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• Yani Jane Costelo

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Situation 1: The homogeneous bar AB weighs 400 lb. End B leans against a vertical wall and end A is supported by a balland-socket joint.

Situation 2: From the given figure, the mass of the block is 200 kg and the coefficient of static friction in all contact surfaces is 0.40.

200 kg 12˚ P

1.

2.

3.

Determine the tension in Cable BC. A. 126.4 lb * B. 343.3 lb

C. 134.1 lb D. 829.8 lb

Determine the wall reaction at B. A. 34.02 lb B. 159.5 lb

C. 164.2 lb * D. 257.9 lb

Determine the total reaction at A. A. 390.4 lb * B. 387.0 lb

C. 376.4 lb D. 763.7 lb

4.

Determine the reaction between the wall and the block if it is raised due to P. A. 1794 N C. 1932 N * B. 1036 N D. 1116 N

5.

Determine the reaction between the wedge and the block. A. 2994 N C. 3225 N * B. 1729 N D. 1862 N

6.

Determine the value of the horizontal force P. A. 2866 N * C. 2680 N B. 1655 N D. 2886 N

Solution: 1962 N

0.40 N1 Solution: TBC sin30º

0.40 N2 (sin12º)

N1 TBC

0.40 N2 0.40 N2 (cos12º) N2 (sin12º)

RB 12˚

TBC cos30º

N2 (cos12º)

E

AX

AY

D AZ

ΣM AE = 0] TBC sin30º ( 4 ) + TBC cos30º ( 5 ) – 400 ( 2 ) = 0

TBC = 126.38 lb → answer Calculating for the value of R B :

ΣM AD = 0] (126.38 ) sin30º ( 6 ) + R B ( 5 ) – 400 ( 3) = 0 R B = 164.17 lb → answer

Calculating for the reaction at A

F F F

x

= 0: A x - 126.38 cos30o = 0 → A x = 109.45 lb

y

= 0: 164.17 - A y = 0 → A y = 164.17 lb

z

= 0: 126.38 sin30o - 400 - A z = 0 → A z = 336.81 lb

R A = 109.452 + 164.17 2 + 336.812 R A = 390.35 lb → answer

Using the Figure above: ΣFH = 0]

N1 – 0.40 N 2 ( cos12º ) - N 2 ( sin12º ) = 0

400 lb

Calculating for the value of TBC :

N2

N1 – 0.599 N 2 = 0 → equation 1 ΣFV = 0]

-0.40 N1 + N 2 ( cos12º ) – 0.40 N 2 ( sin12º ) – 1962 = 0 -0.40 N1 + 0.895 N 2 = 1962 → equation 2 Solving the two equations: N1 = 1,793.91 lb N 2 = 2,993.98 lb Calculating for the reaction between the wall and the block: R1 = 1793.912 + 0.4 (1793.91)  R1 = 1,932.10 N → answer

2

Calculating for the reaction between the wedge and the block: R 2 = 2993.982 + 0.4 ( 2993.98 )  R 2 = 3,224.62 N → answer

2

N2 = 2,993.98 N 2,993.98 sin12º 1,197.59 cos12º 0.40 N2 = 1,197.59 N

2,993.98 cos12º

RB

Cx

12˚

P

1,197.59 sin12º

Cy

RD

0.40 N3 N3

Using the free body diagram: Using the Figure above: ΣFV = 0] N 3 + 1197.59 sin12º – 2993.98 cos12º = 0

ΣM D = 0] 300 ( 60 ) + C y ( 250 ) - 480 (150 ) = 0 C y = 216 N

N 3 = 2,679.56 N ΣFH = 0]

0.40 ( 2679.56 ) + 1197.59 cos12º + 2993.98 sin12º - P = 0 P = 2,865.73 N → answer Situation 3: In the frame shown, members ACE and BCD are connected by a pin at C and by the link DE. For the loading shown:

 80  ΣFy = 0] 216 - 480 + R D   =0  170  R D = 561 N → answer  150  ΣFx = 0] - C x + 300 + 561   =0  170  C x = 795 N R C = Cx 2 + C y 2 = 7952 + 2162 = 823.82 N → answer Situation 4: The force system shown consists of the couple C and four forces. If the resultant of this system is a 500 lb-in. counterclockwise couple, determine:

7.

Determine the reaction at B. A. 800 N B. 300 N *

C. 1200 N D. 480 N

8.

Determine the force in link DE. A. 264 N C. 495 N B. 561 N * D. 216 N

9.

Determine the reaction at C. A. 216 N B. 795 N

C. 824 N * D. 495 N

Solution: RAY RAX

10. the value of P, in lb. A. 200 * B. 130

C. 260 D. 50

11. the value of Q, in lb. A. 200 B. 130

C. 260 * D. 50

12. The value of C, in lb-in. A. 1440 * B. 1944

C. 852 D. 180

RB Solution:

Using the given figure: ΣM A = -500]

Using the free body diagram: ΣM A = 0] 480 (100 ) - R B (160 ) = 0 R B = 300 N → answer

3 4 -80 ( 4 ) + 20 ( 3) + C - P   ( 6 ) - P   ( 6 ) = -500 5 5 42 CP = -240 → equation 1 5

ΣM B = -500]

10000 lb

W

 12  5 80 ( 2 ) - 20 ( 3) + C - Q   ( 6 ) - Q   ( 6 ) = -500  13   13  102 CQ = -600 → equation 2 13

B

θB H

To

ΣM O = -500]  12  80 ( 2 ) + 20 ( 3 ) + C - Q   ( 6 ) - P  13  18 72 CPQ = -720 → equation 3 5 13 Solving the three equations: C = 1,440 lb-in → answer Q = 600 lb → answer

3   ( 6 ) = -500 5

x

Using the free body diagram: ΣFx = 0] 10000 cos ( 62.98

) -T

o

=0

To = 4,542.84 lb → answer P = 200 lb → answer

Calculating for the value of x: ΣFy = 0] 10000 sin ( 62.98

Situation 5: The cable shown carries the uniformly distributed load w0 = 80 lb/ft, where the distance is measured along the horizontal. If the cable tension does not exceed 10 000 lb,

) -W=0

W = 8,908.57 lb W = w o ( x ) → 8,908.57 = 80 x x = 111.36 ft

200 -x = 88.64 ft

Calculating for the value of H: ΣM B = 0] 4,542.84 ( H ) - 8,908.57 ( 0.5 ) (111.36 ) = 0 H = 109.19 ft → answer Calculating for the length of the cable: 2

w x 1 +  o  dx = x1  To  S = 280.09 ft → answer S= 

13. Determine the tension at the lowest point. A. 4543 lb * C. 7620 lb B. 8909 lb D. 6476 lb

x2



111.36

-88.64

2

 80 x  1 +   dx  4542.84 

Situation 6: A truss is loaded as shown: `

14. Determine the length of the shortest cable. A. 253.1 ft C. 307.1 ft B. 334.1 ft D. 280.1 ft * 15. Determine the corresponding vertical distance H. A. 68.4 ft C. 109.2 ft * B. 90.8 ft D. 131.6 ft Solution: 10000 lb 80 (200)

θB 16. Determine the reaction at A. A. 2000 lb B. 4000 lb

Using the free body diagram: ΣM A = 0]

10000 cosθ B ( 40 ) - 10000 sinθ B ( 200 ) + 80 ( 200 ) (100 ) = 0 θ B = 62.98o

C. 4472 lb * D. 3236 lb

17. Determine the force in member FI. A. 2500 lb (Compression) * C. 3610 lb (Compression) B. 2500 lb (Tension) D. 3610 lb (Tension) 18. Determine the force in member JC. A. 2500 lb (Compression) C. 3610 lb (Compression) B. 2500 lb (Tension) D. 3610 lb (Tension) *

Solution:

Using the free body diagram: ΣM A = 0] 3000 ( 8 ) + 1000 (12) - PDE (18 ) = 0 PDE = 2000 lb ΣFx = 0] A x - PDE = 0 → A x = PDE → A x = 2000 lb

Situation 7: As shown in the figure, a flexible cable which supports the 100-kg load is passed over a fixed circular drum and subjected to a force P to maintain equilibrium. The coefficient of static friction μ between the cable and the fixed drum is 0.30.

19. If α = 0, determine the maximum value which P may have in order not to raise or lower the load. A. 2518 N C. 612 N B. 1572 N * D. 382 N 20. If α = 0, determine the minimum value which P may have in order not to raise or lower the load. A. 2518 N C. 612 N * B. 1572 N D. 382 N

ΣFy = 0] A y - 3000 - 1000 = 0 → A y = 4000 lb R A = A x 2 + A y 2 = 20002 + 40002 = 4,472.14 lb → answer

21. For P = 500 N, determine the minimum value which the angle α may have before the load begins to slip. A. 52.62º C. 128.72º B. 38.72º * D. 142.62º Solution: Using the principle of belt friction:

T2 = eμsβ ; if α = 0, β = 90o = 0.5π T1 P β

Using section 1: 3 4 ΣM G = 0] -3000 ( 4 ) - PFI   ( 4 ) - PFI   ( 3) = 0 5 5 PFI = -2500 lb = 2500 lb (compression) → answer

100 kg Figure 1

Considering the 100 kg load will go down (Figure 1): (100) (9.81) = e(0.3) (0.5π) → P = 612.37 N P P β

Using section 2:  2  ΣFx = 0] 2000 - PJC   =0  13  PJC = 3610 lb (tension) → answer

100 kg Figure 2

Considering the 100 kg load will go up (Figure 2): P = e(0.3) (0.5π) → P = 1,571.54 N (100) (9.81)

Based from the above calculations: 612.37 N  P  1,571.54 N  Pmin = 612.37 N → answer

Solution:

200 mm

 Pmax = 1,571.54 N → answer

TU

α

α

300 mm P = 500 N

90°

540 N 100 kg

TL

Figure 3

Considering the 100 kg load will go down (Figure 3): (100) (9.81) = e(0.3) β → β = 2.247 radians 500  180o  o β = 2.247 radians   = 128.72 π radians   α = 128.72o - 90o = 38.72o → answer

Calculating for the angle: 300 tanθ = → θ = 56.31o 200

Considering the joint on the right hand:

F

V

a b c

= = =

300 mm

Calculating the force exerted on the right arm: FR = 540 N → answer

Situation 8: The figure below shows the archer aiming to release the arrow. As the archer pulls the cord with his right arm, he applies a force of 540 N with his left arm. Given:

θ θ 540 N

= 0: TU sin(56.31o ) - TL sin(56.31o ) = 0 TU = TL

200 mm 300 mm 300 mm

F

a

but TU = TL :

H

= 0: TU cos(56.31o ) + TL cos(56.31o ) - 540 = 0

TU cos(56.31o ) + TU cos(56.31o ) - 540 = 0 TU = TL = 486.75 N → answer

b

c

22. Calculate the resulting tension in the upper cord, in N. A. 309 C. 270 B. 487 * D. 556 23. Calculate the resulting tension in the lower cord, in N. A. 487 * C. 270 B. 556 D. 309 24. How much force (N) does the archer exert with his right arm? A. 540 * C. 270 B. 259 D. 518