Question

A cube of mass $2kg$ is held stationary against a rough wall by a force $F=40N$ passing through centre C. Find perpendicular distance of normal reaction between wall and cube from point C. Side of the cube is $20cm$. Take $g=10m/s^2$

• A. $5cm$
• B. $10cm$
• C. $3cm$
• D. $20cm$

Answer 1

The correct option is A $5cm$

$\text{Step 1: Free Body diagram (Refer fig.)}$

f = frictional force

Here block is stationary so it will be in complete equilibirium.

$\text{Step 2: Translational equilibirium}$

From FBD

$\sum F_x=0$

$\Rightarrow N=F=40N$

$\sum F_y=0$

$\Rightarrow f=mg=2\left(kg\right)10\left(m/s^2\right)=20N$

$\text{Step 2: Rotational equilibirium}$

For Rotational equilibrium, net torque about $C=0$

$\Rightarrow {\tau }_{net}=0$

$\Rightarrow f\times 10-Nx=0$

$\Rightarrow 20\times 10-40x=0$

$\therefore$ $x=5cm$

Hence Option $A$ is correct.