# The Tension Force and Equilibrium

by Gannon
Tags: equilibrium, force, tension
 P: 73 Another one... 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 $$\mu$$s = 0.41 Unknown force F Frictional Force Ff Normal Force Fn 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 = $$\mu$$Fn Fnetx = F - $$\mu$$Fn = 0 F = $$\mu$$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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 Quote by Gannon Another one... 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 $$\mu$$s = 0.41 Unknown force F Frictional Force Ff Normal Force Fn No acceleration, so Fnet = 0 in both x and y directions. Fnetx = F - Ff = 0 F = Ff Unknown F equals frictional force Ff.
Very good.
 Fnety = Fn - W = 0
here is where you are missing a force.

 I'm assuming that when the clown pulls downward, the downward force becomes force opposing the static friction.
yes, good. Correct your Fnety equation to include all forces in your free body diagram of the clown.
 P: 73 Ok... so when he pulls down, force F also occurs in the y direction?
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## The Tension Force and Equilibrium

 Quote by Gannon Ok... so when he pulls down, force F also occurs in the y direction?
Yes, do you see why?
 P: 73 I think so. He pulls down with force; this puts tension in the rope that creates an equal force to oppose the static friction.
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 Quote by Gannon I think so. He pulls down with force; this puts tension in the rope that creates an equal force to oppose the static friction.
Yes, this is the part you got correct when you looked in the x direction. What does your equation
look like in the y direction?
 P: 73 Would it be Fnety = Fn - W - F? Using this I got Fnetx = F - $$\mu$$(W + F) = 0 Fnetx = F - $$\mu$$W + $$\mu$$F = 0 Adding weight to both sides, I get F + $$\mu$$F = $$\mu$$W F(1 + $$\mu$$) = $$\mu$$W so F = $$\mu$$W/(1 + $$\mu$$) Is this right?
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 Quote by Gannon Would it be Fnety = Fn - W - F? Using this I got Fnetx = F - $$\mu$$(W + F) = 0 Fnetx = F - $$\mu$$W + $$\mu$$F = 0 Adding weight to both sides, I get F + $$\mu$$F = $$\mu$$W F(1 + $$\mu$$) = $$\mu$$W so F = $$\mu$$W/(1 + $$\mu$$) Is this right?