A circus clown weighs 875 N. The coefficient of static friction between the clown's feet and the ground is 0.630. He pulls vertically downward on a rope that passes around three pulleys and is tied around his feet. What is the minimum pulling force that the clown must exert to yank his feet out from under himself
To determine the minimum pulling force required to yank the clown's feet out from under himself, we need to consider the forces at play.
1. Weight of the clown: The weight of the clown is given as 875 N and acts vertically downward.
2. Static friction force: The static friction force between the clown's feet and the ground opposes any attempted motion. It is calculated using the coefficient of static friction, which in this case is 0.630.
To calculate the static friction force, we use the equation:
Frictional Force (F) = Coefficient of Static Friction (μ) * Normal Force (N)
The normal force is equal to the weight of the object in this case, so the normal force (N) acting on the clown is 875 N.
Now, we can calculate the static friction force:
F = 0.630 * 875 N
F = 550.125 N
This is the maximum force of static friction that can be exerted on the clown's feet before they start to move.
However, since the clown wants to yank his feet out from under himself, he needs to overcome this static friction force. Therefore, the minimum pulling force required would be equal to the static friction force:
Minimum Pulling Force = 550.125 N
Thus, the clown must exert a minimum pulling force of 550.125 N to yank his feet out from under himself.