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(C) For the block to be in equilibrium,the net force acting on it must be zero.$1$. The horizontal force pressing the block against the wall is $F_h =

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$0.5$

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(C) For the block to be in equilibrium,the net force acting on it must be zero.$1$. The horizontal force pressing the block against the wall is $F_h = mg$. This force provides the normal reaction $N = mg$.$2$. The maximum static friction force available is $f_{max} = \mu N = \mu (mg)$.$3$. Considering the vertical forces,the weight of the block $mg$ acts downwards,and an upward force of $\frac{mg}{2}$ is applied. Let $f$ be the friction force acting on the block.$4$. For equilibrium in the vertical direction: $f + \frac{mg}{2} = mg$,which gives $f = mg - \frac{mg}{2} = \frac{mg}{2}$.$5$. For the block to remain in equilibrium,the required friction force must be less than or equal to the maximum static friction: $f \le f_{max}$.$6$. Therefore,$\frac{mg}{2} \le \mu (mg)$.$7$. Solving for $\mu$,we get $\mu \ge \frac{1}{2}$,or $\mu \ge 0.5$. The minimum coefficient of friction is $0.5$.

$A$ block of mass $m$ is pressed against a vertical wall by a horizontal force of magnitude $mg$. An upward force of magnitude $\frac{mg}{2}$ is also applied to the block. If the block is in equilibrium,the minimum coefficient of friction $\mu$ between the block and the wall is:

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