Question

A block of mass $10\text{kg}$ is pressed against a vertical wall. If the coefficient of friction between the wall and the block is $0.2$, then what is the minimum force that should be applied on the block so that does not fall to the ground?

• A. $500\text{N}$
• B. $400\text{N}$
• C. $600\text{N}$
• D. $555.5\text{N}$

Answer 1

The correct option is A $500\text{N}$

By the FBD of the block we get,

The force $F$ is applied against the wall, which produces a normal reaction $N$. Frictional force arises due to this reaction force and counters the weight of the block.
Hence for equillibrium,

$N=F$
$\mu N=mg$
$\mu F=mg\Rightarrow F=\frac{mg}{\mu }=\frac{100}{0.2}=500\text{N}$
Hence a force of $500\text{N}$ has to be applied to keep the block in equilibrium.

Hence option A is the correct answer