# Newton's Third Law Explanation

1. Nov 1, 2012

For every force exerted by object A on object B, an equal and opposing force is exerted by object B on object A.

I need help understanding the principle. I understand it when dealing with a completely stationary object: pushing on a solid wall, it exerts an equal force back to your hand, resulting in no net movement. But what about when I push something and move it? It seems to me that the force of my hand is greater than the force of the pushed object, since it "gives" under the force and moves. How are the two forces equal?

2. Nov 1, 2012

### mikeph

You feel a force acting on your hand when you push the object- the force does exist and it is equal in magnitude to the force felt by the object. If you were floating in space you would accelerate away from the object.

Since you're standing on a planet, you can transmit the force through your shoes to the Earth, and the effective mass of you as one large object is huge. So the Earth will slightly accelerate away from the object, but the motion it will be immeasurably small in magnitude.

3. Nov 3, 2012

### Jespermj

Consider Newtons second law F=ma. Imagine you exerting a pushing force on, say, a balloon. The balloon will of course exert the same magnitude of force on you. But because the mass of the balloon is relatively small it must "compensate" by having a greater acceleration. You, having much greater mass, will feel a very small unnoticeable acceleration.

4. Nov 7, 2012

### nicklasaau

If the object you are pushing arent giving you the same amount of force back, you would be compressing the object, the object needs to excert the same force to your push, if it dosent compress. (without talking about friction) an example if a ball lies on the table, the gravitational force acts on the ball, if the ball needs to stay on the table, the table have to create a normal force that has the same magnitude and opposite direction of the gravitational force, else the ball would have a net internalforce thru the table and the table would break, same principle for the push, no compression.

I think you are mixing the internal forces with the external forces, the external forces is things like the gravitation and a push, but the external forces is the forces that are inside an object (newtons 3 law) which is the force that the object need to deliver, for not to be compressed. Internalforces must be 0 for no deformation/compression (newtons 3 law), and the external forces is the ones that cause the motion.

5. Nov 7, 2012

### Fredrik

Staff Emeritus
When no other forces are relevant, like when you and the object are on a frictionless surface or floating in empty space or something, it's impossible for you to push something without changing your own velocity. If your mass and acceleration are denoted by m1 and a1 respectively, and the object's mass and acceleration are denoted by m2 and a2 respectively, Newton's third is telling you that m1a1=-m2a2. As you can see, it's impossible for a2 to be non-zero while a1 is zero.

When other forces (like friction) are relevant, Newton's third still holds, but that last equality doesn't. Suppose e.g. that you put your hands against your kitchen table and push it gently. Nothing happens, so you slowly increase the force. For a while nothing is happening, and then suddenly the table begins to accelerate. What happens here is that as you're increasing the force that you exert on the table, the force exerted by the floor on the table also increases, and remains equal in magnitude to the force you exert on the table until you finally push harder than the floor is able to push back. Then the table will experience an acceleration given by Ma=F-f where M is the mass of the table, F is the force that you exert on the table, and f is the force that the floor exerts on the table once it has started moving.

This idea applies both to you and the object you're pushing. If you push the table and it moves while you remain stationary relative to the floor, what has happened is not that Newton's third has failed. It's just that you're pushing with a force F that's greater in magnitude than the maximum force that the floor can exert on the table, but smaller in magnitude than the maximum force that the floor can exert on you. Because of this, the sum of the forces acting on you (-F+f1=m1a1) is still zero, while the sum of the forces acting on the object (F+f2=m2a2) isn't.