solid sphere of uniform density starts from rest and rolls without slipping down an inclined plane with angle è = 30o. The sphere has mass M = 8 kg and radius R = 0.19 m . The coefficient of static friction between the sphere and the plane is ì = 0.64. What is the magnitude of the frictional force on the sphere?

To determine the magnitude of the frictional force on the sphere, we need to consider the forces acting on it:

1. The force of gravity (weight): The weight of the sphere is given by the formula W = M * g, where M is the mass of the sphere and g is the acceleration due to gravity (9.8 m/s^2). In this case, W = 8 kg * 9.8 m/s^2 = 78.4 N.

2. The normal force (N): The normal force is the force exerted by the inclined plane on the sphere perpendicular to the plane. It is equal in magnitude but opposite in direction to the vertical component of the weight. In this case, N = W * cos(θ), where θ is the angle of the inclined plane and cos(θ) = cos(30°) = √3/2. Therefore, N = 78.4 N * √3/2 ≈ 67.8 N.

3. The frictional force (F_f): The frictional force exists to prevent the sphere from slipping down the inclined plane. The magnitude of the static frictional force can be calculated using the formula F_f = μ * N, where μ is the coefficient of static friction. In this case, μ = 0.64 and N = 67.8 N. Thus, F_f = 0.64 * 67.8 N ≈ 43.39 N.

Therefore, the magnitude of the frictional force on the sphere is approximately 43.39 N.