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In an ideal and linearly stretched rope, the tension is the same throughout its length, so that the rope won't have infinite acceleration. This means that if I draw a free body diagram of a infinitely small piece of the rope, I will have to show two tension forces of same magnitude and in opposite directions, cancelling each other.

My question is how to use this reasoning to explain that the free body diagram of a pulley in problems where the rope is stretched around one is usually drawn like the one of pulley A in this image:

enter image description here

Based on the assumptions that the ideal pulley has a circular shape and that the tensions continue to cancel each other on every point of the segment of the rope that is around the pulley, I made the following reasoning: for every point of the rope that is in contact with the pulley, there are two tension forces that cancel each other and that have a slope that is equal to the tangent of the circle determined by the pulley at that point.

However, this simply means that the rope continues to have zero net force and doesn't explain how it exerts a force of $2T$ on the pulley, which is cancelled by the force of $2T$ exerted by the clamp. So, again, how does the rope interact with the pulley and exert this $2T$ force? Can I show it exerts this force of $2T$ by some form of integration over all points of the rope that are in contact with the pulley? Am I wrong in my assumptions? I considered the possibility of the pulley exerting a normal force on the rope since this would mean, by Newton's Third Law, that the rope exerts a force on the pulley, but can a normal force exist if the rope has negligible mass?

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  • $\begingroup$ The image is from this video: youtube.com/watch?v=bCRWpw_qinY. It is just a screenshot of the free body diagrams drawn in the video for the pulleys. I am interested in the one drawn for pulley A. $\endgroup$ – ahmg Feb 28 '17 at 21:37
  • $\begingroup$ The key to doing a pulley problem is to temporarily forget about the dynamics at first, and to initially focus instead on the kinematics. Are you saying that the ropes are extensible? $\endgroup$ – Chet Miller Feb 28 '17 at 22:36
  • $\begingroup$ Well, I am considering ropes of fixed lenght. Does this means that they are not extensible? $\endgroup$ – ahmg Feb 28 '17 at 22:46
  • $\begingroup$ Yes. Now you need to determine how the upward acceleration of mass m is related to the downward acceleration of pulley B. And you have to determine how the downward acceleration of pulley C (and mass ?) is related to the downward acceleration of pulley B. $\endgroup$ – Chet Miller Feb 28 '17 at 22:50
  • $\begingroup$ Actually, the acceleration of everything in the system is zero since it is an equillibrium situation. Mass m is 10 kg as written in the image and the other mass is 40kg (the "?" mark is written before the mass is determined, then "40kg" is written after it is calculated, sorry if it caused confusion), in the video this value is found by determining that the whole system is static and that it has mechanical advantage (two movable pulleys). That is why the 10kg mass holds the 40kg mass. I apologize for the fact that the image contains more things than what I want to analyse. $\endgroup$ – ahmg Feb 28 '17 at 23:21
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If the pulley is not accelerating, or sliding against the rope, then the rope exerts a purely radial normal force on it at each point of contact. The resultant force 2T is the vector sum of these forces around the arc of contact.

enter image description here

In the diagram on the left, when the tension in the rope $T$ changes direction by angle $d\theta$ there is a resultant force $dF=2T\sin(\frac{d\theta}{2})\approx Td\theta$. This force acts on the pulley, and has components $dF_x=T\cos\theta d\theta, dF_y=T\sin\theta d\theta$.

In the diagram on the right, we see the rope continually changing direction. Over the element of arc length shown, it changes direction by $d\theta$, which is the same as the angle subtended from the centre. The force $dF$ from each element of the rope of length $ds$ has to be integrated along the arc of contact :
$F_x= \int T\cos\theta d\theta = T\sin\theta$
$F_y= \int T\sin\theta d\theta = T(1-\cos\theta)$
taking $\theta=0$ at the starting point.

If we are integrating round a semicircle then $\theta=\pi$ so $F_x=0, F_y=2T$. If we are integrating round a quarter circle then $\theta=\frac12 \pi$ so $F_x=T, F_y=T$.

These results are obtained far more simply by looking at the pulley and the rope in contact with it as a single Free Body and considering the external forces acting on it, as in your diagrams.

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I think you have two concepts wrong that might be causing a problem.

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In an ideal and linearly stretched rope, tension is the same throughout its length, so that the rope won't have infinite acceleration.

You make the assumption the tension is the same throughout the rope. This occurs when the rope is light. To understand this if a rope is heavy if it hangs from the ceiling then points higher up the rope have greater tension because they are holding up the segments of the rope below.

2. In your freebody diagrams I think your diagrams suggest all forces are equilibrium. * comment about non-equilibrium removed *

Added below

Maybe if you don't think of the tension in the rope as "One thing", say T=98.1 but instead model it as two forces T that pull in from the ends of the rope towards it centre and these are forces that the string exerts on the attached objects, the pulleys. Then transfer all the forces from the various freebody diagram back onto the original diagram, you should see that A is acted on by 2 x T down and 1 x 2T up (it's in equilibrium) and similarly for pulleys B and C. What you will end up with is 2 x 2T from the rope acting upwards on pulley C. So if the system is in equilibrium the mass (labeled "D" in the free-body diagram but not labelled in the original diagram !) exerts a force of 4T down. The magic of pulleys. Here's a standard follow up. How can the pulleys "convert" T into 4T? (Warning, it's the wrong way to think about this second question).

Personally, I put the tensions (pointing inwards on the rope) on the diagram first (It can get messy) then isolate the individual objects and the forces that act on it, and only those forces that act on it - what you call the "free-body diagram".

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  • $\begingroup$ The rope is light, that is why I used the term "ideal rope". And by "ideal pulley", I mean a massless and frictionless pulley. The diagrams are not mine, they are from a video showing a mechanical advantage system, that's why the 10kg block holds the 40kg block and the system remains stationary. What I am trying to understand is how the rope (the blue rope in the drawing) interacts with the pulley and exerts a force of 2T on it that is countered by the force of 2T from the "clamp" (the red line in the drawing). That is, why the FBD of the pulley is drawn like this? $\endgroup$ – ahmg Feb 28 '17 at 22:17
  • $\begingroup$ Ok. I think I understand this a bit better now. Is the following correct? 1) The system is in equilibrium, in fact stationary 2) Although not labelled as such in the diagram "body D" is the unknown mass 3) The aim of the exercise is to find the mass (or weight) D such that the system is in equilibrium. << Is this correct ? $\endgroup$ – Clive Long Feb 28 '17 at 22:52
  • $\begingroup$ 1) Correct. 2) "body D" has a mass of 40kg. (it is determined but the "?" is maintained in the drawing so it can cause confusion as I said in a comment below my question) 3) Yes, but I'm interested only in the FBD of the fixed pulley A. I know how to solve the exercise. I'm only using the image of the exercise because it shows the 2T from the clamp. $\endgroup$ – ahmg Feb 28 '17 at 23:46
  • $\begingroup$ I want to know how the rope exerts a force of magnitude 2T in the pulley (this force is countered by the 2T from the clamp). I'm trying to understand this in a miscroscopic level. As I said in another comment below my question, if I were to draw a FBD of every point of the rope in contact with the pulley, would I find that the forces exerted by the pulley on rope add up to 2T? And then by Newton's third law, if pulley applies 2T on rope then the rope applies 2T on the pulley. $\endgroup$ – ahmg Feb 28 '17 at 23:51
  • $\begingroup$ Maybe we have a language issue here. I say there are 3 pulleys, the objects the ropes to around. The is no force in the pulley, there are forces on the pulley. Look carefully at the diagram, there are two ropes, not one. Maybe I am misunderstanding what you are actually asking. $\endgroup$ – Clive Long Mar 1 '17 at 7:49

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