# Mechanics Question- Massive Rope in a Pulley System

Homework Statement:
A load Q of 60lbs is held in equilibrium by a counterbalance P attached to the end of a rope ABC weighing 10lbs which passes around a pulley. Determine the weight of the counterbalance P, the tensions $$F_{A}$$ and $$F_{C}$$ in the end sections, and the tension in the middle section B of the rope, if (1) the ends A and C are at the same level; (2) the end A is at the highest position; (3) the end A is at the lowest position. Neglect the stiffness of the rope, the radius of the pulley, and friction.

The answers and diagram are in the following photo.
Relevant Equations:
$$\sum \vec{F}=0$$
$$\sum \vec{F}=m\vec{a}$$ I understand how they might have got to these answers but I'm still kind of shaky on how the mass of the rope plays a role in the tension at point B, and how to mathematically represent the tension at any point along the rope; I know the tension varies because the rope has mass. If I was to consider friction, what would should be done differently?

For part 1, I know intuitively that Q and P have to be the same weight because they're at the same level and the system is in equilibrium. I also know that $$\vec{F_{A}}=\vec{F_{_{C}}}=60lbs(-\hat{y})$$ because the weights Q and P are not holding up the weight of the rope.

## Answers and Replies

haruspex
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how to mathematically represent the tension at any point along the rope;
Consider a point at height y from A. What total weight is being held up by the rope at that point?

Wrt friction, I assume you mean static friction in the pulley axle, and assuming enough friction between pulley and rope that that does not slip.
This will lead to a range of values of counterbalance weight which are stable, but axle friction in a pulley is a tricky subject. There is some friction even without a load because of the tightness of fit of the pulley on its axle. Adding load means the normal force is not constant around the arc of contact between pulley and axle. Unbalanced load makes it even more complicated.
Couldn't find any good references on this.

Last edited:
• cwill53
Consider a point at height y from A. What total weight is being held up by the rope at that point?

Wrt friction, I assume you mean static friction in the pulley axle, and assuming enough friction between pulley and rope that that does not slip.
This will lead to a range of values of counterbalance weight which are stable, but axle friction in a pulley is a tricky subject. There is some friction even without a load because of the tightness of fit of the pulley on its axle. Adding load means the normal force is not constant around the arc of contact between pulley and axle. Unbalanced load makes it even more complicated.
Couldn't find any good references on this.
The total weight would be the weight of the rope at that point plus the weight of the load Q.

haruspex
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the weight of the rope at that point
Weight of what section of rope at that point?

Weight of what section of rope at that point?
That’s where I’m shaky on. I would think it would be the section below the point of interest.

haruspex
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the section below the point of interest.
Yes. It doesn't matter how long the section is above that point, it won't change the tension there. To see this, just consider the section below the point as a free body and write out the static balance of forces on it.

Yes. It doesn't matter how long the section is above that point, it won't change the tension there. To see this, just consider the section below the point as a free body and write out the static balance of forces on it.
Why would the tension at point B of the rope be 65lbs instead of 130lbs?

haruspex
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Why would the tension at point B of the rope be 65lbs instead of 130lbs?
This is a common mistake.
Tension is not a force, exactly. It is a pair of equal (or nearly so) and opposite forces acting either side of an element of the rope. The "nearly so" case is where the element has mass and is accelerating, which is not the case here.
If the tension all along a rope is T then it exerts a pull of T at whatever it is connected to at each end. If you anchor a rope to a wall and pull on one end with force F the tension along the rope is F and the rope pulls on the wall with force F.

This is a common mistake.
Tension is not a force, exactly. It is a pair of equal (or nearly so) and opposite forces acting either side of an element of the rope. The "nearly so" case is where the element has mass and is accelerating, which is not the case here.
If the tension all along a rope is T then it exerts a pull of T at whatever it is connected to at each end. If you anchor a rope to a wall and pull on one end with force F the tension along the rope is F and the rope pulls on the wall with force F.
Alright. So if I was to consider the force on the ceiling from which the pulley hangs, THAT would be 130lbs plus the negligible weight of the pulley, correct?
I’m still a bit confused.

haruspex
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Alright. So if I was to consider the force on the ceiling from which the pulley hangs, THAT would be 130lbs plus the negligible weight of the pulley, correct?
I’m still a bit confused.
Yes.

Yes.
Lol I’m sorry if I’m asking too many questions, but what you’re saying is tension isn’t the sum of the weights acting...it’s just that because the system is not accelerating, the weight on each end might as well be a wall because the rope isn’t moving because it is in equilibrium. Thus the total weight on one side is the force acting at point B at the top of the pulley. Am I getting closer?

haruspex
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Lol I’m sorry if I’m asking too many questions, but what you’re saying is tension isn’t the sum of the weights acting...it’s just that because the system is not accelerating, the weight on each end might as well be a wall because the rope isn’t moving because it is in equilibrium. Thus the total weight on one side is the force acting at point B at the top of the pulley. Am I getting closer?
Yes, but as I wrote tension is a pair of forces, not a single force. So you could consider that pair of forces acting to tear the rope apart at the top of the pulley. Each force is then equal to the sum of weights on its side.

• cwill53