HW 12-solutions

HW 12-solutions - kuruvila (lk5992) HW 12 opyrchal (11113)...

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Unformatted text preview: kuruvila (lk5992) HW 12 opyrchal (11113) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points The system shown in the figure is in equilib- rium. A 13 kg mass is on the table. A string attached to the knot and the ceiling makes an angle of 69 with the horizontal. The coeffi- cient of the static friction between the 13 kg mass and the surface on which it rests is 0 . 43. 13 kg m 6 9 What is the largest mass m can have and still preserve the equilibrium? The accelera- tion of gravity is 9 . 8 m / s 2 . Correct answer: 14 . 5625 kg. Explanation: Let : M = 13 kg , m = 14 . 5625 kg , and = 69 . For the system to remain in equilibrium, the net forces on both M and m should be zero, so the tension in the rope has an upper bound value T max , where T max cos = M g (1) T max = M g cos = (0 . 43) (13 kg) (9 . 8 m / s 2 ) cos 69 = 152 . 865 N . For m to remain in equilibrium T max sin = m max g (2) m max = T max sin g = (152 . 865 N) sin69 9 . 8 m / s 2 = 14 . 5625 kg . Alternate Solution: Equations 1 and 2 come directly from the free-body diagram for the knot. Dividing Eq. 2 by Eq. 1, tan = m max M m max = M tan = (0 . 43) (13 kg) tan 69 = 14 . 5625 kg . 002 (part 1 of 2) 10.0 points The uniform diving board has a mass of 31 kg . A B . 9 m 3 . 3 m Find the force on the support A when a 75 kg diver stands at the end of the div- ing board. The acceleration of gravity is 9 . 81 m / s 2 . Correct answer: 3 . 10323 kN. Explanation: Let : m = 31 kg , M = 75 kg , = 4 . 2 m , x 1 = 0 . 9 m , and x d = 3 . 3 m . kuruvila (lk5992) HW 12 opyrchal (11113) 2 Consider rotational equilibrium about B : A F A F B B mg M g x 1 x 2 x 3 x d x 2 = 2- x 1 = 4 . 2 m 2- . 9 m = 1 . 2 m Applying rotational equilibrium at point B , summationdisplay vector = F A x 1- mg x 2- M g x d = 0 F A = parenleftbigg x 2 m + x d M x 1 parenrightbigg g = bracketleftbigg (1 . 2 m) (31 kg) + (3 . 3 m) (75 kg) . 9 m bracketrightbigg (9 . 81 m / s 2 ) 1 kN 1000 N = 3 . 10323 kN . The direction of F A is downward, so it is tension. 003 (part 2 of 2) 10.0 points Find the force on the support B at that same instant. Correct answer: 4 . 14309 kN. Explanation: Applying translational equilibrium, summationdisplay F y = F B- F A- mg- M g = 0 F B = F A + ( m + M ) g = 3 . 10323 kN + (31 kg + 75 kg) (9 . 81 m / s 2 ) 1 kN 1000 N = 4 . 14309 kN ....
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This note was uploaded on 12/03/2010 for the course PHYS 111 taught by Professor Moro during the Fall '08 term at NJIT.

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HW 12-solutions - kuruvila (lk5992) HW 12 opyrchal (11113)...

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