A cubical block of mass M and edge 'a' slides down a rough inclined plane of inclination θ with a uniform velocity. The torque of the normal force on the block about its centre has a magnitude of K Mga sinθ. The value of K is______

The problem involves a cubical block of mass m and edge a sliding down a rough inclined plane at an angle θ with a constant velocity.

1. Uniform Velocity: Since the block moves with a uniform velocity, there is no net acceleration. This means that the downhill force due to gravity (m g sin θ) is exactly balanced by the opposing frictional force.
2. Normal Force and Torque: The normal force is the perpendicular force exerted by the surface of the inclined plane on the block. For a cubical block of mass m and edge a, the normal force does not act through the center of the block because the block is rough. Instead, it creates a torque about the center of the cube.
3. Torque of the Normal Force: Torque (τ) is given by the formula: τ = Force × Perpendicular Distance from the axis of rotation For a cubical block of mass m and edge a, the force is the normal force, and the distance is related to how far the normal force acts from the center. The problem tells us that the torque magnitude is proportional to M g a sin θ, and the constant of proportionality is K.

Solution

For a cubical block of mass m and edge a, the normal force does not pass through the center. Instead, it acts at a distance from the center, creating a torque about the center.

The torque is given by: τ = (a/2) × M g sin θ

• a/2 is the perpendicular distance from the center to the line of action of the normal force.
• M g sin θ is the component of the gravitational force along the inclined plane.

From the equation: τ = K M g a sin θ

Comparing this with τ = (a/2) × M g sin θ, we find: K = 1/2

Final Answer:

The value of K is 1/2.