1. Jul 30, 2004

### kdc6794

I'm kind of new to physics, so this may be a dumb question, but here it is...

I understand that a greater distance from the axis produces greater torque, but I can't understand why. I just can't seem to picture it. Can anyone explain this to me? I'm really confused.

Thanks for the help.

2. Jul 30, 2004

### Gza

Welcome to physicsforums kdc6794. From a qualitative standpoint, imagine a door allowed to swing freely on its hinges. What causes a greater angular acceleration of the door, pushing close to the hinges or pushing close to the edge furthest from the hinges? If you can't picture it, get up and try it on a real door by applying the same force to the different points of the door mentioned above (and make sure to knock before trying it on a door that isn't yours ).

3. Jul 30, 2004

### kdc6794

Hehe.

I realize that's it true from things like using a longer wrench to tighten a bolt or something. But how does the same force, just further away create more torque? I'm sorry if I'm not stating this clearly.

4. Jul 30, 2004

### gerben

If you push further from the axis of rotation you will have to move over a larger distance to rotate the door a certain amount (an angle of 45 degrees for example) than when you you push closer to the axis. The amount of work done by the force, which is the force times the distance over which the force acted, will be the same both times. Either you deliver a strong force over a shorter distance or a weaker force over a longer distance.

The torque is the amount of work that has to be done to rotate the door over some angle.

5. Jul 30, 2004

### Gza

I can understand what you are getting at. To simplify things, just remember that torque isn't some sort of new fundamental physical quantity that you must understand independently of anything else. It's just a plain old linear force that the smarties tinkered around with to make calculations of rotational dynamics much easier. The reason why the same force applied at a greater distance has a larger effect has to do with a definition. Torque (T)itself is defined as:

$$\tau = \vec{r} \times \vec{F}$$

which is the cross product of r (the vector representing the distance from the axis to the point of application of the force) with the force causing the torque. What this torque does is causes a change in the angular momentum (L) of an object about its axis.

$$\tau = \frac {dL} {dt}$$

Now when you increase r, the torque (T) increases, and when the torque increases, the rate the angular momentum changes increases (bigger angular acceleration.) Apologies for the math, I really tried to explain it without it, but i'm sure there is someone here who can do a better job without it.

6. Jul 30, 2004

### kdc6794

I think I get it now, actually. Thank you both for the help.

It's interesting how physics can take things you don't pay much attention to from day to day and show just how fundamental they are to everything else.

7. Jul 30, 2004

### robphy

If you want to "picture it", imagine a parallelogram formed by
a segment from the axis to the point where you apply the force (that's $$\vec r$$ above)
and
a vector $$\vec F$$ representing the force you apply.
The magnitude of the torque is equal to the "area" of that parallelogram.

8. Aug 5, 2004

### Cheman

This is an excellent question to have been asked - moments and torque are always a topic simply taken for granted and I've never found an explanation for the equation. However, may I just ask you to clarify one thing - you say that the "same amount of work must be done" whether you are nearer or further away from the pivot. Since work by definition is F*d may I ask why this must be the case?
Thanks.

9. Aug 5, 2004

### quarkman

It doesn't have to be. It all depends on what is going on with your system. With the door, in one case you need to apply a large force over a small distance and in the other case you need to apply a small force over a large distance. In either case the quantity F*d is the same and that's all that matters for work. If the work was different, say greater for the case of the long moment arm (ie the doorknob) the person pulling the door from the handle would knock himself out because the distance the door moved would be the same as in the case with less work being done. Hope this helps.

10. Aug 6, 2004

### gerben

A same amount of work will be done if the force times the distance over which the force acted (F*d) is the same. If you rotate a door by 45 degrees while pushing close to the hinges you will have to move a shorter distance to do so (but you will have to apply more force) than when you push further from the hinges.

In both cases the work done is: Force*distance = Torque*angle

---------------------------------------------------------------------
Torque = Force*DistanceFromRotationAxis
and
DistanceFromRotationAxis*RotationAngle = DistanceMoved
so we have the following:
Torque*RotationAngle = Force*DistanceFromRotationAxis*RotationAngle = Force*DistanceMoved = Work

(the force must be in the direction tangent to the circular path of the point you push, because the distance moved is in that direction)
---------------------------------------------------------------------

Last edited: Aug 6, 2004