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Unformatted text preview: kuruvila (lk5992) HW 12 opyrchal (11113) 1 This printout should have 11 questions. Multiplechoice 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 freebody 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.
 Fall '08
 moro

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