Question

Find the acceleration of $B$ if the acceleration of $A$ is $4m/s^2$ . ($B$ is only allowed to move in vertical direction)

• A. $\frac{2}{\sqrt{3}}m/s^2$
• B. $\frac{4}{\sqrt{3}}m/s^2$
• C. $2\sqrt{3}m/s^2$
• D. $4\sqrt{3}m/s^2$

Answer 1

The correct option is B $\frac{4}{\sqrt{3}}m/s^2$

By the condition of wedge constrained, acceleration along the perpendicular direction to the common interface should be equal.

From the figure shown above, we can write the components of acceleration of $A$ and $B$ as
The component of acceleration of $A$ along the perpendicular to the incline is $a_A\sin \theta$
Similarly, the component of acceleration of $B$ along the perpendicular to the incline is $a_B\cos \theta$

$\because$ $a_B\cos \theta =a_A\sin \theta$ and $\theta ={30}^{\circ }$
$\Rightarrow a_B\cos {30}^{\circ }=a_A\sin {30}^{\circ }$
$\Rightarrow a_B\left(\frac{\sqrt{3}}{2}\right)=4\left(\frac{1}{2}\right)$
$\Rightarrow a_B=\frac{4}{\sqrt{3}}m/s^2$