Pulley with friction, how to get the friction in the pulley?

In summary, the tension in the rope connected to the block A equals the static friction ##\sum{}^{} F_x =0 \rightarrow T=N_A*0.5=100N##, then the W weight or force equals to the tension in the pulley and the tension T ##W=100+\frac{2W}{\pi} \rightarrow W=275.2N##.
  • #1
Homework Statement
Blocks A and B have a weights of 200N and 100N respectively. If the coefficient of static friction is the same for the ground and blocks ##\mu_s =0.5##. If the pulley is rough with coefficient of friction ##\mu =\frac{2}{\pi}##. Determine the the largest value of the weight W without causing motion.
Relevant Equations
##F_s=\mu_s*N##
##\sum{}^{} F_y =0##
##\sum{}^{} F_x =0##
First, I assumed that the tension in the rope connected to the block A equals the static friction ##\sum{}^{} F_x =0 \rightarrow T=N_A*0.5=100N##, then the W weight or force equals to the tension in the pulley and the tension T ##W=100+\frac{2W}{\pi} \rightarrow W=275.2N##
It's the first problem I have with friction in the pulley so was that correct? I feel uncomfortable with it.
 

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  • #2
I am unsure what the question means by "pulley friction".
If it means friction between rope and pulley we should ignore it. It is only axle friction that will matter.

If it means axle friction then we can think of the axle as a fixed cylinder with the pulley wheel as a rigid ring of slightly larger radius sliding around it. Ignoring the weight of the ring, where will the point of contact between ring and axle be? What will the normal force be there?

A third possibility is that they don’t mean pulley. Rather, it is just a fixed cylinder with the rope sliding over it. The normal force and tension vary around the curve. Are you familiar with capstan analysis? https://en.wikipedia.org/wiki/Capstan_equation. The given coefficient of friction seems crafted to suit this meaning.

You should also note that the question specifies no motion. That is not just no sliding.
 
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  • #3
haruspex said:
I am unsure what the question means by "pulley friction".
If it means friction between rope and pulley we should ignore it. It is only axle friction that will matter.

If it means axle friction then we can think of the axle as a fixed cylinder with the pulley wheel as a rigid ring of slightly larger radius sliding around it. Ignoring the weight of the ring, where will the point of contact between ring and axle be? What will the norm force be there?

A third possibility is that they don’t mean pulley. Rather, it is just a fixed cylinder with the rope sliding over it. The normal force and tension vary around the curve. Are you familiar with capstan analysis? https://en.wikipedia.org/wiki/Capstan_equation. The given coefficient of friction seems crafted to suit this meaning.

You should also note that the question specifies no motion. That is not just no sliding.
This seems much more complicated than what I thought
But this should be solved simply my curriculum has no utterance about that stuff
 
  • #4
The question asks for the maximum W before motion occurs. The possibility of tipping before sliding should also be examined.
 
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  • #5
kuruman said:
The question asks for the maximum W before motion occurs. The possibility of tipping before sliding should also be examined.
I tried it but it will only affect the numbers, won't it?
 
  • #6
YehiaMedhat said:
I tried it but it will only affect the numbers, won't it?
Which numbers? What matters is the maximum tension in the part of the string that is attached to the block. Calculate what it should be (a) if the block is to slide before tipping and (b) if the block is to tip before sliding given the coefficient of static friction and block dimensions. Pick the smaller of the two and then find how much weight ##W## you should hang at the other end of the string in order to get that much tension.
 
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  • #7
YehiaMedhat said:
I tried it but it will only affect the numbers, won't it?
But it's the 'numbers' that you want!!! To spell it out a bit...

The question says ‘without causing motion’. There are three ways 'motion' could start:
1. Block A tips (pivots about its bottom right corner).
2. Block A slips alone.
3. Blocks A and B slip together.

Your first task is to calculate the smallest force (tension in the horizontal part of the rope) for each of the above and decide which one is relevant.

Your second task is to find W. I think that using the capstan equation (as already mentioned by @haruspex) is expected.
 
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  • #8
Steve4Physics said:
But it's the 'numbers' that you want!!! To spell it out a bit...

The question says ‘without causing motion’. There are three ways 'motion' could start:
1. Block A tips (pivots about its bottom right corner).
2. Block A slips alone.
3. Blocks A and B slip together.

Your first task is to calculate the smallest force (tension in the horizontal part of the rope) for each of the above and decide which one is relevant.

Your second task is to find W. I think that using the capstan equation (as already mentioned by @haruspex) is expected.
Ok, I'll do
Here's what I got:in the first case (Block A tips):##\sum{}^{} M_{right\ bottom\ corner} = 0 \rightarrow 200*1=4*T\ (tension\ in\ the\ rope\ connected\ to\ the\ block) \rightarrow T=50N##
The second case (block A slips alone): ##\sum{}^{} F_x =0 \rightarrow 0.5*200=T \rightarrow T= 100N ##
The third case (both blocks slip): ##\sum{}^{} F_x=0 \rightarrow T=0.5*N_A+0.5*(N_A+N_B) \rightarrow T= 250N##
The third case is the right one because the T is the largest, isn't it?
But for the capstan equation, I'm about to study it.
 
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  • #9
I have not verified your numbers but there is a flaw in your logic. Assuming that your numbers are correct, say you start increasing the weight ##W## slowly from zero to a value that causes motion. Will the tension ever reach 250 N?
 
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  • #10
kuruman said:
I have not run the numbers but there is a flaw in your logic. Say you start increasing the weight ##W## slowly from zero to a value that causes motion. Will the tension ever reach 250 N?
This what the numbers got me, so, what's wrong?
 
  • #11
What is that thing shown on top of block A?
Wouldn’t that block tip over before sliding?
Why is pi in the value of the pulley’s coefficient of friction?
 
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  • #12
Lnewqban said:
What is that thing shown on top of block A?
Some hinge from another problem
Lnewqban said:
Wouldn’t that block tip over before sliding?
Ok, so that's my mistake it should tip first because the force required for making tip is much smaller
Lnewqban said:
Why is pi in the value of the pulley’s coefficient of friction?
Something circlish, more precisely to let the angle which the rope is touching the pulley over fit in the capstan law well, as the coefficient is multiplied by the angle, Am I wrong?
 
  • #13
Ok, the friction in the pulley is going to be (by capstan equation & assuming block_A will tip over): ##W=T*e^{\frac{2}{\pi}*\frac{\pi}{2}} \rightarrow W=50*e=135.91N##
Please, tell me this is correct :')
Note: I edited the angle which I substituted in the calculations as Mr. @Steve4Physics had made me notice, how embarrassing!
 
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  • #14
YehiaMedhat said:
The third case (both blocks slip): ##\sum{}^{} F_x=0 \rightarrow T=0.5*N_A+0.5*(N_A+N_B) \rightarrow T= 250N##
For information, that's wrong. If A an B slide together, A doesn't move relative to B. We must treat A and B as a single (30kg) object, ‘AB’.

In that case, the frictional force of A on B is exactly equal in magnitude and opposite in direction to the frictional force of B of A (Newton's 3rd Law). These two internal forces cancel so they do not affect the overall motion of AB.

P.S. What is 90º in radians?
 
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  • #15
YehiaMedhat said:
Ok, the friction in the pulley is going to be (by capstan equation & assuming block_A will tip over): ##W=T*e^{\frac{2}{\pi}*\frac{\pi}{2}} \rightarrow W=50*e=135.91N##
Please, tell me this is correct :')
It has too many significant figures. Especially as you have used an approximate value for g. Is g = 10m/s² satisfactory or are you meant to use a more precise value?
 
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  • #16
YehiaMedhat said:
Some hinge from another problem

Ok, so that's my mistake it should tip first because the force required for making tip is much smaller

Something circlish, more precisely to let the angle which the rope is touching the pulley over fit in the capstan law well, as the coefficient is multiplied by the angle, Am I wrong?
Perhaps it would tip over before sliding, I don’t know, but there must be a reason for the 4 m dimension from surface to string point of connection and the 2 m base width.

I can’t make sense of the pi, because the perimetral pulley throat-string friction would consume energy but would not reduce the force transferred by the pulley, as the pulley, being a pulley rather than a fixed cylinder against which the string would rub, is free to rotate (considering the axle friction provided in the problem).
 
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  • #17
Steve4Physics said:
It has too many significant figures. Especially as you have used an approximate value for g. Is g = 10m/s² satisfactory or are you meant to use a more precise value?
It was stated to use g as approximation, it's not me. But the rest now is correct?!
 
  • #18
YehiaMedhat said:
It was stated to use g as approximation, it's not me. But the rest now is correct?!
EDIT - see next post
Numerically correct. But (in

In an exam' say) I would deduct 1 mark for an inappropriate number of significant figures. You have given the answer to 5 sig. figs. - but the data have a precision of only 1 sig. fig. What do you suggest as a better answer?

Without seeing the working for the whole problem in full, I am unable to to tell you if you would receive full marks for method and clarity of explanation (if the mark-scheme allocates marks for these).
 
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  • #19
@YehiaMedhat, when using the capstan equation, care is needed to get the 'load' and 'hold' tensions the right way round. I'm not sure you have. but I need to think about it (as should you)!

Edit - sorry, I think it's ok.
 
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  • #20
Steve4Physics said:
@YehiaMedhat, when using the capstan equation, care is needed to get the 'load' and 'hold' tensions the right way round. I'm not sure you have. but I need to think about it (as should you)!
Is that also wrong? How the W on the other sould be greater so it makes sense to put on that side.
 
  • #21
YehiaMedhat said:
Is that also wrong? How the W on the other sould be greater so it makes sense to put on that side.
Sorry for any confusion. I confused myself (not hard!). Your Post #15 look ok.
 
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  • #22
Lnewqban said:
I can’t make sense of the pi, because the perimetral pulley throat-string friction would consume energy but would not reduce the force transferred by the pulley, as the pulley, being a pulley rather than a fixed cylinder against which the string would rub, is free to rotate (considering the axle friction provided in the problem).
Not sure what you are saying there.

As I noted in post #2, there are three possible interpretations of the given pulley friction.
  • Judging from the form of the coefficient quoted, they don't mean a pulley at all. Rather, it is just a capstan.
  • If it is for static friction between pulley and rope, the axle being frictionless, we can ignore it.
  • If it means axle friction we need a model for how the pulley's annulus contacts its axle. I would opt for a normal force at ##\pi/4## to the vertical.
I see no interpretation in which the tension is unaffected by static friction and yet, if motion of the block occurs, the friction would consume work.
 
  • #23
haruspex said:
Not sure what you are saying there.

As I noted in post #2, there are three possible interpretations of the given pulley friction.
  • Judging from the form of the coefficient quoted, they don't mean a pulley at all. Rather, it is just a capstan.
  • If it is for static friction between pulley and rope, the axle being frictionless, we can ignore it.
  • If it means axle friction we need a model for how the pulley's annulus contacts its axle. I would opt for a normal force at ##\pi/4## to the vertical.
I see no interpretation in which the tension is unaffected by static friction and yet, if motion of the block occurs, the friction would consume work.
Exactly!
The text of the problem states “the pulley is rough”.
The diagram shows some hatching along the perimeter of rope contact (as well as around the axle).
I understand that as the second case you have listed, which makes the value of in and out tensions alike.
But then, the same text gives a value of μ that includes π - why?
 
  • #24
Lnewqban said:
Exactly!
The text of the problem states “the pulley is rough”.
The diagram shows some hatching along the perimeter of rope contact (as well as around the axle).
I understand that as the second case you have listed, which makes the value of in and out tensions alike.
But then, the same text gives a value of μ that includes π - why?
It seems very likely that the question is simply badly written: ‘pulley’ should be ‘fixed pulley-wheel’ or ‘capstan’.

If you are not familiar with the ‘capstan equation’, it applies to a rope wrapped around a capstan (fixed cylinder). If the subtended angle of contact is ##θ##, then the equation relates the ‘load tension’ and the ‘hold tension': ##T_{load} = T_{hold} ~e^{μθ}##.

In this question ##μ = 2/π## and ##θ = π/2## look like deliberate choices to facilitate use of the equation.
 
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