Question

A block of mass $m$ is placed on a wedge of mass $m$ and inclination $\theta$. All surfaces are smooth. Find the acceleration of the wedge.

• A. $\frac{mg{\cos }^2\theta }{M+m{\sin }^2\theta }$
• B. $\frac{mg{\sin }^2\theta }{M+m{\cos }^2\theta }$
• C. $\frac{mg\sin \theta \cos \theta }{M+m{\sin }^2\theta }$
• D. $\frac{mg\sin \theta \cos \theta }{M+m{\cos }^2\theta }$

Answer 1

Answer: (C)

$\frac{mg\sin \theta \cos \theta }{M+m{\sin }^2\theta }$

From FBD,
$ma=mg\sin \theta +m{a}^{\prime }\cos \theta$
or, $a=g\sin \theta +a^{\prime }\cos \theta ...\left(1\right)$
$N+m{a}^{\prime }\sin \theta =mg\cos \theta ...\left(2\right)$
and $N\sin \theta =M{a}^{\prime }....\left(3\right)$

using (3), (2) becomes, $\frac{M{a}^{\prime }}{\sin \theta }+m{a}^{\prime }\sin \theta =mg\cos \theta$
or $a^{\prime }(M+m{\sin }^2\theta )=mg\cos \theta \sin \theta$
$\therefore a^{\prime }=\frac{mg\cos \theta \sin \theta }{M+m{\sin }^2\theta }$