# Box pushing against another box (conceptual)

1. Sep 8, 2008

### subwaybusker

Let's say Box A is 4 kg. It is accelerating and pushing against Box B, which is 20kg, is the force of A on B equal to the force of B on A?

My gut instinct tells me the answer should be yes, but I don't understand why the boxes haven't stopped since the action-reaction forces cancel each other out. Also, does it make any difference if Box A is accelerating or not?

2. Sep 8, 2008

### tiny-tim

Hi subwaybusker!

(you're right … this is a reaction force … so good ol' Newton's third law always applies)

Whether the boxes stop (or, to be precise, stop accelerating ) depends on the total force …

how much force is on the other side of A, making it accelerate?

3. Sep 8, 2008

### subwaybusker

but if there is more force coming from the 4kg, then wouldn't the action force be greater than the reaction force?

4. Sep 8, 2008

### tiny-tim

The box won't accelerate unless there is more force on the left.

Write out the F = ma equations for the first box, and for both boxes together, and you'll see how it works.

5. Sep 8, 2008

### subwaybusker

let's say the acceleration of both boxes is 4m/s$$^{2}$$
there is a force that is moving both boxes, so F=(24)(4)=96N
and for 20kg, F=(20)(4)=80N
so, the action and reaction forces are 80N and you can only determine it from Box B?
and the extra 16N is what is moving the boxes? and this 16N wouldn't be counted as the action force?

6. Sep 8, 2008

### Staff: Mentor

Action-reaction forces never "cancel out" since they act on different bodies.

Good. That's the net force on both boxes taken together. It's also the force that is applied to one side of box A.
This is the net force on box B, which happens to be the force exerted on it by box A.
The action-reaction forces between A and B are 80N. You could also determine it from the net force on box A. That net force must be (4)(4) = 16N. Since the applied force to one side of box A is 96N, the force from box B must be 80N in order to make the net force come out right.
The 16N is just the net force on box A. Since there's no object exerting a 16N force, it's not part of an action-reaction pair.

You can think of it like this. You need to push with 96N to accelerate those boxes. But since only 16N (net) is needed for box A, there's a force of 80N "transmitted" to box B.

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