HW4 - 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

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 There are two accelerations. a is the acceleration of the m c-pulley system, and a is the acceleration of the m a- m b system: T 2- 2 T 1 = 0 4.56.1 T 1- m a g = m a a 00 = m a ( a + a ) 4.54.2 T 1- m b g =- m b a 000 = m b ( a- a ) 4.54.3 T 2- m c g =- m c a 4.54.4 Adding 4.54.2 and 4.54.3: 2 T 1- ( m a + m b ) g = ( m a + m b ) a + ( m a- m b ) a 4.54.5 T 2 = 2 T 1 = ( m a + m b ) ( a + g ) + ( m a- m b ) a 4.54.6 Subbing into 4.54.4 ( m a + m b ) ( a + g ) + ( m a- m b ) a- m c g =-...
View Full Document

This note was uploaded on 02/17/2010 for the course PHYSICS 7A taught by Professor Lanzara during the Fall '08 term at University of California, Berkeley.

Page1 / 13

HW4 - 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

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