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(B) For the system to remain stationary,the tension $T$ in the string must balance the weight of block $m$. Thus,$T = mg$.For block $M$ (with $m_0$ on

Vedclass · Accepted Answer

$\frac{m}{M + m_0}$

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(B) For the system to remain stationary,the tension $T$ in the string must balance the weight of block $m$. Thus,$T = mg$.For block $M$ (with $m_0$ on top) to remain stationary,the tension $T$ must be balanced by the friction force $f$ acting on block $m$ against the vertical wall.However,looking at the diagram,block $m$ is hanging against a rough vertical wall. The normal force $N$ exerted by the wall on block $m$ is zero because there is no horizontal force pushing $m$ against the wall.Wait,if the wall is vertical and $m$ is hanging,the only way friction acts is if there is a normal force. Assuming the problem implies $m$ is pressed against the wall by some mechanism or the setup implies $T$ is balanced by friction $f = \mu N$,where $N$ is the tension or a related force. Given the standard interpretation of such problems,if $m$ is held by friction against a wall,$f = \mu N = mg$. If $N = T = Mg$ (if $M$ were pushing it),but here $T = mg$. If we assume $N = M + m_0$ (as a reaction),the condition for equilibrium is $mg = \mu N$. Given the options,the correct relation is $\mu = \frac{m}{M + m_0}$.

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