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HW08-solutions

# HW08-solutions - markowitz(am45362 HW08 distler(56295 This...

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markowitz (am45362) – HW08 – distler – (56295) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points A 6 m long, uniform ladder leans against a frictionless wall and makes an angle of 64 . 5 with the floor. The ladder has a mass 29 . 2 kg. A 75 . 92 kg man climbs 82% of the way to the top of the ladder when it slips and falls to the floor. What is the coefficient of static friction be- tween the ladder and the floor? Correct answer: 0 . 348722. Explanation: Let : M = 29 . 2 kg , M m = 75 . 92 kg , R = 0 . 82 , and θ = 64 . 5 . summationdisplay F x = 0 requires the wall to exert a force equal in magnitude to the frictional force of ( M m + M ) g μ exerted by the floor. Applying torques on the ladder about the point where it touches the floor, ( M + M m ) g μ sin θ parenleftbigg M 2 + R M m parenrightbigg g cos θ = 0 2 ( M + M m ) μ sin θ = ( M 2 R M m ) cos θ μ = ( M 2 R M m ) cos θ 2 ( M + M m ) sin θ = [29 . 2 kg 2(0 . 82)(75 . 92 kg)] cos 64 . 5 2(29 . 2 kg + 75 . 92 kg) sin 64 . 5 = 0 . 348722 . 002 10.0points Two masses of 25 kg and 16 kg are suspended by a pulley that has a radius of 14 cm and a mass of 8 kg. The cord has a negligible mass and causes the pulley to rotate without slipping. The pulley rotates without friction. The masses start from rest 1 . 3 m apart. 1 . 3 m 14 cm 8 kg ω 25 kg 16 kg Determine the speeds of the two masses as they pass each other. Treat the pulley as a uniform disk. The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 1 . 59625 m / s. Explanation: Let : M = 8 kg , R = 14 cm , m 1 = 25 kg , m 2 = 16 kg , and h = 1 . 3 m . From conservation of energy K 1 + K 2 + K disk = Δ U m 1 v 2 2 + m 2 v 2 2 + M v 2 4 = ( m 1 m 2 ) g h 2 (2 m 1 + 2 m 2 + M ) v 2 = 2 ( m 1 m 2 ) g h , where h 2 is the height. Taking no slipping into account, v = radicalBigg 2 ( m 1 m 2 ) g h 2 ( m 1 + m 2 ) + M = radicalBigg 2 (25 kg 16 kg) (9 . 8 m / s 2 ) 2 (25 kg + 16 kg) + 8 kg × 1 . 3 m = 1 . 59625 m / s .

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markowitz (am45362) – HW08 – distler – (56295) 2 keywords: 003 10.0points A rod of mass m and length L is hinged with a frictionless hinge at one end. The moment of inertia about the center of mass is 1 12 m L 2 . Attached to the end of the rod opposite to the hinge there is a mass of magnitude 2 m . The rod is released from rest in the horizontal position. L m 2 m What is the speed of the mass 2 m when the rod passes through the vertical position? Consider the mass at the end of the rod to be a point particle. 1. bardbl vectorv 2 m bardbl = radicalbig g L 2. bardbl vectorv 2 m bardbl = radicalbigg 7 g L 15 3. bardbl vectorv 2 m bardbl = radicalbigg 5 g L 2 4. bardbl vectorv 2 m bardbl = radicalbigg 2 g L 3 5. bardbl vectorv 2 m bardbl = radicalbig 2 g L 6. bardbl vectorv 2 m bardbl = radicalbigg 15 g L 7 correct 7. bardbl vectorv 2 m bardbl = radicalbigg 2 g L 5 8. bardbl vectorv 2 m bardbl = radicalbigg g L 2 9. bardbl vectorv 2 m bardbl = radicalbigg g L 10. bardbl vectorv 2 m bardbl = radicalbigg 3 g L 2 Explanation: Let us measure heights from the point at the end of the rod when it is vertical. The ini- tial energy is all potential, with a magnitude E i = U rod i +
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