Question

In the situation as shown in figure if acceleration of $B$ is $a$ then find the acceleration of $A$ ($B$ always remains horizontal),

• A. $2a\tan \theta$
• B. $a\sin \theta$
• C. $a\cot \theta$
• D. $2a\cos \theta$

Answer 1

The correct option is D $2a\cos \theta$

Components of accelerations of $A$ and $B$ along normal $\hat{n}$ to the incline of wedge should be same

$A\sin \theta =a\cos (2\theta -{90}^{\circ })$

$A\sin \theta =a\sin 2\theta$

$A=2a\cos \theta$

Final answer : (d)