Hw04_Solutions

Hw04_Solutions - Problem 4.32 Part A F m B g N = m B a...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This note was uploaded on 03/13/2009 for the course PHYSICS 7A taught by Professor Lanzara during the Fall '08 term at Berkeley.

Page1 / 13

Hw04_Solutions - Problem 4.32 Part A F m B g N = m B a...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online