =−6. .25

F – Dy= Ky

( 53 )+22.5

∑ F h=0 CD=−7.211 cos 33.69 .+ 18.75

¿ 12.75 N

ΣFy= 0 Ky= -4500KN (J) DY+CD+JK+KY= 0 Σ Fx= 0 CD= -7500KN (T) -Kx – 1000 – 1000 – 1000= 0 KX= -3000 (T)

4.160 If PCD = 6000 N and PGD = 1000 N (both compression), find P and q.

4.162 Determine the forces members AC, AD and DE.

in

Ans: P = 5170 N Q = 370 N

∑ F v =0 6000=1000 ( sin 56.31 ) + P P=5167.95 N (T ) Ans:

∑ M R =0

PDE = 40 kN (T)

6000 ( 2 )+ 3Q ( 1.5 )=5167.95 (2)

PAD = 0

Q=369.8(C )

4.161 Determine the forces members BC, CE, and FG.

PAC = 50 kN (C) in

∑ M L=0 CD (12)=60(8)

CD=40 (T )

∑ F h=0 40=−AC

( 45 )

AC =50 ∑Mc=0 (1.33)FG(cos12.529)+560(1.5)=0 FG=645.37 N (C) ∑Mh=0 -560(1.5)-(2100)-3(1/1.8)CE=0 CE=1764 N (C) ∑Mf=0 -BC(2)+3(560)+4.5(560)=0 BC=2100 N (T)

4.163 Determine the forces GI, FH and GH.

4.164 Determine the forces in members CD, IJ, and NJ of the K-truss in terms of P.

Ans:

Sin= -(5P)/CA

PCD = 4.0 P (T)

∑fy=0 -2P+CAcos=0 Cos=2P/CA

PIJ = 4.0 P (C) PNJ = 0.559 P (C)

∑ M G =0 A y ( 6 ) ( a )=P ( 5 a ) + P ( 4 a ) + P ( 3 a ) + P ( 2 a )+ P(a)

4.166 Find the forces in members BC and BG.

A y =2.5 P

∑ M I =0 2.5 P ( 2 a ) =CD ( a )+ P CD=4 P (T )

∑ F h=0 4 P+ IJ =P

Ans:

IJ =4 P (C)

PBC = 4.47 P (C)

∑ F v =0

PBG = 2.0 P (C)

1 2 2.5 P+ NJ =0 √5 2

1 MA (clockwise) = BG ( 2 ¿ + P (L)

NJ=5.59(C )

L - PL = BG ( 2 ¿

()

4.165 Determine the largest allowable θ value for the angle if the magnitude of the force in member BC is not exceed 5 P.

BG =

−2 PL L

BG = 2P (C) MB (clockwise) = 1.5 (L) – AG 4 L 1 √17 2 + AG √17 (L)

( )( )

@ JOINT C ∑Fx=0 -5P-CAsin=0

∑Fy/∑Fx= 2/-5 tan^-1=21.8

( )

= 1.5 (L) - AG

( √1717 )

- 1.5L = - AG

( √ 1717 L )

L

6400-(4/√41)BE=0 BE=10245 N (T)

AG = 6.18 P (T)

( √117 )

Fy = 1.5 P – P + AG

+ BC

( √15 ) = 1.5 P – P + 6.18

1 √17

( )

+ BC

1 √5

( ) = 2P + BC

( √15 )

- 2P = BC

( √15 )

4.168 A couple acting on a winch at G slowly raises the load W by means of a rope that runs the pulley attached to the derrick at A and B. Determine the forces in member EF and KL of the derrick, assuming the diameters of the pulleys and the winch are negligible.

BC = 4.47P (C)

4.167 Determine the forces in members BC and BE and the horizontal pin reaction at G.

Ans: PEF = 1.828 W (T) PKL = 2.83 W (C)

+∑Me=0 -KL(L)-4(L)(L/(L√a))W=0 KL=2.83W (C) +∑Mf=0 (16000)(8)+(12000)(16)-GY(16)=0 GY=20000 N +∑Me=0 -(5)BC+(1200)(4)-(20000)(4)=0 BC=6400 N (C) ∑Fh=0

+∑Ml=0 EF(L)-W(L/(L√a))(4L)+W(L)=0 EF=W(4L/√a)-W(L)/L EF=1.828W (T)