Two masses A and B of 10, kg and 5, kg respec | vedclass

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(A) For the system to be in equilibrium,the tension $T$ in the string must be equal to the weight of block $B$,so $T = m_B g = 5g$.For block $A$ (with

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$15$

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(A) For the system to be in equilibrium,the tension $T$ in the string must be equal to the weight of block $B$,so $T = m_B g = 5g$.For block $A$ (with block $C$ on top) to remain stationary,the limiting friction $f_L$ must be at least equal to the tension $T$.The normal force $N$ on block $A$ is $(m_A + m_C)g$.The limiting friction is $f_L = \mu N = \mu (m_A + m_C)g$.Setting $f_L = T$,we get $\mu (m_A + m_C)g = m_B g$.Substituting the given values: $0.2 	imes (10 + m_C) = 5$.$2 + 0.2 m_C = 5$.$0.2 m_C = 3$.$m_C = \frac{3}{0.2} = 15\, kg$.

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