# Why is the moment of forces "torque" 0 here?

Why is the moment of forces "torque" 0 here?

Let's say we have this situation

A spool of mass $$M$$ rests on an inclined plane a distance $$D$$ from the bottom. The spool has a maximum radius $$R$$ and a minimum radius $$r$$, and a moment of inertia I about its axis. A long string of negligible mass is wound several times around the center of the spool and the other end is attached to the top of the incline. Find the sufficient amount of friction force so that it doesn't slip.

My thought was that it's sum of forces would be $$\sum F=ma$$ but it turns out that the acceleration in this situation is 0. Why is that? And also and my most important doubt, is that I thought that the sum of torques would never be 0 since there are external forces like the friction force, but it actually turns out that $$\sum \vec{\tau}=0$$ how is this?

Note: I'm not interested in the numeric or any way of solving the problem, just having issues seeing how the sum of forces is 0 and the sum of torques is 0.

• Why do you believe sum(torques)=0? If it were so, this thing would never move. Nov 11, 2022 at 12:03
• @GeorgeMenoutis because the solution says so, that's what I'm saying
– user348222
Nov 11, 2022 at 12:14
• Then solution is wrong. There must be a net torque, because gravity pushes down the hill COM of disk, but static friction tries to push back up-the-hill contact points of disk. Hence by definition static friction produces torque about COM,- $\vec \tau = \vec R \times \vec F_{fr}.$ Nov 11, 2022 at 12:16
• Thank you, that was all I was asking, I don't know why my question is closed since I'm asking for a concept but Oh well....
– user348222
Nov 11, 2022 at 12:21
• Voting to re-open. It seems like there are users on this site who will vote to close a question because it "looks like a homework problem" at a cursory glance, rather than actually reading the question and figuring out what's being asked. Nov 11, 2022 at 15:13

If the spool does not slip, then it does not move. Therefore the sum of the all forces on the spool must be zero and, in addition, the sum of all the torques must be zero. The friction produces a torque, as does the tension in the string, but they add to zero

• I'm having trouble imaginating how it wouldn't move, does it not rotate then as well? I thought that "not slipping" meant that $v=r\omega$ or $a=\alpha r$ would be true and that's all, essentially meaning that there is rotation but only stational friction, and not that it doesn't move at all, does the fact that it doesn't move only is due to this specific situation or not?
– user348222
Nov 11, 2022 at 22:10
• The problem says "rests" That means that the spool is not not moving. It is being held in place by the string and by friction, both applying forces that oppose the component of gravity down the slope. If the spool slipped it would slide down the slope with the string unwinding off the spool. The sppol would be rotating in the opposite direction to what it would do if there were no string and it rolled. Nov 11, 2022 at 22:49

I thought that the sum of torques would never be 0 since there are external forces like the friction force

I don't understand this sentence. External forces may produce torques. But the sum of them can be zero.

In the diagram above gravity and the normal force from the plane act, but both operate through the center of mass and produce zero torque.

Remaining are the force of friction and the force from the string. Since the problem says the total torque is zero (the spool doesn't change the spin rate), then the torques from these two force must be equal and opposite to sum to zero.

• which part of the problem suggests you that "the total torque is zero"? I'm really interested in learning from which parts of the problem I should get that idea
– user348222
Nov 11, 2022 at 22:06
• If you have a non-zero torque, then the angular momentum must be changing. but the spool is resting on the plane and not slipping. So it's not turning. The angular momentum must not be changing. Nov 12, 2022 at 0:45