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 $F=5\text{N}$ in $+x$ direction as shown. Find the pseudo force acting on a block of mass $m=2kg$.

• A. $1\text{N},-x$ direction
• B. $2\text{N},-x$ direction
• C. $1\text{N},+x$ direction
• D. $2\text{N},+x$ direction

Answer 1

The correct option is A $1\text{N},-x$ direction

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

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

To find Pseudo force acting on block of mass $2\text{kg}$ we have from the FBD


$\overrightarrow{F_s}=-ma\hat{i}=-2\times 0.5\hat{i}=-1\hat{i}\text{N}$
The magnitude of pseudo force acting on block of mass $2\text{kg}=1\text{N}$
The direction of pseudo for $2\text{kg}$ mass is $-x$ direction .