## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 3 - Problems - Page 118: 103

#### Answer

(a) $T = (m_1+m_2)~g~\mu$ (b) $T = (m_1+m_2)~g~\mu~cos~\theta$

#### Work Step by Step

(a) We can find the maximum acceleration that the force of friction $F_f$ can provide for the block $m_2$: $F_f = m_2~a$ $m_2~g~\mu = m_2~a$ $a = g~\mu$ We can find the tension which provides this acceleration to the system of both blocks: $T = (m_1+m_2)~g~\mu$ (b) We can find the maximum acceleration that the force of friction $F_f$ can provide for the block $m_2$: $\sum F = m_2~a$ $F_f - m_2~g~sin~\theta = m_2~a$ $m_2~g~cos~\theta~\mu - m_2~g~sin~\theta = m_2~a$ $g~cos~\theta~\mu - g~sin~\theta = a$ $a = g~(\mu~cos~\theta-sin~\theta)$ We can find the tension which provides this acceleration to the system of both blocks: $\sum F = (m_1+m_2)~a$ $T-(m_1+m_2)~g~sin~\theta = (m_1+m_2)~g~(\mu~cos~\theta-sin~\theta)$ $T = (m_1+m_2)~g~(\mu~cos~\theta-sin~\theta+ sin~\theta)$ $T = (m_1+m_2)~g~\mu~cos~\theta$

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