A block of mass M is placed on a horizontal s | vedclass

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(A) For the block $m$ to remain stationary,the tension $T$ in the string must be equal to the weight of the block $m$,so $T = mg$.For the block $M$ (w

Vedclass · Accepted Answer

$\frac{m}{\mu} - M$

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(A) For the block $m$ to remain stationary,the tension $T$ in the string must be equal to the weight of the block $m$,so $T = mg$.For the block $M$ (with $m_0$ on top) to remain stationary,the tension $T$ must be balanced by the maximum static friction force $f_{max}$ acting between block $M$ and the horizontal surface.The normal force $N$ acting on the surface is $N = (M + m_0)g$.The maximum static friction force is $f_{max} = \mu N = \mu(M + m_0)g$.For equilibrium,$T = f_{max}$,which implies $mg = \mu(M + m_0)g$.Dividing both sides by $g$,we get $m = \mu(M + m_0)$.Rearranging for $m_0$,we have $\frac{m}{\mu} = M + m_0$,which gives $m_0 = \frac{m}{\mu} - M$.

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