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In Chapter 19 of Feynman's Lectures on Physics, Volume I, it says

...where can we apply a single force to balance the gravitational force on the whole thing, so that the entire object, if it is a rigid body, will not turn? The answer is that this force must go through the center of mass..."

What I don't understand is, what does it mean for the force to "go through the center of mass"? For example, if we are holding up an object, we exert the force on the side of the object instead of its center. But the object does not rotate, so I assume that the gravitational force on it should be balanced. Then, what exactly counts as "going through the center of mass"?

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A force "going through the centre of mass" is acting along a line that passes through the centre of mass.

The weight pulls in the centre of mass as well, and thus these two forces cancel each other out (if they have the same magnitude) and cancel each other's torques out about every point.

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For an object, mass is usually evenly spread out. But for the purposes of theoretical calculations, we take there to exist a centre of mass, where all the mass is at.

Without anything holding it, it will fall down due to its weight, which originates from its centre of mass, vertically downwards.

Hence, if we are able to exert a force exactly opposite to this, which is also vertical, but upwards. Notice that it will cut through the centre of mass.

Now when we use our hands to hold a object, while it may seem like you are exerting a force on the sides only, your palm is also exerting a force through the object’s centre of mass, such that it is stationary. Other forces such as friction exist to balance out the forces acting on the object too.

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  • $\begingroup$ "your palm" is not relevant. It is possible to hold an object stationary on your finger tips in such a way that each of your fingers actually does exert its force "on the sides only." $\endgroup$
    – Solomon Slow
    Commented Jul 22, 2018 at 12:34
  • $\begingroup$ That’s why I raised the possibility of friction that can hold it in place too. $\endgroup$
    – QuIcKmAtHs
    Commented Jul 22, 2018 at 13:18
  • $\begingroup$ Unfortunately, it's more complicated than that. Imagine a horizontal beam, resting on two knife edges, one at either end. There is no friction to speak of, because no surface slides against any other surface. I am not a physicist, but I think that if you want a simple analytical model of how the forces hold the beam in place, you might start here: en.wikipedia.org/wiki/Rigid_body_dynamics $\endgroup$
    – Solomon Slow
    Commented Jul 23, 2018 at 11:33
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For ease in solving problems.

IRL, gravitational force is acting b/w the earth and every single particle of the object (although the individual magnitudes of each of these forces will be very small).

But, when solving problems, it will be difficult to deal with such a large number of small forces. So, we replace them all by a single force which has an equivalent effect.

For an analogy in electricity — let's say we have millions of small resistors connected in series and we want to find the voltage across one end of the first one and the other end of the last/millionth one. We can do this in two ways :

  1. Find the p.d. across the first resistor, then across the second one and so on until we reach the final one, and then finally add all of the small voltages.

  2. Somehow replace them all by a single, equivalent resistor, and then find the p.d. across it in one step.

Which method is easier?

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The quotation is addressing the case of a single applied force to counteract gravity. If you are holding an object by its edge against gravity you are applying more than one force - for example you are pulling down at the edge and pushing up closer to the centre of mass.

You cannot hold the object against gravity using the tip of a single finger at the edge of the object. You can only do so if the centre of gravity is balanced above or below your finger. Then you are pushing or pulling through the centre of mass.

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