Question

Find the height by which the sphere rises when the wedge is moved to touch the wall. The distance between the wedge and the wall is $10\text{m}$ and the wedge is moved to the left until it touches the wall.

• A. $10\text{m}$
• B. $10\sqrt{3}\text{m}$
• C. $\frac{10}{\sqrt{3}}\text{m}$
• D. $5\sqrt{3}\text{m}$

Answer 1

The correct option is C $\frac{10}{\sqrt{3}}\text{m}$

By condition of wedge contrained the displacement, velocity and acceleration along normal to the common interface is same.
Let $S_w$ be the displacement of the wedge.
And $S_s$ be the displacement of the sphere.
$S_w\sin \theta =S_s\cos \theta$
$\Rightarrow S_s=S_w\tan \theta$
$\Rightarrow S_s=10\tan {30}^{\circ }$
$S_s=\frac{10}{\sqrt{3}}\text{m}$