# Could some people help to solve an arguement?

1. Oct 20, 2005

### MooMansun

Could some people help to solve an arguement???

A question came up on the GameDev forums about a diagram with a force being applied to it. I provide the link, but be warned, there is 6 pages of posts to go through, although some are quite funny, quite a heated battle ensues:

http://www.gamedev.net/community/forums/topic.asp?topic_id=352000

At first I was messing about with the people posting, but they then settled on a solution of f=F for the diagram; the resulting force at the center of mass would be the same as the input force.

I then stopped joking and provided a solution of f=F-e, where e is the force lost (for various reason detailed in the above link). The main basis of my arguement is the conservation of Energy and that for f=F to be true it would need to defy that law. Also, as an arguement to real world applications the basic principles of elastic and in-elastic collisions would also prohibit f=F.

Could the good people of the physics forums help resolve this?

Who is correct?

2. Oct 20, 2005

### KingOfTwilight

F doesn't create any force. Actually it makes the object accelerate, so f should be a = F/m.

3. Oct 20, 2005

### whozum

Is F an instantaneous collision or a continued applied force?
F does not equal f in any case. The way to find f would be to do an energy analysis on the system depending on details that arent stated in the orginal problem.

Last edited: Oct 20, 2005
4. Oct 20, 2005

### MooMansun

http://www.gamedev.net/community/forums/topic.asp?topic_id=352000

F is a linear force and f is a linear force.

F/m is incorrect. We are trying to find the equation for f.

As I said earlier, the arguement comes down two options f=F or f=F-e.

5. Oct 20, 2005

### MooMansun

There is no mention of instantaneous or applied, only F.

6. Oct 20, 2005

### MooMansun

So whozum, you say f=F is wrong.

The energy analysis would be drawn from the equation f=F-e, where e is the force lost.

Would you say that's fair?

7. Oct 20, 2005

### whozum

Neither of those two works. Let E be the energy of the entire system, with the bar having no kinetic energy. Then the energy will be conserved, and the KE of the object colliding with the bar (with contact force F) will be split into linear and angular energy for the bar, as well as some energy left for the object to deflect and continue a path.

8. Oct 20, 2005

### MooMansun

At this point we are just calculating the linear force.

Basically, I say that the resultant force (f) is based on the input force (F) minus any loss incurred (e) irrespective of how it is lost.

We are not considering any other aspect than linear force.

9. Oct 20, 2005

### whozum

Theres no such thing as force loss, f=F-e is wrong because it is using forces, you need to look at energy or momentum. Lets say its a trailer thats parked and a car runs into the edge. Then the energy system would look something like this:
$KE_{car}$ = energy of the car.
$KE_{trailer}$ = trailer energy.
$KE_t = KE_i = KE_f$ = total + initial and final energy of the system.
Before the collision, all the systems energy is in the car ($KE_t = KE_{car}$. At the moment of (elastic) collision, the car loses energy which is placed into moving the trailer. At that moment the trailer will begin to spin a bit as well as get displaced in a linear fashion. The energy then will look something like ths:
$$KE_t = KE_i = KE_f = \frac{1}{2}\left(mv_{car}^2 + mv_{trailer}^2 + I\omega_{trailer}^2\right)$$

The linear force would be the linear component of the trailer's energy ([itex]KE_{trailer} = \frac{1}{2}mv_{trailer}^2) [/tex]divided by how far the trailer moves. Note the car's velocity component above is AFTER the collision.

Theres no simple answer to the posted problem because it is way too vague.

Last edited: Oct 20, 2005
10. Oct 20, 2005

### MooMansun

That would present a fundemental flaw. Force is energy in another form, if energy must be conserved, so then must a force.

So force loss is a real thing.

11. Oct 20, 2005

### KingOfTwilight

F does not create any force f. F creates acceleration a, which you think is caused by some force f.

12. Oct 20, 2005

### MooMansun

If a linear force (F) is applied to a system a corresponding linear force will be found at the center of mass (f).

Whilst I am sure it would accelerate, we are not considering that. Only the tranfer of F to f, linear force.

13. Oct 20, 2005

### whozum

The problem is too vague to be discussed. There is no right answer, all the solutions you both provide dont make any sense, either.

14. Oct 20, 2005

### MooMansun

Perhaps not...

I thought that too at the start, however, all the necessary componants are there.

Basically, an unspecified force is inputed at F and a force is produced at f. If this force is linear what is the relationship between F and f.

So its crystal clear. Is the relationship f=F, where there is a complete transfer of force, or is it f=f-e, where e represents the loss of force?

Its a basic physics question...

Look at the diagram and think of a spaceship in deep-space.

15. Oct 20, 2005

### vanesch

Staff Emeritus
I have to say that the question in the problem is not entirely clear, but I think I can guess it.
It might be that what the problem wants to solve is to find THE EQUIVALENT SET OF FORCE/TORQUE that will describe the entire motion of the rigid body. Indeed, in the mechanics of rigid bodies, there are EQUIVALENCE THEOREMS of force systems: these are different sets of forces, applied at different points, which result in exactly the same motion of the rigid body. Mind you, that's only valid for a rigid body: it is NOT valid for an elastic body, or to calculate mechanical stresses in materials.
One of these theorems tells you that ANY force system can be replaced by:
1) a force applied to the center of gravity
2) a torque applied around the center of gravity
The force 1) is simply the sum of all forces everywhere.
The torque 2) is simply the sum of all perpendicular lever arms times the forces.
Me thinks this is the problem that is presented, but it should be made clear.
When I apply that theorem, then the EQUIVALENT force f = F, simply because there is only one force (so its sum is just the one force).
The EQUIVALENT TORQUE tau is D x F.
So this means you can dispense with force F, and replace it with force f and torque tau, and this will result in exactly the same motion of the bar.
It doesn't mean of course that something is pulling at the CM with force f, it simply means that you can PRETEND (as far as the rigid body motion goes) that something is pulling there, and giving it a torque too.
The acceleration of the CM will be f/m = F/m
You will also start rotating.

16. Oct 20, 2005

### pervect

Staff Emeritus
As Vanesch points out, the question "as-is" is confusingly worded. There is an archived thread here on physics forum where a similar question was discussed

note that the problem has been reformulated somewhat, along the lines that Vanesch has mentioned.

I hope that you'll be able to determine who was correct and who wasn't in the linked thread -- if you need a hint, look at the "ratings" of the posters with different viewpoints (no rating, homework helper, science advisor, PF super-mentor, etc.).

17. Oct 20, 2005

### krab

Force is not conserved; only energy is. If force were conserved, a lever would not be useful.

18. Oct 20, 2005

### pervect

Staff Emeritus
We actually need to consider two systems, the problem as originally stated is confusingly described.

System #1
Code (Text):

|
|
|
|
|
|
|-----> force F

will move in an identical manner to System #2

Code (Text):

|
|
|
|-----> Force f, torque T (counterclockwise)
|
|
|

when f=F and T=F*L/2

Energy is not equal to force. Energy equals force*distance, and is called "work".

For a given time interval, the work done by force F in system #1 will be greater than the work done by force f in system #2.

This is true because force F will be applied over a longer distance than force f.

The work per unit time (aka power) done by force F will be F*(v+L*w/2), where v is the velocity of the center of mass, w is the angular velocity of the rod, and L is the length of the rod.

The work per unit time done by force f will be f*v

The work per unit time done bye torque T will be T*w = F*L/2*w

Conservation of energy tells us that the total work done by force F in system #1 will be equal to the work done by force f and the work done by torque T in system #2. This is true becase

F*(v+L*W/2) = f*v + T*w, given that f=F and T=F*L/2

19. Oct 20, 2005

### MooMansun

Conservation of Energy.

Hi Krab,

No, I do not agree with this, the lever principle principle whilst attributed to distance, has more to with arcs, surface area and basic pressure, that is force over a given square area. Also, the conservation of Energy works perfectly along with the scenario.

Force is energy in a another form, therefore an absolute must is that it obey the laws of conservation. There is no way to beat this.

As for the rest of you posters, well, I do understand that it is somewhat vague, however, it simply comes down to the transfer of F to f and whether that relationship is expressed as:

1. f=F
2. f=F-e

That's it...all that is left now is for someone not to evade the question...:rofl:

20. Oct 20, 2005