Question

A block of mass 2 kg is to be kept at rest against a rough vertical wall (p = 0.4) by applying a horizontal force F. The minimum value of Fis (g = 10 m/s2

Answer 1

Given:

Mass of the block m = 2 kg

Coefficient of friction $\[\mu\]$ = 0.4

Acceleration due to gravity g = 10 $m/s^{2}$

To find:

We have to find the minimum horizontal force F required to keep the block at rest.

Solution:

In order to solve this question, we have to refer to the free body diagram as shown in the attachment.

Now,

Friction force is given by the product of the coefficient of friction and normal force, here the normal force is the horizontal force.

Then,

For the block to be in rest the friction force must be equal to the weight of the block.

So, we get,

Weight of the block = Friction force

⇒$\[m \times g = \mu \times F\]$

⇒$\[2 \times 10 = 0.4 \times F\]$

⇒$F = \frac{2 \times 10}{0.4}$

∴$F=50N$

The minimum horizontal force required to keep the block at rest is 50N.