# Why is the work done by tension in a pulley always zero?

Consider a pulley with a rope around it connecting two masses. According to this post the work done by the rope on the two mass system is 0. The explanation of this post makes sense to me; the blocks move in opposite directions so the work is 0.

However, I am struggling with what, to me, is an equally compelling solution but does not say the work is 0.

The center of mass of the two mass system moves down. Furthermore, both tension force vectors point up. So shouldn't there be negative work done by the rope?

• @AaronStevens sorry I made a typo when I said "two pulley system". I have edited to include an image for reference. – Serendipitous Epiphany May 8 '19 at 1:49
• It is not true that the work done is always zero. If your rope has mass or a non zero spring constant, then work will be done. – Paul Childs May 8 '19 at 2:19
• It depends upon the idealization of pulley and string. – Shreyansh Pathak May 8 '19 at 7:51

The center of mass of the two mass system moves down.

If you take your system to be the masses and rope, then this is due to gravity and the vertical force the pulley exerts on the system. The tension force cannot effect the center of mass because it is an internal force.

If you take your system to just the masses then there still isn't an issue because even though both tension forces have an "upward effect" on the center of mass, the total work done by these forces isn't determined by the motion of the center of mass. You have to go back to the definition of work done on each mass by each tension force separately, as covered below.

Furthermore, both tension force vectors point up.

You should look at the first part of your post:

the blocks move in opposite directions so the work is 0.

Work is given by $$W=\int\mathbf F\cdot\text d\mathbf x$$ so the work done on each mass depends on the direction of the force as well as the direction of the displacement. Each mass experiences the same upward tension force and the same magnitude of displacement but in opposite directions. Hence the sum of the work done on each mass by the rope is $$0$$.

• Why is tension necessarily internal? What if my system only contains the two blocks and not the rope? – Serendipitous Epiphany May 8 '19 at 1:58
• @TypicalHighschooler True. Then the tension force would effect the COM, but I don't see how that means the net work done by the tension force is not $0$. It's kind of a weird question here because you are technically looking at two different tension forces, even though they are equal and are related to the same string – BioPhysicist May 8 '19 at 2:00
• My reasoning for the work by the tension force not being 0 is the following. Consider this free body diagram: sketchtoy.com/68941808 which represents the forces on the center of mass. Since the center of mass moves down, the work done by the two upward force vectors should be negative (in particular, no 0). My goal with this question is to find mistake in this apparent contradiction. I think both explanations make sense. – Serendipitous Epiphany May 8 '19 at 2:13
• @TypicalHighschooler Each tension force acts on each mass separately. There are some instances where you can treat the system as a single particle at the center of mass, but this is not one of those instances. You need to think of the work done by each tension force on each mass separately, as is thought of in the second part of my answer – BioPhysicist May 8 '19 at 2:52

Don't get hung up on the movement of the center of mass. The work done by each force depends on the distance, through which, each of the forces acts.

The center of mass can move up, or down, or not at all. That doesn't matter. The work by each force is equal to that force multiplied by the distance, through which, that force acts.

The center of mass of the two mass system moves down. Furthermore, both tension force vectors point up. So shouldn't there be negative work done by the rope?

Let $$\hat y$$ be the unit vector in the upwards direction.

Consider mass $$m$$ as the system.
Mass $$m$$ has two external forces acting on it.
Force due to rope (tension) $$\vec T = T\,\hat y$$
Force due to gravitational attraction of the Earth $$m\, \vec g =- m\,g\,\hat y$$
If the displacement of mass $$m$$ is $$\vec y = y\,\hat y$$ then the work done on the mass $$m$$ by the two external forces is $$\vec T \cdot \vec y+ m\,\vec g \cdot \vec y = T\,y - m\,g\,y.$$
The tension force due to the rope does work on mass $$m$$.

Consider mass $$M$$ as the system.
Mass $$m$$ has two external forces acting on it.
Force due to rope (tension) $$\vec T = T\,\hat y$$
Force due to gravitational attraction of the Earth $$M\, \vec g =- M\,g\,\hat y$$
The displacement of mass $$M$$ is $$-\vec y = -y\,\hat y$$ and the work done on the mass $$M$$ by the two external forces is $$\vec T \cdot (-\vec y)+ M\,\vec g\cdot (-\vec y) = -T\,y + M\,g\,y.$$
Again the tension force due to the rope does work on mass $$M$$.

Consider mass $$m$$, mass $$M$$, the rope and the pulley as the system.
This system has the following external forces acting on it.
Force due to the pulley support (eg ceiling) $$2\vec T = 2T\,\hat y$$
Forces due to gravitational attraction of the Earth $$M\, \vec g + m\,\vec g=- M\,g\,\hat y- m\,g\,\hat y$$
The displacement of mass $$M$$ is $$-\vec y = -y\,\hat y$$, the displacement of mass $$m$$ is $$\vec y = y\,\hat y$$ and the displacement of the support is $$\vec 0$$.
The work done on the system by the external forces is $$2\vec T \cdot \vec 0+ M\,\vec g\cdot (-\vec y)+m\,\vec g\cdot (\vec y) = M\,g\,y - m\,g\,y$$
[Alternatively the centre of mass of the two masses undergoes a displacement of $$\dfrac{M-m}{M+m} \vec y$$ and the work done by the external gravitational force is $$(M+m)\vec g \cdot \dfrac{M-m}{M+m} \vec y=(M-m)\,g\,y$$ as before.]

It is the work done by the gravitational attractive forces due to the Earth on the masses which changes the kinetic energy of this system and in doing so the centre of mass of this system changes position.

If you want the system to be just the two masses then there are two external tension forces acting on the masses $$\vec T = T\,\hat y$$ but their displacements are $$\vec y$$ and $$- \vec y$$ so the work done by these two tension forces is $$\vec T \cdot \vec y + \vec T \cdot (-\vec y)=0$$.
The work done on the system by the other two external forces is $$M\,g\,y - m\,g\,y$$.

• Why do you do 2T * 0 and Mg* -y and mg*y when considering m, M, the rope and the pulley as the system. Shouldn't you be concerned with the displacement of the entire system, not individual components? – Serendipitous Epiphany May 8 '19 at 5:36
• @TypicalHighschooler I have added the work done in terms of the displacement of the centre of mass of the system. – Farcher May 8 '19 at 6:12

The tension, $$T ,$$ on both sides of the pully is: $$T= mg+m \alpha \,,$$ where $$\alpha =\frac{M-m}{M+m} g \,.$$

The pully is just changing the direction of $$T$$ which is equal on both sides and is being pulled down by a net force of $$2 T$$ which is supported by the ceiling.

As long as the pully is frictionless and has no mass and the rope infinitely flexible and massless with no friction, no work is done by the pully. But say you attach a spinning paper fan to the axis of pully; then, the tension $$T$$ will not be equal on both sides anymore and work will be done by the pully. Or if pully has mass and angular inertia, or friction.