# Torque (really confused)

1. Nov 27, 2007

### kevinr

Hello.
I just learned torque and i am really confused about one thing.

T = r x F.

Now in the above Torque, we need to find a perpendicular force part which contributes to the torque compared to the r.

Hence: T = r X Fsin(theta).

But i get confused in the following example:
http://img187.imageshack.us/my.php?image=physicslu1.png

Basically i am so confused of how to find r x F (i know you try to find perpendicular but it doesnt make sense because i saw some problems in my book use even cosine).

Could anyone please clarify which angle you use (outside / inside) compared to r and whether you need cosine or sin.

(In the diagram, if the force was instead pushing up on r, would the same angle be used compared to it pushing down on r? )

You can take clockwise angle or counterclockwise angle. So im lost.

Last edited: Nov 27, 2007
2. Nov 27, 2007

### Staff: Mentor

When computing $rF\sin\theta$, $\theta$ is the angle between $\vec{r}$ (which is a position vector from the pivot to the point of application of the force) and $\vec{F}$. With a bit of trig, you can often use another angle to get an equivalent answer (for example, $\sin\theta = \sin(180 - \theta)$ or $\sin\theta = \cos(90 - \theta)$).

3. Nov 27, 2007

### kevinr

True but what i dont understand is which $\theta$ to take: $\theta$ can be clockwise / counterclockwise between r and F.

(Say you start at force and go to r which gives you the angle between them. If you go clockwise you get a different angle than counter clockwise).

4. Nov 28, 2007

### Staff: Mentor

Always take the shortest path (the smallest angle) between r and F. That will give you the correct direction for the torque. But it doesn't really matter, since the $\sin\theta$ will correct you. If you choose the larger angle, $\sin\theta$ will be negative, which tells you to reverse your initial answer.

5. Nov 28, 2007

### kevinr

Ah thank you!