Question

A block of mass $m$, is kept on a wedge of mass $M$, as shown in figure such that mass $m$ remains stationary w.r.t wedge. The magnitude of force $P$ is

• A. $g\tan \beta$
• B. $mg\tan \beta$
• C. $(m+M)g\tan \beta$
• D. $mg\cot \beta$

Answer 1

The correct option is C $(m+M)g\tan \beta$

If mass $m$ remains stationary w.r.t $M$, means it is moving with same acceleration $a$ as that of $M$.

So, considering $(M+m)$ as a system, apply Newton's second law

$F_{ext}=M_{system}a$

$P=(M+m)a...........\left(1\right)$

From the FBD of $m$


Here, $ma$ is a pseudo force acting on block $m$ opposite to acceleration.

Applying equilibrium condition along the inclined plane

$mg\sin \beta =ma\cos \beta$

$\therefore a=g\tan \beta$

From eq. $\left(1\right)$,

$P=(M+m)g\tan \beta$

Thus, required force $P$ is $(M+m)g\tan \beta$.

$\begin{array}{c}Ans.C\end{array}$

$\begin{array}{c}\text{Why this question}?\\ \text{When a body is stationary w.r.t another}\\ \text{body, it means both have same acceleration.}\\ \text{For such problems we solve from}\\ \text{accelerated frame by including a pseudo}\\ \text{force in the FBD of body.}\end{array}$