What is the minimum pulling force that the clown must exert to yank his feet out from under himself?

Question

The drawing shows a circus clown who weighs 850 N. The coefficient of static friction between the clown's feet and the ground is 0.40. 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?

Answer 1

Cute problem.

Let W be the weight of the clown
Let P be the pulling force.

Then the downward force on the surface holding up the clown is D = W - P

The horizontal force needed to overcome friction is C x D, where C is the coefficient of friction (i.e. 0.4)

Since this is the same as the pulling force (assuming no friction in the pulleys)

P = C x D = C x (W - P)

You know the mass and the acceleration due to gravity so you can compute W. You also know C. So you can solve the above for P.