Will the 2 dimensional sphere rotate?

  • #1

Homework Statement


A 200kg sphere is in touch with two walls. The horizontal wall has no coefficent of friction and the vertical has μ=0.25. If we apply a force F=400N will the sphere rotate?

Homework Equations




The Attempt at a Solution

[/B]
What I can't understand is, if there is balance in the y and x axis beacuse if so the problem is easily solved. I proved it theoritically but I can not find a mathematical solution. Please help me.
 

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Answers and Replies

  • #2
BvU
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I am missing a force (*) in your drawing. Can you post your theoretical proof ?

[edit] (*) unless the B has a specific meaning
 
  • #3
I am missing a force (*) in your drawing. Can you post your theoretical proof ?

[edit] (*) unless the B has a specific meaning
Sorry my fault, B is representing a point not a force. The theoritical proof is that in order for the sphere to rotate, the torque of F should be bigger than the torque of T. Its clear that both T and F are in the same distance from the center K which means we can now compare F and T as forces. The biggest value of T is always less than F since T=μ*N=0.25*N and N the reaction of the verticall wall is not big enough to cause balance.
 
  • #4
BvU
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torque of T
Good you mention it. I can now even distinguish a T in the picture.

I agree with your reasoning and wonder why you have difficulty with the force balances: if you write then in full (i.e. ##\ \vec a = \displaystyle \sum \vec F\ \ ##) there should be no problem. Clearly there is no acceleration to the right, so the reaction force from the wall to the left is equal to F. That gives you the magnitude of T as you used it.

[edit] the next line is based on a wrong assumption:
Note that even when ##\mu = 1 \;##, nothing happens since the sum of vertical forces won't exceed 0.

[edit] the wrong assumption being: nothing happens with ##\mu = 0.25 \;##
 
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  • #5
Good you mention it. I can now even distinguish a T in the picture.

I agree with your reasoning and wonder why you have difficulty with the force balances: if you write then in full (i.e. ##\ \vec a = \displaystyle \sum \vec F\ \ ##) there should be no problem. Clearly there is no acceleration to the right, so the reaction force from the wall to the left is equal to F. That gives you the magnitude of T as you used it.

Note that even when ##\mu = 1 \;##, nothing happens since the sum of vertical forces won't exceed 0.
Yeah
 

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