Question

The block of mass m is in equilibrium relative to the smooth wedge of mass M which is pushed by a horizontal force F. Find pseudo force acting on (a) m, (b) M as viewed by the observer sitting on the wedge. Will these pseudo forces (c) equal and opposite, action reaction pairs? Explain.

Answer 1

Taking block and wedge as system.
Net force on the system horizontally $=F$
Total mass $=M+m$
Therefore, acceleration of the system in +ve x-direction
$a=\frac{F}{M+m}\Rightarrow \overrightarrow{a}=\frac{F}{M+m}\hat{i}$
$a$. Pseudo force acting on block m is in opposite direction of observer acceleration and magnitude equal to multiplication of mass of the object and the acceleration of the observer
$\overrightarrow{F_{PS}}=-ma\hat{i}=-\frac{mF}{M+m}\hat{i}$
$b$ Pseudo force acting on block M $\overrightarrow{{F^{\prime }}_{PS}}=Ma\hat{i}$
$\overrightarrow{{F^{\prime }}_{PS}}=-\frac{MF}{M+m}\hat{i}$
$c$ Pseudo force is not a real force. It is applied to solve the problems of non-inertial frame. It does not have action-reaction pairs.