A wooden wedge of mass M and inclination angle α rests on a smooth floor. A block of mass m is kept on wedge. A force P → is applied on the wedge as shown in figure, such that a block remains stationary with respect to wedge. The magnitude of force P → is

Since, P = ( M + m ) a Now as in free body diagram of block, m a cos ⁡ α = m g sin ⁡ α ∴ a = g sin ⁡ α cos ⁡ α = g tan ⁡ α or P = M + m g tan ⁡ α

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PhysicsNLMChallenging application of NLMEasy2 minQB

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A wooden wedge of mass $M$ and inclination angle $α$ rests on a smooth floor. A block of mass $m$ is kept on wedge. A force $\vec{P}$ is applied on the wedge as shown in figure, such that a block remains stationary with respect to wedge. The magnitude of force $\vec{P}$ is

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Answer:

Solution:

Since, $P = (M + m) a$

Now as in free body diagram of block,

$m a \cos α = m g \sin α$

$∴ a = g \frac{\sin α}{\cos α} = g \tan α$ or $P = (M + m) g \tan α$

Stream:JEESubject:PhysicsTopic:NLMSubtopic:Challenging application of NLM

⏱ 2mℹ️ Source: QB

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