Question

The two blocks m = 20 kg and M = 96 kg are not attached to each other. The coefficient of static friction between the blocks is 0.40, but the surface beneath the larger block is frictionless. What is the minimum magnitude of the horizontal force F required to keep the smaller block from slipping down the larger block?

Answer 1

The minimum force required to prevent the smaller block from slipping down the larger block, given that the larger block is on a frictionless surface, is found using the static frictional force formula and is 78.4 N.

Step-by-step explanation:

To determine the minimum magnitude of the horizontal force F required to keep the smaller block from slipping down the larger block, we need to consider the forces acting on the blocks and apply Newton's laws of motion. The force of static friction f_s between the two blocks provides the centripetal force necessary to keep the smaller block from sliding down. This force can be calculated using the equation f_s = μ_s N, where μ_s is the coefficient of static friction, and N is the normal force. Since the larger block is frictionless underneath, the only horizontal force acting on it will be applied force F. The force of static friction is also what accelerates the smaller block together with the larger one.

Therefore, F must be equal to or greater than the maximum static frictional force f_s to prevent slipping. This implies that F must be greater than or equal to μ_s(mg), where g is the acceleration due to gravity.

Using the given values, the minimum force required is given by F = μ_s(mg) = 0.40 (20 kg)(9.8 m/s²) = 78.4 N.

Answer 2

F = 8 N

Explanation:

Given,

The mass of the smaller block, m = 20 Kg

The mass of the larger block, M = 96 Kg

The coefficient of the static friction, μ = 0.4

The static friction force acting on a body is the minimum horizontal force that is required to keep the smaller block from slipping down the larger block. It is given by the formula,

F = μ · η N

Where, η - normal force acting on the body

F = 0.5 x 20

= 8 N

Hence, the minimum horizontal force that is required to keep the smaller block from slipping down the larger block is, F = 8 N.