HW_5_Solutions_F05_ed

HW_5_Solutions_F05_ed - Physics 112 - HW #5 Solutions Fall...

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Physics 112 - HW #5 Solutions Fall 2005 5-24 [Monkey Business] (a) If crate is stationary on a horizontal surface and no horizontal force is applied, then friction is zero. (b) Maximum possible static friction is F s,max = µ s N = µ s W. Since surface is horizontal, N = W, and F s,max = (0.40) (40.0 N) = 16.0 N. When monkey pushes with a force of 6 N, this is less than the maximum possible static friction. So crate does not move and the friction force is 6.0 N. (c) To start box in motion, monkey must exert a force equal to the maximum static friction, 16.0 N. (d) To keep box moving at constant speed, need a force equal in magnitude to the kinetic friction, F k = µ k N = µ k W = (0.20) (40.0 N) = 8.0 N. (e) If applied force is F a = 18.0 N, the box is sliding, so the magnitude of the friction force is F k = µ k N = 8.0 N. (same as (d)). The acceleration is found by applying N’s 2 nd law: Σ F x = F a - F k = M a = g W a. Then a = g W (F a - F k ) = 9.81 m/s 2 40.0 N (18.0 N – 8.0 N) = 2.45 m/s 2 . 5-47 [Airplane in Banked Turn] The free-body diagram shows the forces acting on the plane while turning, as viewed from the front or rear. The plane's acceleration is a = v 2 /R in the +x direction ( ). v = 240 km/h 10 3 m 1 km 1 h 3600 s = 66.7 m/s Write Newton's 2nd Law, Σ F = m a : +x: F lift sin θ = ma x = m v 2 R (1) +y: F lift cos θ - mg = ma y = 0 F lift cos θ = mg (2) Divide equation (1) by equation (2) to get: F lift sin θ F lift cos θ = tan θ = v 2 Rg = (66.7 m/s) 2 (1200 m)(9.8 m/s 2 ) = 0.378 θ = 21° θ +x +y θ mg F lift m
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- 2 - 5-89 [Block on Front of Cart] For the block to not slide down, the (horizontal) normal force exerted by the cart on the block must be large enough to give a (vertical) friction force that equals the weight of the block. First draw an FBD for the block. Apply Newton’s 2 nd Law to the block. Σ F x = N = ma (1) Σ F y = f - w = 0 (2) At the minimum acceleration for no slipping to occur, we also have f = µ s N (3) and we also know that w = mg From (1) and (3): a = N m = f µ s m From (2): f = w = mg, So a = f µ s m = mg µ s m = g µ s , a = g µ s An observer would perceive the block to be “stuck” to the front of the cart. 5-112 [Motorcycle Inside Sphere] (a) To maintain contact with the sphere, the normal force N exerted by the sphere on the tires must be greater than zero. When at the top of the sphere, the radial component of the acceleration of the bike is a r = - mv 2 R , pointing vertically downwards. The normal force N also points downwards. Apply Newton’s 2 nd Law to bike: Σ F y = - N - w = - ma y N – mg = - mv 2 R At the minimum speed, N = 0 Then g = v min 2 R , and v min = Rg = (13.0 m)(9.81 m/s 2 ) = 11.3 m/s (b) At the bottom of the sphere, v = 2v min = 2 Rg Now the normal force N and the radial component of the acceleration both point upwards , so applying Newton’s 2
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This homework help was uploaded on 04/21/2008 for the course PHYS 1112 taught by Professor Leclair,a during the Fall '07 term at Cornell.

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HW_5_Solutions_F05_ed - Physics 112 - HW #5 Solutions Fall...

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