Question

A smooth block of mass $m$ is held stationary on a smooth wedge of mass $M$ and inclination $\theta$ as shown in figure. If the system is released from rest, then the normal reaction between the block and the wedge is:

• A. $mg\cos \theta$
• B. less than $mg\cos \theta$
• C. greater than $mg\cos \theta$
• D. may be less or greater than $mg\cos \theta$ depending upon whether $m$ is less or greater than $M$

Answer 1

The correct option is B less than $mg\cos \theta$

Given, surface between wedge and block is smooth.

Therefore, block will gain a tendency to move downwards and wedge will move backwards. But since it is held stationary, we will change our frame and solve this question by sitting on the wedge.

Draw free body diagram of block and wedge respectively as shown in image,

equating forces of block in vertical direction, we will have,

$N+\left(Ma\right)sin\left(\theta \right)=\left(mg\right)cos\left(\theta \right)$

therefore , $N=\left(mg\right)cos\left(\theta \right)-\left(Ma\right)sin\left(\theta \right)$

Ans : B, less than $\left(mg\right)cos\left(\theta \right)$