8
$\begingroup$

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

enter image description here

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:

enter image description here

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

$\endgroup$
2
  • 1
    $\begingroup$ Every post on this site is a question, so there's no need to put "question" in the title. We have a FAQ on writing good question titles. Please take a look. $\endgroup$
    – DanielSank
    Commented Oct 24, 2015 at 22:48
  • $\begingroup$ A late comment, but: Can't you just consider the rope a part of your body? In that case, the FBD contains the pulley and the upper part of the rope, positioned on top of the pulley. The whole particle system will then have two tension forces acting on it (all tension forces inside the rope are internal forces and cancel each other out, so these are the only external forces, caused on the rope by the surrounding rope), and you can then claim that the sum of the forces is $0$, since the CM has acceleration $0$. $\endgroup$
    – Max
    Commented Nov 27, 2016 at 22:19

3 Answers 3

1
$\begingroup$

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, $T_1'=T_2'$.

$\endgroup$
3
  • $\begingroup$ Thanks. The $\vec{T}$ that I drew supposed to represent the force applied by the rope on the pulley (not the rope which connects the pulley with the ceiling but the rope which is connected to the masses). The rope acts on the pulley not just from left and right but also in every point of their contact, am I wrong? Also, aren't $\vec{T_1}'$ and $\vec{T_2}'$ actually forces of friction? The rope and the pulley aren't actually connected. As I understand it - the rope just slides on the pulley, so these are friction forces (which contradicts the fact that there's no friction between them). $\endgroup$
    – user86253
    Commented Jul 19, 2015 at 9:11
  • $\begingroup$ In the statement of the problem, it is mentioned whether the friction is to be taken into account, i.e., if it is negligible or not? And also, you are interested in the static solution (balance of forces, the system does not move) or to the dynamic solution (the masses move)? $\endgroup$
    – sintetico
    Commented Jul 19, 2015 at 9:52
  • 1
    $\begingroup$ yes, the ropes and the pulleys are massless and there's no friction anywhere. I just fail to understand why the tension force acts on the pulley from the sides (tangentially) if there's no friction. Shouldn't the rope push the pulley just in one point (from the top)? $\endgroup$
    – user86253
    Commented Jul 19, 2015 at 9:59
0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ I think that user86253 wants to know on which body the forces $\vec{T_1}'$ and $\vec{T_2}'$ act and why (he claims that these forces can't act on the pulley because the system is frictionless). $\endgroup$
    – cth
    Commented Jul 19, 2015 at 13:55
0
$\begingroup$

Working on the principle that the tension is the same everywhere in the string of a frictionless pulley system makes the analysis easy. If the mass on the left is such that that it exerts a force of 10N then $$T_1 = T_1' = T_2 = T_2' = T_3 = T_3' = 10 N.$$Looking at the first pulley on the left, there is a total tension of 20N acting downward ($T_1'$ and $T_2'$). This is balanced by a tension of T = 20N in the string supporting that pulley from the roof. The second pulley has a total force of 20N acting upwards on it due to the tension of the strings either side of it ($T_2 +T_3$). This means for the system to be in equilibrium, the weight hanging off the second pulley on the right has to 20N or twice that of the weight on the left. If the right hand weight is equal to the weight on the left, then there is 20N acting upward from the tension on the stings and only 10N acting downward, so there is a nett force of 10N acting upward and the right hand pulley will have to move upward.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.