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The drawing shows a circus clown who weighs 875 N. The coefficient of static friction between the clown's feet and the ground is 0.41. 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? The pulleys and rope are assumed to be massless and frictionless.

Picture: http://www.webassign.net/CJ/04_58.gif

W= 875N

[tex]\mu[/tex]s = 0.41

Unknown forceF

Frictional ForceFf

Normal ForceFn

No acceleration, so Fnet = 0 in both x and y directions.

Fnetx =F-Ff= 0

F=Ff

Unknown F equals frictional force Ff.

Fnety =Fn-W= 0

Fn=W= 875N

Normal Force equals 875N.

By definition,Ff= [tex]\mu[/tex]Fn

Fnetx =F- [tex]\mu[/tex]Fn= 0

F= [tex]\mu[/tex]Fn= 0.41(875N)

F= 359N

This is not correct.

I'm assuming that when the clown pulls downward, the downward force becomes force opposing the static friction. Thanks for your help.

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# The Tension Force and Equilibrium

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