Question

A block with mass m1m1m_1 is placed on an inclined plane with slope angle ααalpha and is connected to a second hanging block that has mass m2m2m_2 by a cord passing over a small, frictionless pulley. The coefficient of static friction is μsμsmu_s and the coefficient of kinetic friction is μkμkmu_k. Find the smallest value of m2 when the blocks will remain at rest if they are released from rest.

Answer 1

Answer:

Explanation:

m₂ is hanging vertically and m₁ is placed on inclined plane . Both are in limiting equilibrium so on m₁ , limiting friction will act in upward direction as it will tend to slip in downward direct . Tension in cord connecting the masses be T .

For equilibrium of m₁

m₁ g sinα= T + f where f is force of friction

m₁ g sinα= T + μsx m₁ g cosα

m₁ g sinα -  μs x m₁ g cosα = T

For equilibrium of m₂

T = m₂g

Putting this value in equation above

m₁ g sinα -  μs x m₁ g cosα = m₂g

m₂ = m₁ sinα -  μs x m₁ cosα