# Force between movable objects versus one stationary one

1. Jun 27, 2010

### lenfromkits

Hi. I'm having trouble Googling this question.

Take two equal sized objects and call them 'left' and 'right'

a) If a force is applied for 1 second between them and they move apart at equal speeds then:

b) what happens if we double the mass of the left one? Will it move at half the speed while the right one continues at the same speed it did in scenario 'a' (relative to the center point)? (in other words, are the left and right forces are independent of each other?)

c) what happens when the left one is the Earth (ie, 'almost' doesn't move at all and ignoring gravity)? What will be the speed of the 'right' object? Will it still be the same speed as scenario 'a' or will it be travelling twice as fast as in scenario 'a'?

"intuitively" I would expect that if the left object didn't move then the right one will accelerate off at twice the speed, but "theoretically" the Earth moves a tiny amount and still absorbs that side of the force.

Last edited: Jun 27, 2010
2. Jun 27, 2010

### Staff: Mentor

You specified the forces are equal - and they have to be. By f=ma, the objects accelerates at "a" in the first scenario, the right object accelerates at a and left object accelerates at .5a in the second.
That's unclear - what does being on earth have to do with anything? What direction is the force in?
Again, you specified that the force is the same in every scenario, so the force is the same in every scenario - and it is always the same between the two objects.

3. Jun 27, 2010

### lenfromkits

Thanks, I think. There is only 'one' force involved. The 'masses' are equal. One force pushes in both directions (ie, Newton: equal and opposite).

The Earth has everything to do with it. If two objects are pushed away from each other by one force, and one of those objects is huge and 'effectively' not moveable, then how fast does the other move away compared to when the two objects are the same size (given same force)?

When I use my legs to push my boat away from a dock, I move a lot faster than when I push with the same force away from another boat that has give to it, so I don't believe you are correct in saying that the the acceleration would be the same (ie, "the right object accelerates at a ").

Last edited: Jun 27, 2010
4. Jun 27, 2010

### Staff: Mentor

f=ma. If f is the same and m is the same, then a is the same. It doesn't matter what else is being pushed.
You've changed the scenario. In all of your previous scenarios, you've specified that the force is the same in each (and said nothing at all about distance). But If you are pushing against a dock vs against another boat, then you are capable of pushing with a much higher force and for a longer distance.

5. Jun 27, 2010

### lenfromkits

I think my words are getting minced. The dock-and-boat situation is identical to an-object-pushing-away-from-earth and the two-boat-scenario is identical to one-force-pushing-two-moveable-objects.

And yes, the force is the same in all scenarios. It's not impossible to have a constant force, for example, a big spring pushing the two objects apart. I didn't really need the need to know about being 'capable of pushing with a much higher force and for longer'

Please, if you want to respond, can you private message me instead. I'm hoping to not clutter this up too much. I'm hoping someone else might be able to help.

6. Jun 27, 2010

### pallidin

OK. Take a 10lb. ball and a 5lb. ball close together.
Now separate them with x amount of force.

The 5lb. ball will move at twice the speed of the 10lb. ball.
But, guess what... the 5lb. ball will have 1/2 the force of the 10lb.ball, equaling everything out.

7. Jun 27, 2010

### lenfromkits

Thank you for such a clear response. What I'm not sure about though is if you meant that they move apart at the same speed (ie, twice as fast BUT half the force, therefore same speed). If so, that makes sense to me until we make one of those balls the size of the Earth because the Earth doesn't move when I push on it.

Why is it that one ball would have 1/2 the force? I realize that is the case with gravity but this is just a big spring pushing them apart. I'm not sure the same case applies.

Last edited: Jun 27, 2010
8. Jun 28, 2010

### Staff: Mentor

9. Jun 28, 2010

### pallidin

Hmmm... perhaps I'm using the wrong term here. My bad.
Still, there is no net gain of energy by having one of the balls flying-off faster than the other in this scenario.

10. Jun 28, 2010

### lenfromkits

Thanks Pallidin. That makes sense that they'd maintain the same energy. Ultimately I'm still wondering what speed each ball would travel at given different masses. I'm starting to think this is actually a fairly involved equation. I just can't find a word to search on to describe it (on Google).

11. Jun 28, 2010

### pallidin

Sure it does.
When you jump off the floor the earth does indeed move in the opposite direction.
Not much though, barely if even instrumentally detectable at all.

12. Jun 28, 2010

### lenfromkits

Oh right, (pallidin). I did mention in the first comment that it 'almost' doesn't move. But, I meant, it doesn't continue to move away at the same speed as a small ball, it essentially stops moving while the other smaller ball does what? same speed or double in speed or something in between?

13. Jun 28, 2010

### pallidin

For myself I must retire for the evening, but look forward to a continued discussion.

14. Jun 28, 2010

### lenfromkits

I think I might have found a clue. "elastic collisions of unequal masses." It seems to me that as two masses bounce off each other, that is pretty much the same thing as an equal force pushing them apart. This wikipedia site explains that quite well.
http://en.wikipedia.org/wiki/Elastic_collision

15. Jun 28, 2010

### Staff: Mentor

Ok, then lets use a spring instead of your legs. Your legs don't work the way you are describing. They don't work like a spring. Please understand - I'm trying to help, but all I have to go on is what you write so when I see an inconsistency, I need to clear it up.

If you are using a spring, the spring will apply the same amount of energy to the system in every scenario - same force, same distance (to the spring). But when you apply the force to a larger object (the earth), the larger object will accelerate slower and the two objects will move away from their common center of gravity. So for the boat-earth scenario, since earth barely moves (even though the force is the same as in the other scenarios) the distance through which you apply the force to the boat [almost] doubles, which means the energy you apply [almost] doubles.

That would defeat the purpose of having a forum.

16. Jun 28, 2010

### AJ Bentley

The original question needs a bit of clarification.

Take two objects and apply a force between them...
Usually that means the same force is applied (in opposite directions) to each object. Like Samson between the pillars, although it's possible that maybe you got two Samsons - one pushing each pillar?

Lets assume it's the former. And let the pillars be two round balls.
If one is heavier than the other it will be harder to move, and slower. That's what F=ma means.

If one of the balls is the earth itself - it's hardly going to move at all. But it WILL move.
If the other is a pea, then neglecting air resistance, it will fly off like a bullet.

Actually - it's difficult for Samson to apply equal force to the pea because it will fly out of his reach at the first little push and he will fall over trying - it hasn't enough inertia to let him really push hard. (Like trying to stab a pin through a piece of paper floating in the air). It's best not to involve very large or small objects - it gets complicated.

If you look at it like that, you can see that the forces left and right must be the same - can you think of a way (without using a third object) that Samson could apply more force to one than the other?

Stand in a doorway and try to push the walls apart - can you push harder with one hand than the other?

The last part (c) of your question -
If you apply a force to an object it moves. The speed depends on how hard you push. It doesn't care if you are pushing something else at the same time.
It doesn't need the earth or anything to 'absorb' the other push - it simply doesn't care.
Push is push - F=ma. Push me - I fall over. Bracing against a wall might make it easier for you to get a really good shove in - but all that does is increase the force you are able to apply.

17. Jul 1, 2010

### lenfromkits

Thanks AJ Bently. You put it very nicely with Samson. In my case it is 'one' Samson. And let's just say, one short quick force. This means that the forces on each object are not independent of each other.

It appears that as the masses of the two objects become more and more unequal, the larger one will move less from the 'center point' while the other moves more. If the heavier mass reaches such a large amount that it practically doesn't move, the other will move double what it would have from the center point compared to when the 2 masses were the same (relative to the object's point of view, it always moves away from the other at the same speed). The equation to figure out the ratios of movement based on mass, seems to be similar to the formula on this site:
http://en.wikipedia.org/wiki/Elastic_collision

I am still not exactly sure what that exact formula is though. I imagine some creative physicists somewhere have figured it out at some point since the scenario of one free-floating object pushing off another is a pretty common occurrence.

18. Jul 4, 2010

### The Dark One

If you consider the two balls to be made of the same elastic material (they get back to their original shape after the deforming force is gone) or some rigged material then the collision between them can be considered elastic with a very small and tolerable error (no error for the rigged case) and the formula is:
m1V1=m2V2

In the first case, m1 and m2 are equal and thus V1 and V2 are also equal.

In the second case, m1= 2*m2 and thus V2=2*V1.

In the third case, the elastic collision approach is invalid and a better approach would be the conservation of energy:
E= (1/2)m V2
In case one, the balls share the energy equally:
E= (1/2)m1(V1)2+(1/2)m2(V2)2=m2(V2)2 (since m1 and V1 is equal to m2 and V2 respectively)
In case three, the ball gets almost all the energy as the earth doesn’t move:
E= (energy gained by the Earth almost=0)+(1/2)m2(v2)2
And since it is the same force:
m2(V2)2=(1/2)m2(v2)2
Dropping the m2 from both sides and taking the square root of the equation:
v2= (2)1/2 V2
Which means that (the speed of the ball in case three) = (the speed of the ball in case one multiplied by the square root of 2).

Some important things to keep in mind:
1- If the two balls were not made of the same material or a not very elastic material then some of the energy will be lost in the less elastic material and the elastic collision approach may become invalid.
2- If the two balls were made of inelastic material, then they’ll probably just mesh into each other like we see in some car accidents.
3- With all respect to everyone who said that the earth will move a little, YOU ARE WRONG. The earth isn’t a rigged body and even if you hit it with force equal to that generated by an atomic bomb, it’ll still not move from its original path, instead, it will vibrate in its path causing earthquakes or terrain deformations. If you punch the ground in the garden, you’ll notice that the place where you hit sinks a little and the surrounding dirt will raise a little from its original level.