# Is it really the Tension in the rope that pulls the pulley?

We always solve questions by drawing two tension forces at the diametrically opposite end, to show that tension is pulling the pulley, but from the research I have done, it is clear that it is not exactly the tension force but the summation of normal forces between rope and pulley that pulls it. Then why do we not use this in real life and put 2 arrows along opposite points of the pulley? And how can I prove that the summation of normal forces is actually equal to $$2T$$?

For simplicity, assume the pulley is not rotating but the rope is slipping on it and is massless (nearly) and frictionless, and that the rope also has negligible mass.

It would be great if you were able to clear my concepts.

Consider a segment of the rope with length $$r\,\mathrm d\theta$$, where $$r$$ is the radius of the pulley: The tensions on either side of this segment are not quite antiparallel, so there is a net force with magnitude $$T\,\mathrm d\theta$$ directed towards the pulley’s center.

If we split this force into its horizontal and vertical components, and integrate over the upper half of the pulley, the total force on the pulley is

\begin{align} T_x &=-T\int_\cap\cos\theta\,\mathrm d\theta =0 \\ T_y &=-T\int_\cap\sin\theta\,\mathrm d\theta =-2T \end{align}

(If you haven’t done these integrals before, the trick is that $$\sin\theta\,\mathrm d\theta = \mathrm d(\cos\theta)$$.) You can also use this method to find the net force on the pulley if the two ends of the rope make some angle other than a half-turn.

• Wouldn't the term "coaxial" be more appropiate than "antiparallel"? Jun 15 at 5:25
• No. Neither of these force vectors has an axis. (I usually think of “coaxial” as a special case of parallel cylinders, like a coaxial cable.)
– rob
Jun 15 at 5:36
• But they do! Their axis is precisely the rope. Coaxial simply means that they share a common axis, and certainly not only cylinders have an axis. Jun 15 at 16:50
• If your were in one of my courses, I would remind your of our first-week discussion that a vector has a magnitude and a direction, but not a location. That discussion takes about eleven minutes, and doesn't fit in a comment. Furthermore, my point is that the angle between the two force vectors is nearly, but not quite, a half-turn. "Not quite antiparallel" says this efficiently.
– rob
Jun 15 at 17:19
• @ahpoblete Yes, which is why you would have another vector denoting the point of application, usually denoted as $\mathbf r$. But neither $\mathbf r$ nor $\mathbf F$ have a physical location. Just because the point of application could be important does not mean vectors have a physical location. Jun 15 at 18:21

Yes, if we have a smooth rope and a frictionless pulley, the pulley is acted on by normal forces between the pulley and the rope. These normal forces will be different all the way along the part of the rope the touches the pulley. From Newton's 3rd law there are equal and opposite normal forces on the rope caused by the pulley

The normal forces on the rope, plus the two tension forces, must add up (vectorially) to zero. If they didn't, there would be an unbalanced force on the rope, and because the rope is very light this would mean the rope would have a large acceleration.

Likewise the normal forces acting on the pulley, plus the force of the pulley mount on the pulley, must be zero, or the pulley will accelerate, moving up, down or sideways, not just turning.

It is good to see what is really happening in a system, especially in a real system. For most purposes though, we can ignore the intermediate stage and just say that the tension forces, plus the force on the pulley mount, add to zero.

Consider a small segment of rope $$AB$$ located at an angle of $$\theta$$ is in the left digram. The two tension forces $$T$$ do not act along the same line of action as shown in the middle diagram.
The segment of rope $$AB$$, which subtends an angle $$\delta \theta$$ at the centre of the pulley, is acted on by three forces, the two tension forces and the normal reaction $$N(\theta)$$ and since there is no net force on the rope segment the three forces can be drawn as a triangle of forces as shown in the right hand diagram.

If $$\delta \theta$$ is small then $$N(\theta) \approx T \delta \theta$$.

To find the net normal force on the pulley on must add together all the $$T\delta \theta$$ contributions from $$\theta = -\pi/2$$ to $$\theta = +\pi/2$$

By symmetry the $$x$$-component of the net normal force is zero.

The $$y$$-component of the net normal force is given by $$\displaystyle \int^{+\pi/2}_{-\pi/2} T \,d\theta\,\cos \theta = 2T$$ which is equal to the magnitude of the sum of the two tension forces pulling the down.