
#1
Oct707, 08:33 PM

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 [tex]\mu[/tex]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 = [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. 



#2
Oct707, 09:27 PM

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#3
Oct707, 09:37 PM

P: 73

Ok... so when he pulls down, force F also occurs in the y direction?




#4
Oct707, 09:41 PM

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The Tension Force and Equilibrium 



#5
Oct707, 09:46 PM

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.




#6
Oct707, 10:02 PM

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look like in the y direction? 



#7
Oct707, 10:21 PM

P: 73

Would it be Fnety = Fn  W  F? Using this I got
Fnetx = F  [tex]\mu[/tex](W + F) = 0 Fnetx = F  [tex]\mu[/tex]W + [tex]\mu[/tex]F = 0 Adding weight to both sides, I get F + [tex]\mu[/tex]F = [tex]\mu[/tex]W F(1 + [tex]\mu[/tex]) = [tex]\mu[/tex]W so F = [tex]\mu[/tex]W/(1 + [tex]\mu[/tex]) Is this right? 



#8
Oct707, 10:36 PM

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#9
Oct807, 05:25 PM

P: 73

Ah! I see now. Thanks for your help.



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