First slide

Static friction

Question

A block pressed against the vertical wall is in equilibrium.if $μ$ = 0.3, the acceleration of the block will be:

Moderate

A

zero
B

$\frac{g}{10} ↑$
C

$\frac{g}{4} ↓$
D

$\frac{g}{5} ↓$

Solution

If $μ$ = 0.3
Maximum value of resisting force will be
$F_{\lim} = μN = 0.3 mg$
But have driving force is mg/2.Therefore the friction will be kinetic nature; the block will slid down. Hence acceleration of the block
$a = \frac{\frac{mg}{2} - 0.3 mg}{m} = 0.2 g = \frac{g}{5} ↓$

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