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(C) For the block to be in equilibrium,the net vertical force must be zero.The forces acting on the block in the vertical direction are:$1$. Weight of

Vedclass · Accepted Answer

$0.5$

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(C) For the block to be in equilibrium,the net vertical force must be zero.The forces acting on the block in the vertical direction are:$1$. Weight of the block acting downwards: $mg$$2$. External upward force: $\frac{mg}{2}$$3$. Frictional force $f$ acting on the block.Since the block has a tendency to slide downwards (because $mg > \frac{mg}{2}$),the frictional force will act upwards.For equilibrium: $f + \frac{mg}{2} = mg$$f = mg - \frac{mg}{2} = \frac{mg}{2}$The normal force $N$ exerted by the wall on the block is equal to the horizontal pressing force: $N = mg$.The maximum possible static friction is $f_{\max} = \mu N = \mu mg$.For equilibrium,the required friction must be less than or equal to the maximum static friction:$f \leq f_{\max}$$\frac{mg}{2} \leq \mu mg$$\mu \geq \frac{1}{2} = 0.5$Therefore,the minimum coefficient of friction is $0.5$.

$A$ block of mass $m$ is pressed against a vertical wall and is in equilibrium. An external upward force of $\frac{mg}{2}$ is applied to the block. The minimum coefficient of friction $\mu$ required is:

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