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Physics - NLM Question with Solution | TestHub

PhysicsNLMFBD & EquilibriumEasy2 minQB

PhysicsEasysingle choice

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 $θ$ is

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Answer:

Solution:

Pulling force is $m^{'} g$ , while mass in motion is $m + M + m^{'}$ . So acceleration of the system will be given by
$a = \frac{m^{'} g}{(m + M + m^{'})} ....(i)$

Mass $m$ will not slide over $M$ if

a cosθ = g sinθ ....(ii)

From equation (i) in (ii), we get

$\frac{m^{'} g cos θ}{(m + M + m^{'})} = g sin θ or m^{'} = [\frac{m + M}{cot θ - 1}]$

Stream:JEESubject:PhysicsTopic:NLMSubtopic:FBD & Equilibrium

⏱ 2mℹ️ Source: QB

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