What angle do I use to find the perpendicular force in torque calculations?

[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 doesn't 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 I am lost.

THanks for your time!

 

[Doc Al]

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 $\sin\theta = \cos(90 - \theta)$).

 

[kevinr]

True but what i don't 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).

 

[Doc Al]

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.

 

[kevinr]

Ah thank you!