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 calculate the minimum pulling force, we need to consider the forces acting on the clown and use the concept of static friction.
First, let's identify the forces acting on the clown. We have the weight of the clown acting vertically downward, which is 875 N. We also have the normal force acting upward, which is equal in magnitude to the weight of the clown (since the clown is not accelerating vertically). Additionally, there is the pulling force applied to the rope.
The clown will be able to pull his feet out from under himself when the static friction force between his feet and the ground is overcome. The static friction force equation is given by:
Fs = μs * N
Where:
Fs is the static friction force
μs is the coefficient of static friction
N is the normal force
In this case, the normal force is equal to the weight of the clown, which is 875 N.
Substituting the given values into the equation, we have:
Fs = 0.630 * 875 N
Now we can calculate the static friction force:
Fs = 551.25 N
Therefore, the minimum pulling force that the clown must exert to yank his feet out from under himself is equal to the static friction force, which is 551.25 N.