Question

Consider the shown arrangement. Assume all surfaces to be smooth. If $N$ represents magnitudes of normal reaction between block and wedge, then acceleration of $M$ along horizontal is equal to:

• A. $\frac{N\sin \theta }{M}$ along +ve x-axis
• B. $\frac{N\cos \theta }{M}$ along -ve x-axis
• C. $\frac{N\cos \theta }{M}$ along +ve x-axis
• D. $\frac{N\sin \theta }{m+M}$ along -ve x-axis

Answer 1

The correct option is D $\frac{N\sin \theta }{m+M}$ along -ve x-axis

From the figure,

The value of H is given by

$H=Nsin\theta$

Therefore, the value of acceleration is given by;

$a=\frac{Nsin\theta }{M+m}$ along -ve x axis.