Pulley-rope tension question

Question

I'm learning mechanics and I don't understand the following (simple, I guess) concept. Here's a pulley-rope system:

The force of tension, as I understand it, is defined as the force exerted on an object connected to the rope/string. For example, $\vec{T_1}$ is the force of tension exerted on the object with mass $m_1$. But what are the forces $\vec{T_1}'$ and $\vec{T_2}'$? My textbook says that these are tension forces too, but I don't understand on which body they act - on the pulley? Well, there's no friction here. So the only reasonable explanation is that these are internal tension forces (which is very different from $\vec{T_1}$ for example). In this case:

1. Why do we even care about them? Don't they all cancel out along the rope (when it's massless)?

2. Why are we interested in the forces $\vec{T_1}'$ and $\vec{T_2}'$ which are acting on the rope tangentially to the pulley and not, for example, in the tension force acting on the pulley itself, that is:

Or maybe in the internal tension forces acting on the rope in some other places?

Answer 1

On the pulley on the left, there are 4 forces applied, $T_1'$, $T_2'$, the gravitational acceleration on the pulley (its weight) $m' g$ (directed downwards), and the tension of the rope at the center of the pulley $T$, which is the one that you draw, but directed upwards. Now, the tension $T$ balances the weight $mg$ and the other two tensions $T_1'$ and $T_2'$, and the pulley don't move. However, the toques of the tensions $T_1'$ and $T_2'$ may not balance, and may result in a rotation of the pulley. In fact, if $L=I\omega$ is the angular momentum of the pulley, $I$ the momentum of inertia, and $\omega$ the angular velocity, one has $$ \frac{d L}{dt} =I\frac{d \omega}{dt}=r T_1'-r T_2' $$ where $r$ is the radius of the pulley and the terms at the right side of the equations are the torques of the tension forces applied to the pulley.

If your problem is just to determine the static equilibrium of the system, and not its dynamics, you may want to assume $\frac{d L}{dt}=0$ and therefore balance the two torques $r T_1'=r T'_2$, that is, $T_1'=T_2'$.

Answer 2

If the pulleys are frictionless and the rope and pulleys are massless, then indeed the tension will be the same everywhere along the rope and you can just consider twice the tension acting on the center of the pulley.

It gets more interesting when the pulleys have mass, and there is friction (and even more when the rope also has mass...) . That is when the diagram you drew becomes more useful - there will be net torque on the pulleys so they will have both angular and linear acceleration. But without friction the problem is straightforward - no need to get confused or overthink.

Responding to the comment: $T_1$ and $T_2$ are the same magnitude (massless rope) and as such their torques w.r.t. the center of rotation cancel. But the sum of their forces acts on the center of the pulley - this is true whether there is friction or not: "the center of mass moves as though all forces acting on the body act on the center of mass". The force $T$ to the ceiling balances these two forces. If you imagine the rope making a full turn around a frictionless pulley, then if you pulled it tight you would squish (compress) the pulley but it would not rotate. With the rope halfway around, you are squishing half the pulley - and the force from the suspension to the axle of the pulley provides the counter force.