Math Is Fun Forum

paulb203

Member

Registered: 2023-02-24

Posts: 321

Action and Reaction Forces

Newton’s 3rd Law

For every action there is an equal and opposite reaction.

I push a box with a force of x Newtons, the box pushes me back with a force of x Newtons.

Let’s say both have a mass of 90kg.

Q. If the forces are equal, but in opposite directions, as stated above, why is it that if I try hard enough I can move the box across the floor? If I increase the force of x Newtons on the box, the force back on me from the box will increase to match it, yeah?

Q. How is the above different from me trying to push a man of 90kg across the floor? Is it different? I push the man (who is relaxed, not trying to push me back), he pushes back on me.

Q. How is the (directly) above different from me pushing the man but this time he is pushing back with the same effort?

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Prioritise. Persevere. No pain, no gain.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,626

Re: Action and Reaction Forces

Hhhmmm, it's more complicated than that. The reason you can move a box at all is because of friction between your feet and the floor. If you tried to push a box whilst floating weightless in space you couldn't do it because you've got nothing to push against. What would happen is you and the box would separate and float apart ( conservation of momentum ).  You'd have the same problem if you were standing on ice.

So your force = the boxes reactive force. That is transmitted through your body to the point of contact with the ground, and, if the friction is sufficient to oppose this, then the box moves. Of course, if it's on a friction surface, you've got to overcome that too.

If you're battling someone who is also trying to push you, then it's largely down to each persons friction as to what happens. If you're on a rubber mat and he's on a sheet of ice, then you'd win regardless of who is stronger. If your surfaces are the same then one or both of you might slide, or you might fear you're about to topple backwards and take a step back.

The starting point for any problem in statics is to make a force diagram.  Put in just the forces that are acting on one object. If nothing is moving then the components in two (usually perpendicular) directions can be equated.  If there's not equilibrium then you can use the first and second laws to work out the acceleration.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

paulb203

Member

Registered: 2023-02-24

Posts: 321

Re: Action and Reaction Forces

Thanks, Bob

“If you tried to push a box whilst floating weightless in space you couldn't do it because you've got nothing to push against. What would happen is you and the box would separate and float apart ( conservation of momentum ).”

If I was 100kg and the box was 50kg, and the force was 200N, would the following apply, in space?

Box;
F=ma
200N=50kg(a)
a=200N/50kg
a=4m/s^2

Me;
200N=100kg(a)
a=200N/100kg
a=2m/s^2

“So your force = the boxes reactive force. That is transmitted through your body to the point of contact with the ground, and, if the friction is sufficient to oppose this, then the box moves.”
The reactive force is transmitted through my body to the point of contact with the ground? A force, from the box, transmits through me, to the ground, yeah?
I’ve been told the normal force is an electrostatic force, e.g, the electrons around a mug on a table are being repulsed by the electrons on the table surface( this is why the mug doesn’t fall through the table); does this have anything to do with the reaction force? I’m thinking that the box, in a sense, is just sitting there, so how can it push me back, when I push it.

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Prioritise. Persevere. No pain, no gain.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,626

Re: Action and Reaction Forces

Your acceleration calculations look ok. Just one thing bothers me. Does it make a difference if the force is applied for a measured amount of time. I've searched for an answer to this and made no progress.  Why ask this question?  If you apply a force of 200N for, say, 5 seconds then the impulse is 200 x 5 = 1000 N.sec and this gives that amount of momentum to the two objects.

Momentum before = 0

Momentum after = 1000

If I consider myself ( as the pusher) as at rest with the box moving away at velocity v after 5 seconds then

50 x v = 1000 so v = 1000/50 = 20 m/s

Using v = u + at     20 = 0 + 5a     so a = 4m/s^2

Oh joy! That's what you got before. So maybe all's well in the world of Newtonian mechanics.

I’m thinking that the box, in a sense, is just sitting there, so how can it push me back, when I push it.

Do you drive yet? Don't try to experience this 'box' effect by driving a car into a tree. The tree will definitely exert a big force on you!

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

paulb203

Member

Registered: 2023-02-24

Posts: 321

Re: Action and Reaction Forces

Thanks, Bob.

Glad you mentioned the tree experiment. Good timing. I've crossed it off today's to-do-list

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Prioritise. Persevere. No pain, no gain.