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Unformatted text preview: billing (cab4763) – HW 12 – opyrchal – (11109) 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 61 ◦ with the horizontal. The coeffi cient of the static friction between the 13 kg mass and the surface on which it rests is 0 . 32. 13 kg m 6 1 ◦ 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: 7 . 50485 kg. Explanation: Let : M = 13 kg , m = 7 . 50485 kg , and θ = 61 ◦ . 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 θ = μM g (1) T max = μM g cos θ = (0 . 32) (13 kg) (9 . 8 m / s 2 ) cos 61 ◦ = 84 . 0908 N . For m to remain in equilibrium T max sin θ = m max g (2) m max = T max sin θ g = (84 . 0908 N) sin61 ◦ 9 . 8 m / s 2 = 7 . 50485 kg . Alternate Solution: Equations 1 and 2 come directly from the freebody diagram for the knot. Dividing Eq. 2 by Eq. 1, tan θ = m max μM m max = μM tan θ = (0 . 32) (13 kg) tan 61 ◦ = 7 . 50485 kg . 002 (part 1 of 2) 10.0 points The uniform diving board has a mass of 27 kg . A B 1 m 3 . 2 m Find the force on the support A when a 67 kg diver stands at the end of the div ing board. The acceleration of gravity is 9 . 81 m / s 2 . Correct answer: 2 . 39462 kN. Explanation: Let : m = 27 kg , M = 67 kg , ℓ = 4 . 2 m , x 1 = 1 m , and x d = 3 . 2 m . billing (cab4763) – HW 12 – opyrchal – (11109) 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 = ℓ 2 x 1 = 4 . 2 m 2 1 m = 1 . 1 m Applying rotational equilibrium at point B , summationdisplay vector τ = 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 . 1 m) (27 kg) + (3 . 2 m) (67 kg) 1 m bracketrightbigg × (9 . 81 m / s 2 ) × 1 kN 1000 N = 2 . 39462 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: 3 . 31676 kN. Explanation: Applying translational equilibrium, summationdisplay F y = F B F A mg M g = 0 F B = F A + ( m + M ) g = 2 . 39462 kN + (27 kg + 67 kg) (9 . 81 m / s 2 ) × 1 kN 1000 N = 3 . 31676 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
 Physics, Work

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