ok physic genius', here you go...?

Question

The drawing shows a circus clown who weighs 960 N. The coefficient of static friction between the clown's feet and the ground is 0.52. 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?

like, i can't even find a place to start. it is 11:30 at night and im tired and just want to finish this homework and go to sleep. please, Please help me. you don't even have to fully explain your work. i just need help.

thank you so much!

Answer 1

I'm just guessing at what the diagram looks like. I think the pulleys must redirect the rope so that the rope pulls sideways on his ankles, right?

Okay. The rope pulls sideways with a force of T (for "tension"). To make the clown slip, that sideways force needs to exceed the maximum force of friction between his floppy shoes and the floor. That maximum friction force is:

max_friction = (Fn)μ

where Fn is the normal force, and μ is the coeffiction of friction (0.52). That means, for the clown to fall:

T >= max_friction
T >= (Fn)μ

Now, you might think that "Fn" just equals the clown's weight (960N)--but there's a catch. Since the clown is pulling down on the rope with force T, the rope is pulling UP on the clown with force T. So the actual normal force is:

Fn = 960N−T

Combine this with the previous inequality:
T >= (960N−T)μ

Solve for "T":

T >= (960N)μ / (1+μ)

Answer 2

well i ahve no idea what the hell it looks like so try scanning it w/e, and get his mass, which is 960 N divided by 9.8N/kg and work from there, i would help you if i had a free body diagram but im too lazy. Good luck.

Answer 3

166.4N

3 pulleys right? If so that should be the answer.

It would help if we had a diagram. Good luck with physics.