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  4. A hanging block of mass m' prevents the smaller block of mass textm from slipping over a movable triangular block of mass textM . All the surfaces are frictionless and the strings and the pulleys are light. Value of mass m' in terms of m , M and θ is <img class=img-fluid question-image alt=Question src=https://cdn.tardigrade.in/q/nta/p-6utbig3mqf9a.png />

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Q. A hanging block of mass $m^{'}$ 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 $m^{'}$ in terms of $m$ , $M$ and $\theta $ is

Question

NTA AbhyasNTA Abhyas 2020Laws of Motion

Solution:

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 $$ \frac{ m ^{\prime} g \cos \theta}{\left( m + M + m ^{\prime}\right)}= g \sin \theta \quad \text { or } \quad m ^{\prime}=\left[\frac{ m + M }{\cot \theta-1}\right] $$