Question

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