HW01 - Statics HW PDF

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PROBLEM 2.9 B

A

30° α

A disabled automobile is pulled by means of two ropes as shown. Knowing that the tension in rope AB is 3 kN, determine by trigonometry the tension in rope AC a nd the value of α so that the resultant force exerted at A is a 4.8-kN force directed along the axis of the automobile.

C

SOLUTION

Using the law of cosines:

TAC 2  (3 kN)2  (4.8 kN)2  2(3 kN)(4.8 kN)cos 30° TAC  2.6643 kN

Using the law of sines:

sinB sin 30n  3 kN 2.6643 kN B  34.3n TA C  2.66 kN

34.3n ◀

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PROBLEM 2.37

80 lb 120 lb

α

60 lb a'

α

Knowing that B  40°, determine the resultant of the three forces shown.

20° a

SOLUTION 60-lb Force:

Fx  (60 lb) cos 20n  56.382 lb Fy  (60 lb)sin 20n  20.521 lb

80-lb Force:

Fx  (80 lb) cos60n  40.000 lb Fy  (80 lb)sin 60n  69.282 lb

120-lb Force:

Fx  (120 lb) cos 30n  103.923 lb Fy  (120 lb)sin 30n   60.000 lb

and

R x  4Fx  200.305 lb R y  4 Fy  29.803 lb R  (200.305 lb)2  (29.803 lb)2  202.510 lb

Further:

tan B 

29.803 200.305

B  tan1  8.46n

29.803 200.305

R  203 lb

8.46n ◀

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PROBLEM 2.41 A

65°

Determine (a) the required tension in cable AC, knowing that the resultant of the three forces exerted at Point C of boom BC must be directed along BC, (b) the corresponding ma gnitude of the resultant.

C

25° 35°

75 lb 50 lb

B

SOLUTION

Using the x and y axes shown:

(a)

Rx  4 Fx  TA C sin10n  (50 lb) cos 35n  (75 lb) cos 60n  TA C sin10n  78.458 lb

(1)

R y  4 F y  (50 lb)sin 35n  (75 lb)sin 60n  T A C cos10n R y  93.631 lb T A C cos10n

(2)

Set R y  0 in Eq. (2): 93.631 lb T A C cos10n  0 TA C  95.075 lb

(b)

T A C  95.1 lb ◀

Substituting for T A C in Eq. (1):

R x  (95.075 lb)sin10n  78.458 lb  94.968 lb R  Rx

R  95.0 lb ◀

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PROBLEM 2.43 A

50°

30°

B

Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC.

C

400 lb

SOLUTION Free-Body Diagram

Force Triangle

Law of sines: T AC T BC 400 lb   sin 60 n sin 40 n sin 80 n

(a)

TAC 

400 lb (sin 60n) sin80 n

TAC  352 lb ◀

(b)

T BC 

400 lb (sin 40n ) sin 80n

T B C  261 lb ◀

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A

PROBLEM 2.48

5°

Knowing that B  20n, determine the tension (a) in cable AC, (b) in rope BC.

C 1200 lb

α

B

SOLUTION Free-Body Diagram

Law of sines:

T AC sin 110n

Force Triangle



T BC sin 5n



1200 lb sin 65n

(a)

T AC 

1200 lb sin 110n sin 65n

(b)

T BC 

1200 lb sin 5 n sin 65 n

T A C  1244 l b ◀

T B C  115.4 lb ◀

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PROBLEM 2.50 A

B

Two cables are tied together at C and are loaded as shown. Determine the range of values of P for which both ca bles remain taut.

30° 30°

200 N

C

45° P

SOLUTION Free-Body Diagram

4Fx  0

 TCA sin 30 n  TCB sin 30 n  P cos 45 n  200 N  0

For T CA  0 we have, 0.5T CB  0.70711P  200  0 (1)

4Fy  0

TCA cos30n  TCB cos30n  P sin 45 n  0 ; again setting TCA  0 yields, 0.86603T CB  0.70711P  0 (2)

Adding equations (1) and (2) gives, 1.36603TCB  200 hence T CB  146.410N and P  179.315N Substituting for T CB  0 into the equilibrium equations and solving simultaneously gives, 0. 5T CA  0. 70711P  200  0 0. 86603TCA  0. 70711P  0

And TCA  546.40N, P  669.20N Thus for both cables to remain taut, load P must be within the range of 179.315 N and 669.20 N. 179.3 N < P < 6 69 N ◀

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PROBLEM 2.55

D A α

B C

β

A sailor is being rescued using a boatswain’s chair that is suspended from a pulley that can roll freely on the support cable ACB and is pulled at a constant speed by cable CD. Knowing that B  30° and C  10° and that the combined weight of the boatswain’s chair and the sailor is 200 lb, determine the tension (a) in the support cable ACB, (b) in the traction cable CD.

SOLUTION Free-Body Diagram

4Fx  0: T A CB cos 10n T A CB cos 30n T CD cos 30n  0 TCD  0.137158TA CB

(1)

4Fy  0: T A CB sin 10n  T A CB sin 30n  T CD sin 30n  200  0 0.67365T A CB  0.5T CD  200

(a)

Substitute (1) into (2):

0.67365TA CB  0.5(0.137158TA CB )  200 T A CB  269 l b ◀

T A CB  269.46 l b

(b)

From (1):

(2)

TCD  0.137158(269.46 lb)

T CD  37.0 lb ◀

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PROBLEM 2.61

C

A

B

A movable bin and its contents have a combined weight of 2.8 kN. Determine the shortest chain sling ACB that can be used to lift the loaded bin if the tension in the chain is not to exceed 5 kN.

0.7 m

1.2 m

SOLUTION Free-Body Diagram

tanB 

h 0.6 m

(1)

Isosceles Force Triangle

Law of sines:

sinB 

1 2

(2.8 kN)

T AC T A C  5 kN

sinB 

1 2

(2.8 kN) 5 kN

B  16.2602n

From Eq. (1): tan16.2602n 

h 0.6 m

= h  0.175000 m

Half-length of chain  AC  (0.6 m) 2  (0.175 m) 2

 0.625 m Total length:

1.250 m ◀

 2 q 0.625 m

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PROBLEM 2.65 A cable loop of length 1.5 m is placed around a crate. Knowing that the mass of the crate is 300 kg, determine the tension in the cable for each of the arrangements shown.

SOLUTION Free-Body Diagram



Isosceles Force Triangle



W  300 kg 9.81 m/s2  2943.0 N 1 1500 mm  400 mm  300 mm  300 mm 2 EB  250 mm • 200 mm -¬ -  36.87 n B  cos 1 žž Ÿž 250 mm --® EB 

TAE

TAE  TBE 1 TAE sin B  2943.0 N  2 1 sin36.87n  2943.0 N 2 TAE  2452.5 N Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 66

(a)

T AE  2450 N ◀

Problem 2.65 (Continued)

Isosceles Force Triangle Free-Body Diagram

1 1500 mm 300 mm  400 mm  400 mm  2 EB  250 mm • 150 mm ¬-  41.41n B  cos 1 žžž Ÿ 200 mm -® EB 

TAE

TAE  TBE 1 TAE sin B   2943.0 N 2 1 sin 41.41n   2943.0 N 2 TAE  2224.7 N (b)

T AE  2220 N ◀

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PROBLEM 2.69 A

B

A load Q is applied to the pulley C, which can roll on the cable ACB. The pulley is held in the position shown by a second cable CAD, which passes over the pulley A and supports a load P. Knowing that P  750 N, determine (a) the tension in cable ACB, (b) the magnitude of load Q.

25° D

55°

C

P Q

SOLUTION Free-Body Diagram: Pulley C

(a)

4Fx  0: T A CB (cos25n cos 55n ) (750 N) cos55°  0

Hence:

T A CB  1292.88 N T A CB  1293 N ◀

(b)

4 Fy  0: TA CB (sin 25n  sin 55 n)  (750 N)sin 55 n  Q  0 (1292.88 N)(sin 25 n  sin 55 n)  (750 N)sin 55 n Q  0

Q  2219.8 N

or

Q  2220 N ◀

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