# The non-doubling of forces from compression/tension, is it just convention?

I drew 2 diagrams. A 10-kg ball rests on the ground or is suspended from a rope. Everything is in static equilibrium. (I drew the first one twice just to illustrate that the gravity force acts at the center of the ball, and the normal force acts at the contact point.)

Why does the scale read 10 kg at the blue points? Clearly both cases, there are two compressive forces of 98 N, or two tension forces of 98 N. So shouldn't it read 20 kg?

Is it just convention about how are scales are defined? Is it just an arbitrary historical definition? In other words, we defined weight to be whatever the scales read in these situations?

Two more things about this that I feel compelled to state explicitly, to avoid as much confusion as possible.

1. The two forces illustrated are not Action-Reaction pairs as defined by Newton's 3rd Law. That law says that Action-Reaction acts on two different bodies. (If they both acted on the same body, nothing would ever be able to move). The forces I drew are both acting on the ball, and both acting on the rope, so they cannot be Action-Reaction pairs.

2. The Normal Force is not a Reaction to Action of gravitational force. The reaction to a gravitational force is another gravitational force. The Earth pulls down the apple, and the apple pulls on the Earth. (diagram link) The reaction to the normal force is another normal force. The surface of the ball pushes on the ground, and the surface of the ground pushes on the ball. The confusion here comes from the word "reaction" because you can use it loosely and ofc the normal force would not be there if gravity did not accelerate the ball towards the ground.

P.S., I've seen the question here. It may be a duplicate but I feel like it's too specific about a case of tension. Mine is more abstract about compression or tension. And ultimately what I'm asking here is if the non-doubling of forces read by a scale is just the arbitrary definition of what scales read.

The scale measures the force that is exerted on it by the string.

The part of the scale that measures the force is positioned all the way at the top of the string. This means that, the scale measures the force that is exerted on it by the top-most particle of the string.

Consider the top-most particle on the string. It is pulled downward by tension exerted by the particle just below it, and simultaneously upward by the scale. At equilibrium, these two forces exactly cancel, so the force by scale on top-most particle = tension. And by Newton's Third Law, force by scale on top-most particle = force by top-most particle on scale. That's why the scale measures exactly the tension in the string.

Note that the scale only properly measures the tension if it measures particularly the force on the top-most particle of the string. Because this specific particle doesn't have any neighbouring particle above it that could exert an additional upward force on it, which would affect the reading by the scale.

The scale measures the force that it is applied to its mechanism. You can calculate that force using Newton's second law: $$F-mg=0$$, because the body is at rest (here F is the force done by the scale on the body, which is an action reaction pair with the force the scale feels). From here you get, in both cases, F=mg.

Why does the scale read 10 kg at the blue points? Clearly both cases, there are two compressive forces of 98 N, or two tension forces of 98 N. So shouldn't it read 20 kg?

There is no situation in which we need to separate the two forces. If the scale is not accelerating away, then besides the measured force, there must be some other compensating force or it would leave the area.

Even though they are not "action-reaction" pairs in the sense of Newton's 3rd law, the normal force is a reaction to the weight of the object. It arises from the interaction of the object and the ground/shelf that supports it.

So yes, since it won't be the case that we have 100N from above and 50N from below, we simply calibrate scales that are compressed between a pair of 100N forces to read 100N.

• Okay so what if it's not in static equilibrium anymore? What if it's in free space, no gravity, just some force of 100 N is applied to a string with a 10 kg ball at the other end? What is the tension in the string? 50 N? That seems impossible. Commented Oct 29, 2021 at 0:16
• If you're applying 100N, then the tension must be 100N. Without gravity, that just means the ball is accelerating. a = F/m. "light" strings must have tension equal to the force on each end. Commented Oct 29, 2021 at 2:10
• I'm even more confused now. So we started with 2 opposite forces of 98 N each, but the tension was 98 N. Now we remove one of the forces and yet the tension is still 98 N. That seems to imply that this is not just some arbitrary calibration convention of scales, or else it should have reduced 50% to 49 N. ...So does this have anything to do with the fact that gravity is not a "real force", but rather just some curvature of spacetime? Commented Oct 29, 2021 at 3:27
• For "light" things like a string, you can't have unbalanced forces. If you try to apply one, it will accelerate to remove the unopposed force. In this case you also have a mass that can apply a force to the other end of the string due to its inertia. So you didn't just remove one force. You removed gravity, but then you had to do work (by pulling the rope) to maintain a 100N force. Commented Oct 29, 2021 at 4:06
• For "light" things like a string, you can't have unbalanced forces. If you try to apply one, it will accelerate to remove the unopposed force. In this case you also have a mass that can apply a force to the other end of the string due to its inertia. This makes no sense to me. I said it's not in static equilibrium anymore. It is accelerating to the right so there is obviously not a net force of zero. There is no balanced force for the 100 N accelerating it to the right. Commented Nov 1, 2021 at 2:58