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Unformatted text preview: Problem 4.32 Part A F m B g N = m B a 4.32.1 F m W g + N = m W a 4.32.2 Adding these equations and setting acceleration to zero: 2 F m B g m W g = 0 4.32.3 F = ( m B + m W ) g 2 = 80 kg · 9 . 8 m s 2 2 = 392 N ’ 390 N 4.32.4 Part B Say that force is increased from F to F . From 4.32.1 and 4.32.2: 2 F ( m B + m W ) g = ( m B + m W ) a 4.32.5 a = 2 F ( m B + m W ) g m B + m W = 2 · 1 . 18 · 392 N 80 kg · 9 . 8 m s 2 80 kg = 1 . 76 m s 2 ’ 1 . 8 m s 2 4.32.6 1 Problem 4.49 Part A N mg y = N mg cos θ = ma y = 0 4.49.1 mg x = mg sin θ = ma x 4.49.2 a x = g sin θ 4.49.3 In order to nd the distance d the block will slide up the plane, we use 1D constant acceleration kinematics: v 2 f = 0 = v 2 2 a x d 4.49.4 d = v 2 2 a x = v 2 2 g sin θ = ( 4 . 2 m s ) 2 2 · ( 9 . 8 m s 2 ) · sin(21) = 2 . 51 m ’ 2 . 5 m 4.49.5 Part B x f = 0 = v t 1 2 a x t 2 4.49.6 t = 2 v a x = 2 · 4 . 2 m s ( 9 . 8 m s 2 ) · sin(21) = 2 . 39 s ’ 2 . 4 s 4.49.7 2 Problem 4.54 Since the pulley is massless, the net force on it must be zero: T 2 2 T 1 = 0 4.54.1 T 1 m 1 g = m 1 a 4.54.2 m 2 g + T 1 = m 2 a 4.54.3 Subtracting 4.54.2 from 4.54.3: ( m 2 m 1 ) g = ( m 2 + m 1 ) a 4.54.4 a = m 2 m 1 m 2 + m 1 g 4.54.5 Adding 4.54.2 and 4.54.3 and plugging in 4.54.5: 2 T 1 ( m 2 + m 1 ) g = ( m 2 m 1 ) a = ( m 2 m 1 ) 2 m 2 + m 1 g 4.54.6 T 2 = 2 T 1 = ( m 2 m 1 ) 2 ( m 2 + m 1 ) 2 m 2 + m 1 g = 4 m 1 m 2 m 2 + m 1 g = 4 · 3 . 2 kg · 1 . 2 kg 3 . 2 kg +1 . 2 kg ( 9 . 8 m s 2 ) = 34 . 2 N ’ 34 N 4.54.7 3 Problem 4.56 Assigning each mass and pulley a coordinate, we can see that: x p + x c = const. 4.56.1 ( x a x p ) + ( x b x p ) = const. 4.56.2 Solving 4.56.1 for x c , plugging into 4.56.2, and taking two derivatives: a a + a b + 2 a c = 0 4.56.3 T 2 2 T 1 = 0 4.56.4 T 1 m a g = m a a a 4.54.5 T 1 m b g = m b...
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 Fall '08
 Lanzara
 Physics, Acceleration, Force, Friction, µs

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