Question

Two uniform solid cylinders A and B each of mass 1kg are connected by a spring of constant 200 Newton per metre at their axles and are placed on a fixed wedge. The coefficient of friction between the wedge and cylinder is 0.2. The angle made by the line AB with the horizontal,in equilibrium is​

Answer 1

The angle made by the line AB with the horizontal,in equilibrium is​ θ = 30∘

Explanation:

• At equilibrium Torque = 0
• Therefore neglect frictional force component along the slope and equating it to the component of spring force.

In equilibrium, if θ is the required angle

• Cylinder A:

mgsin(60∘) = kxcos(60∘−θ)

• Cylinder B:

mgsin(30∘) = kxcos(30∘+θ)

= kxsin(60∘−θ) on solving

θ = 30∘

Hence the angle made by the line AB with the horizontal,in equilibrium is​ θ = 30∘

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