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Physics - NLM Question with Solution | TestHub

PhysicsNLMFBD & EquilibriumMedium2 minQB

PhysicsMediuminteger

Three identical blocks are placed at rest on a smooth inclined plane with force acting on block A parallel to the inclined plane. Find the contact force between block B and C.

$F = 15 N$

Answer:

Solution:

The three blocks move together: $a = \frac{F}{3 m} - g sin θ$

For block C along the incline: $N - m_{C} g sin θ = m_{C} a$

$N = m_{C} g sin θ + m_{C} (\frac{F}{3 m} - g sin θ) = \frac{F}{3}$

$N = \frac{15}{3} = 5$ N

Subject:PhysicsTopic:NLMSubtopic:FBD & Equilibrium

⏱ 2mℹ️ Source: QB

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