Question

The block of mass $2\text{kg}$ is in equilibrium w.r.t $8\text{kg}$ wedge. The wedge is pushed by a force of $5\text{N}$ in $+x$ direction as shown in the figure. The pseudo force acting on the wedge is

• A. $4\text{N},+x$ direction
• B. $2\text{N},+x$ direction
• C. $4\text{N},-x$ direction
• D. $2\text{N},-x$ direction

Answer 1

The correct option is C $4\text{N},-x$ direction

Consider block and wedge as one system, the net force $F$ acting on the system horizontally is $5\text{N}$

Total mass of the system $=10\text{kg}$
$\therefore$ Acceleration of the system $\overrightarrow{a}=\frac{5}{10}=0.5\hat{i}m/s^2$

To find the pseudo force acting on wedge, we have FBD of the wedge as

$\overrightarrow{F}=-m\overrightarrow{a}\hat{i}$
$\Rightarrow \overrightarrow{F}=-8\times 0.5\hat{i}=-4\hat{i}\text{N}$
The magnitude and direction of pseudo force on wedge is $4\text{N}$ along $-x$ direction.