HW5Sol - Physics 112 - HW #5 Solutions 5-42 When ball's...

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Physics 112 - HW #5 Solutions 5-42 When ball's speed v = v t /2 = mg/D /2 (half its terminal speed), f d = Dv 2 = mg/4. (a) Going up: mg f d m Σ F = m a ( + ): mg + f d = ma = mg + mg 4 = 5mg 4 a = 5 4 g (b) Going down: mg f d m Σ F = m a ( + ): mg - f d = ma = mg - mg 4 = 3mg 4 a = 3 4 g 5-47 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° 5-75 (a) Draw a free-body diagram for all forces acting on the crate, including the friction force f. We don't yet know whether f is static or kinetic. Write out Σ F = m a : y: N - mg = ma y = 0 N = mg x: f = ma x = ma If f is static friction, then f = ma = f s µ s N = µ s mg , so that . . . a
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This note was uploaded on 04/02/2008 for the course PHYS 1112 taught by Professor Leclair,a during the Fall '07 term at Cornell University (Engineering School).

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HW5Sol - Physics 112 - HW #5 Solutions 5-42 When ball's...

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