Question

A hanging block of mass $m'$ prevents the smaller block of mass $m$ from slipping over a movable triangular block of mass $M$ . All the surfaces are frictionless and the strings and the pulleys are light. Value of mass $m'$ in terms of $m , M$ and $\theta$ is

• A. $\left[\frac{ m + M }{\cot \theta -1}\right]$
• B. $\left[\frac{ m - M }{\cot \theta +1}\right]$
• C. $\left[\frac{ m - M }{\cot \theta -2}\right]$
• D. $\left[\frac{ m + M }{\cot \theta +2}\right]$

Answer 1

Answer: (A)

Pulling force is $m'g$, while mass in motion is $m+M+m'$.
So acceleration of the system will be given by
$a=\frac{m g}{\left(m+M+m^{\prime}\right)}$...(i)
Mass $m$ will not slide over $M$ if
$a \cos \theta=g \sin \theta \ldots$ (ii)
From equation (i) in (ii), we get
$\frac{m' g \cos \theta}{\left(m+ M+ m'\right)}=g \sin \theta$
or $m'=\left[\frac{m +M}{\cot \theta-1}\right]$