Question

A hanging block of mass $m^{{}^{\prime }}$ prevents the smaller block of mass $\text{m}$ from slipping over a movable triangular block of mass $\text{M}$ . 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. $[\frac{m+M}{(cot\theta -1)}]$
• B. $[\frac{m-M}{(cot\theta +1)}]$
• C. $[\frac{m-M}{(cot\theta -2)}]$
• D. $[\frac{m-M}{(cot\theta +2)}]$

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{ m ^{\prime} g }{\left( m + M + m ^{\prime}\right)} {\quad} \ldots \text { (i) } Mass $m$ will not slide over $M$ if a \cos \theta=g \sin \theta From equation (i) in (ii), we get \frac{ m ^{\prime} g \cos \theta}{\left( m + M + m ^{\prime}\right)}= g \sin \theta {\quad} \text { or } {\quad} m ^{\prime}=\left[\frac{ m + M }{\cot \theta-1}\right]