# Force and point of application

1. Jun 10, 2012

### luis20

Hi everyone :). I have one question about classical mechanics, which is illustrated in the attachment.

The question is: for rigid bodies, how can the same external force F produce more overall movement in situation A) than in situation B)? This seems to contradict the conservation of energy!

The same external force F is applied in situations A) and B). Since the point of application in A) is away from the center of mass C, we can move that force to C and add a torque T (it's the same).

But if I had applied this force F in the center C in the first place (situation B) I wouldn't have this extra torque!

Conclusion:
The sum of acceleration/velocity/distance travelled of all points of the body is higher in A) than in B). So we give more cinetic energy if we apply this force F away from the center of mass C?

Thanks in advance and I'm sorry if my english isn't the best ^^

#### Attached Files:

• ###### situation.png
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2. Jun 10, 2012

### haruspex

In A, the applied force has progressed a greater distance, so has put more energy into the system.

3. Jun 10, 2012

### luis20

Yes, but that force was applied in the same amount of time. So that force should give the same amount of velocity to the body's points in both situations. But total velocity is bigger in A), why?

Edit: About the "greater distance", that might be wrong. The distance is bigger sure, but some of that distance has a differente direction from the force (force has constant direction), so that doesn't give any energy to the system.

4. Jun 10, 2012

### haruspex

Energy is force times distance (in the same direction). Force times time is momentum. So in A the force has imparted the same linear momentum, but more energy (since some of the energy has gone into rotation).

It isn't entirely clear from your picture what r2 measures. I took it as distance in the direction of the force.

5. Jun 10, 2012

### ecneicS

In A), you can't move the force to C, because then what force would supply the torque? In other words, torque costs force.

But what I don't know is how much of the force is given to create rotational motion and how much is given to create linear motion. It certainly will depend on time.

EDIT: Take what I said with a grain of salt, I'm confusing myself.

Last edited: Jun 10, 2012
6. Jun 10, 2012

### haruspex

I agree with that part

7. Jun 11, 2012

### Studiot

Diagram A is impossible as drawn.

That is because it is missing the pivot reactions necessary to create rotation.

8. Jun 11, 2012

### luis20

Pivot reactions? What do you mean? :O

9. Jun 11, 2012

### luis20

That would make sense to me. But in classical mechanics, any force in a point A can be moved to a point B and a torque must be added. That's the problem :\

10. Jun 11, 2012

### Studiot

You have shown the rigid body in A rotating, presumably because of the moment of F about some fixed point.
Label the ends and centre of the body a, b and c. F is applied at c and a is the rotation centre.
This rotation cannot happen unless there is a reaction force at this fixed point (a).

It does not matter where you apply a single force F to a rigid body that is in free space, you will get the same effect. The only effect you will get is a linear acceleration in the direction of the force. You will not get a rotation.

Last edited: Jun 11, 2012
11. Jun 11, 2012

### luis20

Are you sure? If I hit a pencil in one of the ends the pencil goes forward and rotates. The only point of the pencil who doesn't rotate and only goes forward is its center of mass!

The same happens even if I do this in the air, with low friction. Where is the reaction force?

Edit: In vacuum there wouldn't be rotation? Also: I didn't fix any point in the illustration, the body moves forward and rotates

Last edited: Jun 11, 2012
12. Jun 11, 2012

### Studiot

There is always a reaction point on Earth, that is why I mentioned space. You can't spin a space rocket with one 'jet'.

So back to your diagrams A and B, in the absence of any other forces it does not matter where you apply F on your body the result will be the same. Linear motion up the paper.

13. Jun 11, 2012

### luis20

Wait, wait!

(begin at 14:40)

He says the body actually rotates. This is from MIT :O

Last edited by a moderator: Sep 25, 2014
14. Jun 11, 2012

### Studiot

To understand how the turning couple arises (on Earth)

Place a book flat on a table with its spine towards you.
Since gravity is a right angles to the table top gravity does not affect this experiment.

No push the book away gently from you by pushing the spine with one finger in the middle of the book.

The book slides away over the table without rotation.

Now repeat moving the point of pushing with the finger a little towards one end of the book.

If you repeat this several times you will come to a point where the book rotates like your pencil as well as sliding forwards.

The closer to one end you push the more rotation and the less translation you will note.

Why?

Well the distance from the push point to one end is equal to the distance from the push point to the other when you push in the middle.

If you imagine the resistance to motion distributed in little arrows all along the spine, the moment of that resistance force to the right of the push force is of the opposite direction to the moment generated by the push force to the left.

When the distance are not equal the moments are not equal. So the book rotates.

#### Attached Files:

• ###### spin1.jpg
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Last edited: Jun 11, 2012
15. Jun 11, 2012

### Studiot

I note your video example applies an impulse not a sustained force?

16. Jun 11, 2012

### luis20

Yes! That's it. What is the difference? If a single impulse rotates the body, a single force (which will have to turn to an impulse to do anything) will also rotate the body.

So maybe a reaction force in the pivot point is not needed. At least is what that professor says. What do you think?

17. Jun 11, 2012

### luis20

I think if I apply the same force/impulse the translation should be the same, regardless the distance to one end. Like that professor stated.

This wouldn't work with round objects (because there is the wheel effect, but I think doesn't matter).

Last edited: Jun 11, 2012
18. Jun 11, 2012

### Staff: Mentor

If the single force creates a torque about the center of mass, then you'll get both rotation and translation.

19. Jun 11, 2012

### Studiot

Yes I was wrong about that.

20. Jun 11, 2012

### luis20

Studiot, read my last reply. See if you can understand what I thought was true but now I see it isn't !!! :P

Edit: I will put it here:

If that professor is right, than I can create more overall movement with less movement. I intuitevely thought this wasn't possible! But it appears to be. For instances, I thought that conservation of linear momentum wouldn't consider the signal of the velocity, but it does!

So imagine the following collision:

Body A: mass=1kg v=9m/s
Body B: mass=2kg v=0

What I call overall movement does not consider if velocity is positive or negative. In the inicial conditions I have 1kg of mass with 9m/s, this is what I call overall movement.

After collision, the real result is:

Body A: v=-3m/s
B: v=6m/s

Linear momentum is -3*1 + 6*2= 9, like before. But wait! I would intuitevely say that this couldn't be true because I should work with modules | | because I don't care if velocity is positive or negative.

What I see in this result is:
1kg with 3m/s and 2 kg with 6m/s. This for me is on average 3kg of 5m/s or 1kg of 15m/s: this is more than the initial 9m/s in 1 kg!!! (I see mass as a number of something, like atoms). This intuitive idea is wrong.
I thought that if body A has lost 6m/s (9-3 and not 9-(-3)), it could only give 3m/s to the body B, because this body doubles the mass. But this is not what happens.
So I can conclude that In a closed system I can create more of this visual "overall movement", I don't know if this gives potential to anything since energy remains the same.

I don't know if I'm making myself clear, all this can be very silly :9

Last edited: Jun 11, 2012