BrainSalad said:
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?
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 m
1 and a
1 respectively, and the object's mass and acceleration are denoted by m
2 and a
2 respectively, Newton's third is telling you that m
1a
1=-m
2a
2. As you can see, it's impossible for a
2 to be non-zero while a
1 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+f
1=m
1a
1) is still zero, while the sum of the forces acting on the object (F+f
2=m
2a
2) isn't.