The Tension Force and Equilibrium

In summary, the drawing shows a circus clown weighing 875 N and pulling downward on a rope tied to his feet. The coefficient of static friction between his feet and the ground is 0.41. To calculate the minimum pulling force needed to yank his feet out from under himself, we need to consider the forces acting on him in both the x and y directions. In the x direction, the tension in the rope equals the frictional force. In the y direction, the normal force equals the weight and the tension force acts in the same direction as the normal force. Using the correct equation, F = \muW/(1 + \mu), we can calculate the minimum pulling force to be 359 N.
  • #1
Gannon
73
0
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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  • #2
Gannon said:
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.
 
  • #3
Ok... so when he pulls down, force F also occurs in the y direction?
 
  • #4
Gannon said:
Ok... so when he pulls down, force F also occurs in the y direction?
Yes, do you see why?
 
  • #5
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
Gannon said:
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?
 
Last edited:
  • #7
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
Gannon said:
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.
 
  • #9
Ah! I see now. Thanks for your help.
 

1. What is tension force?

Tension force is a pulling force exerted by a string, rope, cable, or other object in response to a stretching force applied to it.

2. How is tension force related to equilibrium?

In a system in equilibrium, the tension force acting on an object is equal to the sum of all other forces acting on that object.

3. What is the difference between tension force and compression force?

Tension force is a pulling force, while compression force is a pushing force. Tension force occurs when an object is being stretched, while compression force occurs when an object is being compressed or squeezed.

4. How does tension force affect the stability of structures?

Tension force can help stabilize structures by balancing out other forces and maintaining equilibrium. However, too much tension force can cause structures to collapse or become unstable.

5. How is tension force measured and calculated?

Tension force is typically measured in Newtons (N) and can be calculated using the equation T = F * cos θ, where T is the tension force, F is the applied force, and θ is the angle between the applied force and the direction of tension.

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