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Solution 63GP Introduction We shall have to calculate net

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett ISBN: 9780321570567 2

Solution for problem 63GP Chapter 7

Calculus: Early Transcendentals | 1st Edition

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Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

Calculus: Early Transcendentals | 1st Edition

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Problem 63GP

Solution 63GP Introduction We shall have to calculate net torque first followed by linear acceleration and time.

Step-by-Step Solution:

Step 1: Given that, diameter = 12 cm Radius R = 6 cm R = 0.06 m Mass of the pulley M = 2.0 kg 1 2 2 Therefore, its moment of inertia I = ×2.0 kg×(0.06) kg.m 2 I = 0.0036 kg.m Let T be the tension of the rope attached to the 4.0 kg mass and T be the tension of the 1 2 rope attached to the 2.0 kg mass. For the 4.0 kg mass, the equation of motion can be written as follows. 4gT = 41 ( a is the acceleration) T =14g4a…..(1) For the 2.0 kg mass, the equation of motion can be written as follows. T 2g = 2a 2 T =22g+2a…..(2) Thus, torque on the pulley = (T T1)0.062N.m (from equations (1) and (2)) Now, the net torque excluding the frictional torque = (T T )0.06 N.m0.50 N.m 1 2 …..(3) Step 2: We know that, torque = I Therefore, substituting the values of I = 0.0036 kg.m and from equation (3) 2 (T 1 )0.26 N.m0.50 N.m = 0.0036 kg.m × Now, substituting the values of T an1 T from2equations (1) and (2) and using a = R 2 (4g4a2g2a)0.06 N.m0.50 N.m = 0.0036 kg.m ×a/R 2 (2g6a)0.06 N.m0.50 N.m = 0.0036 kg.m ×a/0.06 2×9.8×0.06 N.m0.36a N.m0.50 N.m = 0.06a N.m 1.18 N.m0.50 N.m0.36a N.m = 0.06a N.m 0.68 N.m = 0.42a N.m a = 1.62 m/s 2

Step 3 of 3

Chapter 7, Problem 63GP is Solved
Textbook: Calculus: Early Transcendentals
Edition: 1
Author: William L. Briggs, Lyle Cochran, Bernard Gillett
ISBN: 9780321570567

The full step-by-step solution to problem: 63GP from chapter: 7 was answered by , our top Calculus solution expert on 03/03/17, 03:45PM. This full solution covers the following key subjects: acceleration, calculate, followed, Introduction, Linear. This expansive textbook survival guide covers 85 chapters, and 5218 solutions. The answer to “Solution 63GP Introduction We shall have to calculate net torque first followed by linear acceleration and time.” is broken down into a number of easy to follow steps, and 17 words. Since the solution to 63GP from 7 chapter was answered, more than 299 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Calculus: Early Transcendentals, edition: 1. Calculus: Early Transcendentals was written by and is associated to the ISBN: 9780321570567.

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