Explanation 1) Concept: We use the concept of Newton’s second law. We find acceleration of the system by applying Newton’s second law for both masses and pulley. We can write three equations because we have three unknowns. Solving these equations simultaneously, we get acceleration. Once acceleration is found, we can find tensions. 2) Formulae: i) F = m a ii) τ = I α = F × r 3) Given: i) Mass of block 1 is m 1 = 400 g ii) Mass of block 2 is m 2 = 600 g iii) Mass of the disk is M = 500 g iv) Radius of disk is R = 12.0 c m 4) Calculations: a) The magnitude of the acceleration of the blocks: First we write the equation for net force on each block and pulley: m 1 a = T 1 - m 1 g … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … ( 1 ) m 2 a = m 2 g - T 2 … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … . . . ( 2 ) T 2 R - T 1 R = I α We know moment of inertia of disk is I = 1 2 M R 2 also α = a R T 2 R - T 1 R = 1 2 M R 2 a R R will get cancelled out from both sides, so we get T 2 - T 1 = 1 2 M a We can write this equation as T 2 = 1 2 M a +

10th Edition

ISBN: 9781119500100

Author: Halliday

Publisher: MCGRAW-HILL HIGHER EDUCATION

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Chapter 10, Problem 83P

To determine

To find:

a) The magnitude of the acceleration of the blocks.

b) The tension in the cord at left.

c) The tension in the cord at right.

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Chapter 10, Problem 82P

Chapter 10, Problem 84P

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a) The magnitude of the acceleration of the blocks. b) The tension in the cord at left. c) The tension in the cord at right.

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To find: a) The magnitude of the acceleration of the blocks. b) The tension in the cord at left. c) The tension in the cord at right.

Want to see the full answer?

Chapter 10 Solutions

To find:

a) The magnitude of the acceleration of the blocks.

b) The tension in the cord at left.

c) The tension in the cord at right.