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We know that the center of gravity of a ring whose mass is uniformly distributed is at the geometrical center. Now if the ring is in a vertical plane and a vertical force against the gravity is applied at any point on the ring, how does the ring remain in equilibrium?( the applied force is directing opposite to the gravity)

As far as I know, the principle of transmissibility of force is applicable if two forces(same direction and same magnitude) have same line of action and their point of action is on the body (in our case its the ring),then both of them will have the same effect individually. But in the above stated case the gravity is working outside the body so we cannot expect that it will have same effect if it were acting at the point where another force,F(force applied to keep the ring in static equilibrium) is acting. So how does the force F is causing the ring to not to fall?

I considered it for a while as something hypothetical that the center of gravity of a ring being at the center is something imaginary. Even if it is something imaginary how does the applied force F is keeping the ring in equilibrium?

(My expression might not be good enough or might be something stupid, but I did not mean to put something stupid in this. I am just curious about the action of forces in this case)

Maybe my understanding is flawed. It would be a great help if someone provides me a clear and easy explanation to figure out this topic.

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2 Answers 2

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There is not one individual force acting at the center of mass. In a simple, first-order model, multiple gravitational forces are acting downward on every bit of mass in the ring. If we add all those vectors together, the net vector is the mass of the ring, $m$, times the gravitational field strength, $g$. The net line of action passes through the geometric center of the ring (uniform mass distribution).

Each bit of mass in the ring interacts with its nearest neighbors through electromagnetic forces (molecular bonding) so that the ring maintains is shape. The fact that bits of the ring are not accelerating relative to each other tells use the net internal forces are zero, so we don't worry about them. In reality, for a vertical ring, the ring distorts slightly and creates local stresses, but we ignore those in introductory physics.

If the ring is sitting on a table, the table exerts an upward force on the ring at the point of contact. If we observed the ring to not be accelerating (constant velocity or constant zero velocity), we know that (in our chosen reference frame) the vector sum of forces is zero. How can that happen? Because the intermolecular forces of the table hold it together and prevent the ring from passing through it, and result in an upward force on the ring.

The lines of actions of the forces are only important in 1) determining how to add the vectors because of the angular relationships and 2) in determining whether there is any net torque because the lines are or are not co-linear. In your case, you have specified that the normal (electromagnetic) force from the table is co-linear with and opposite in direction to the net gravitational force from the Earth.

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  • $\begingroup$ Is principle of transmissibility of forces applicable for this case too? $\endgroup$
    – MSKB
    Commented Mar 31, 2021 at 16:40
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    $\begingroup$ For purposes of calculating net force or net torque, the exact point of "touching" is unimportant. For calculating net force, only directions and magnitudes are important. For calculating torques, line of action and magnitude is important. So, yes. $\endgroup$
    – Bill N
    Commented Mar 31, 2021 at 16:47
  • $\begingroup$ Is it like some extension of the principle,since the principle states that the point of both action should be on the body? $\endgroup$
    – MSKB
    Commented Mar 31, 2021 at 17:12
  • $\begingroup$ You're being too literal about "on the body." As long as there are real forces on the structure, it doesn't matter whether the net force appears to be acting in space. That net vector can be floated and slid around as needed. The real forces, as I said in my answer, are one the body. $\endgroup$
    – Bill N
    Commented Mar 31, 2021 at 17:23
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Idk if I fully understood your question but I'll try to help.

You talked about the center being something imaginary(and it kind of is) but in this case the center is the point in which the weight of the ring lies, so even though the weight is acting upon an "imaginary" point this point is still relevant in our analysis.

So how does the force F is causing the ring to not to fall? Ans: It depends on where the force F is placed, if it is in the same direction and opposite orientation of the weight then you just add the 2 vectors together and it is pretty straight forward to see why F causes the ring not to fall. Now let's discuss another possibility: if F is not applied in the same direction of the weight, then we'll have torque, which will cause the ring to rotate until F is, again, in the same direction and in the opposite orientation of the weight, which will in turn cause the system to be in equilibrium, assuming that F

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  • $\begingroup$ The question was about the force being applied directing exactly at the opposite direction of gravity. But gravity is working at a point which is outside te ring whereas the applied force is acting on the ring so how does both of them cancel each other even though the are acting on the same line of action? $\endgroup$
    – MSKB
    Commented Mar 31, 2021 at 14:06
  • $\begingroup$ This is pretty simple: each little piece (it's up to you to divide in how many pieces you want) that composes the ring has weight, when we sum all the weights of all those little(and also imaginary) pieces we see that they add up to the total weight of the ring which, as you said, lies upon the center of the ring. But this sum of all the "little weights" is just a concept, in reality what we have is that the weight of the ring is distributed along the ring, so gravity isn't working outside the body, it's working along the body itself! $\endgroup$
    – Gabriel Oliveira
    Commented Mar 31, 2021 at 15:27
  • $\begingroup$ Is the applied force being distributed throughot the ring? or suppose that applied force is acting on one molecule of the body the other molecules are also being pulled to maintain intermolecular equilibrium. Which one is correct assumption in this case? $\endgroup$
    – MSKB
    Commented Mar 31, 2021 at 15:34
  • $\begingroup$ I'm a high schooler so take my answers with a grain of salt! I think that the second option you said is the right one, imagine for example a long bar which get's thinner towards it's tips if, now imagine that we place this bar on it's tips, if we add stress to the bar it is probably going to brake where there's less material which means that the force that holds the molecules together is smaller $\endgroup$
    – Gabriel Oliveira
    Commented Mar 31, 2021 at 16:02
  • $\begingroup$ I am also a high schooler $\endgroup$
    – MSKB
    Commented Mar 31, 2021 at 16:04

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