# Chapter 7 - Newton's Third Law - Exercises and Problems: 8

The maximum mass of block A is 12 kg

#### Work Step by Step

Since the angled rope is at an angle of $45^{\circ}$, the horizontal component of the tension is equal to the vertical component of the tension in this rope. At equilibrium, the vertical component of the tension will be equal in magnitude to the weight of block A. Then the tension in the rope attached to block B is equal to the weight of block A. To stay at equilibrium, the weight of block A can not exceed the maximum possible force of static friction acting on block B. $m_A~g = F_f$ $m_A~g = m_B~g~\mu_s$ $m_A = m_B~\mu_s$ $m_A = (20~kg)(0.60)$ $m_A = 12~kg$ The maximum mass of block A is 12 kg.

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