Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 7 - Newton's Third Law - Exercises and Problems: 13

Answer

(a) $F = 32~N$ (b) $T_{1t} = 19.2~N$ (c) $T_{1b} = 16~N$ (d) $T_{2t} = 3.2~N$

Work Step by Step

(a) Let $m_b$ be the mass of a block. Let $m_r$ be the mass of a rope. We can find $F$. $\sum F = (2m_b+2m_r)~a$ $F - (2m_b+2m_r)~g = (2m_b+2m_r)~a$ $F = (2m_b+2m_r)(g+a)$ $F = [(2)(1.0~kg)+(2)(0.25~kg)](9.80~m/s^2+3.0~m/s^2)$ $F = 32~N$ (b) We can find the tension $T_{1t}$ at the top end of rope 1. $\sum F = (m_b+2m_r)~a$ $T_{1t} - (m_b+2m_r)~g = (m_b+2m_r)~a$ $T_{1t} = (m_b+2m_r)(g+a)$ $T_{1t} = [(1.0~kg)+(2)(0.25~kg)](9.80~m/s^2+3.0~m/s^2)$ $T_{1t} = 19.2~N$ (c) We can find the tension $T_{1b}$ at the bottom end of rope 1. $\sum F = (m_b+m_r)~a$ $T_{1b} - (m_b+m_r)~g = (m_b+m_r)~a$ $T_{1b} = (m_b+m_r)(g+a)$ $T_{1b} = (1.0~kg+0.25~kg)(9.80~m/s^2+3.0~m/s^2)$ $T_{1b} = 16~N$ (d) We can find the tension $T_{2t}$ at the top end of rope 2. $\sum F = m_r~a$ $T_{2t} - m_r~g = m_r~a$ $T_{2t} = (m_r)(g+a)$ $T_{2t} = (0.25~kg)(9.80~m/s^2+3.0~m/s^2)$ $T_{2t} = 3.2~N$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.