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The Tension Force and Equilibrium

by Gannon
Tags: equilibrium, force, tension
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Gannon
#1
Oct7-07, 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.
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PhanthomJay
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Oct7-07, 09:27 PM
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Quote Quote by Gannon View Post
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.
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.
Gannon
#3
Oct7-07, 09:37 PM
P: 73
Ok... so when he pulls down, force F also occurs in the y direction?

PhanthomJay
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Oct7-07, 09:41 PM
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The Tension Force and Equilibrium

Quote Quote by Gannon View Post
Ok... so when he pulls down, force F also occurs in the y direction?
Yes, do you see why?
Gannon
#5
Oct7-07, 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.
PhanthomJay
#6
Oct7-07, 10:02 PM
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Quote Quote by Gannon View Post
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?
Gannon
#7
Oct7-07, 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?
PhanthomJay
#8
Oct7-07, 10:36 PM
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Quote Quote by Gannon View Post
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?
Yes, but your equation is wrong and your math error made it right. In the y direction, the normal force acts up, the weight acts down, and the tension in the rope acts up. The tension force, F, which always pulls away from an object (or clown!), acts in the same direction as the normal force here, so its value should be plus, not minus.
Gannon
#9
Oct8-07, 05:25 PM
P: 73
Ah! I see now. Thanks for your help.


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