Coefficient of friction between cylinders

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  • #1
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Homework Statement


3 cylinders of mass m and radius 0.5[m]. what is the minimum coefficient of friction between them.

Homework Equations


Friction: f=μN

The Attempt at a Solution


f, the friction force, balances the torque of the weight:
$$mg\cos30^0\cdot0.5\sin30^0=\mu\cdot mg\cos30^0\cdot 0.5(1+\sin60^0)\rightarrow \mu=0.26$$
It should be 1/3
 

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  • #2
Nathanael
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Your magnitude for the normal force is wrong. This doesn't change your answer though, because the normal force cancels in your equation: [itex]F_nR\sin30°=\mu F_nR(1+\sin60°)[/itex]

I'm not sure the normal force between the bottom two cylinders would be zero. Is the ground frictional?
 
  • #3
haruspex
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Maybe I'm missing something, but I agree with your answer.
 
  • #4
haruspex
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Your magnitude for the normal force is wrong. This doesn't change your answer though, because the normal force cancels in your equation: [itex]F_nR\sin30°=\mu F_nR(1+\sin60°)[/itex]
Not if the top cylinder has mass 2mo_O
I'm not sure the normal force between the bottom two cylinders would be zero. Is the ground frictional?
If there's still a normal force between the lower cylinders, it's not about to slip. If there's no friction on the ground, it will always slip. The minimum froction on the ground is another matter.
 
  • #5
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What is the magnitude for the normal force? why isn't it correct?
The ground is frictional and was asked about it but since i didn't solve right the first stage i didn't ask yet.
Maybe the normal force between the lower cylinders accounts for the difference, i will check
 
  • #6
Nathanael
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Not if the top cylinder has mass 2mo_O
Nope, it's still wrong. That was only half the problem.
(You're saying as the 30° angle gets larger towards 90° the normal force will get smaller towards zero...?)

What is the magnitude for the normal force? why isn't it correct?
The normal force only supports half the weight of the top cylinder. The other half of the weight is supported by the other cylinder.
 
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  • #7
haruspex
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Nope, it's still wrong. That was only half the problem.
(You're saying as the 30° angle gets larger towards 90° the normal force will get smaller towards zero...?)

.
I think you're overlooking the contribution of the frictional force in supporting the weight. Resolve along the normal.
 
  • #8
Nathanael
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I think you're overlooking the contribution of the frictional force in supporting the weight. Resolve along the normal.
Indeed I was overlooking that. But I don't see how that would make the normal force mgsin(30°)? I just get a friction-dependent normal force.
 
  • #9
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I solved with the normal force between the cylinders and got μ=0.31 but the book says μ=1/3
Edit: i think there isn't any normal force between the lower cylinders since the friction between the upper cylinder and the lower ones cancel the torque and the lower cylinders don't press against each other
 
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  • #10
haruspex
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Indeed I was overlooking that. But I don't see how that would make the normal force mgsin(30°)? I just get a friction-dependent normal force.
you're right, it isn't.
 
  • #11
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now i see that i drew it wrong, the cylinders are 1.2 apart (centers), i will calculate again
 
  • #12
Nathanael
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Disregard; my thinking was flawed.

I solved with the normal force between the cylinders and got μ=0.31 but the book says μ=1/3
Edit: i think there isn't any normal force between the lower cylinders since the friction between the upper cylinder and the lower ones cancel the torque and the lower cylinders don't press against each other
For the cylinders to be in equilibrium, the torque must be zero about every point. Consider the point of contact between the top and bottom cylinder; the force of friction between the ground is causing a torque and the only torque to balance that is the normal force between the two bottom cylinders.
I haven't actually calculated anything though, so sorry if I'm misleading you.

I don't know how you calculated the normal force between the two cylinders without using this torque equation.

Also be careful about the normal force between the top and bottom cylinder.
 
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  • #13
haruspex
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For the cylinders to be in equilibrium, the torque must be zero about every point.
.
For each rigid object, if the linear forces are in balance in two different directions and the torques are in balance about some fixed point, then torques are in balance everywhere. It may be necessary to consider torques on different objects.
Karol considered torques on a lower cylinder about its point of contact with the ground. This is sensible because it eliminates the uninteresting frictional and normal force there. Torques on the upper cylinder must be in balance by symmetry.

The normal force between the cylinders is interesting, but only because of the answer (mg/2). As we've observed, its value is not important in solving this question since it cancels out in the torque equation.
 
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  • #14
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I got the same answer as haruspex and Karol: μ = 2 - √3 = 0.27

Here is what I got for the forces:
[tex]N=\frac{mg}{2}[/tex]
[tex]T=\frac{mg}{2}\frac{sin30}{1+cos30}[/tex]

I checked the normal force and the tangential force on the ground. I found that this is not where the slip would occur.

Chet

 

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