How do I determine if a force will cause clockwise or counterclockwise rotation?

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Tsunoyukami
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I'm not exactly sure where to ask this question but I would like someone to please help explain to me how to understand which direction a force a will torque on object. That is, how do I determine whether the force will cause it to rotate clockwise or counterclockwise?

I'm not very good at solving even elementary torque problems because this is a huge element of understanding torque.

This is what I know about toruqe so far:

[itex]\tau[/itex] = [itex]\stackrel{\rightarrow}{F}[/itex] x [itex]\stackrel{\rightarrow}{r}[/itex]

|[itex]\tau[/itex]| = |[itex]\stackrel{\rightarrow}{F}[/itex]| |[itex]\stackrel{\rightarrow}{r}[/itex]| sin [itex]\theta[/itex]
 
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Tsunoyukami said:
This is what I know about toruqe so far:

[itex]\tau[/itex] = [itex]\stackrel{\rightarrow}{F}[/itex] x [itex]\stackrel{\rightarrow}{r}[/itex]
That should be [itex]\tau[/itex] = [itex]\stackrel{\rightarrow}{r}[/itex] x [itex]\stackrel{\rightarrow}{F}[/itex]

Are you familiar with the right hand rule for cross products?

Also, view the animation on this page: Torque

Torque_animation.gif
 
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I have used the right-hand rule for cross products before (namely in terms of magnetism) but have never been particularly comfortable with it. Also, if I used torque as I have defined it would my resultant torque be negative (is that not a property of the cross product?)

I understand that if you apply a force in one direction to some radius (I believe this is called a moment arm?) it will cause a toruqe in one direction and if you apply a force in the opposite direction to torque, in turn will point in the opposite direction.
 
Tsunoyukami said:
I have used the right-hand rule for cross products before (namely in terms of magnetism) but have never been particularly comfortable with it.
This illustrates the version that I use for any cross product:

120px-R_hand_Rule.png


Here's another illustration specifically for torque: http://hyperphysics.phy-astr.gsu.edu/hbase/tord.html"
Also, if I used torque as I have defined it would my resultant torque be negative (is that not a property of the cross product?)
Yes. Your (incorrect) definition of F x r would be in the opposite direction to r x F.

It may be worth your time to explore this site: http://hyperphysics.phy-astr.gsu.edu/hbase/torcon.html"
 
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Thank you for these links; I will explore them and ask again if I need any further clarification.

If I were to use my improper definition of torque when considering each and every torque on a given system it would still yield the same result, correct? That is, at least in any static problem since the net torque would be zero. If the problem were not static and I consistently used this definition my final result would have a negative symbol?
 
Tsunoyukami said:
If I were to use my improper definition of torque when considering each and every torque on a given system it would still yield the same result, correct?
That depends on the specific thing you need to figure out.
That is, at least in any static problem since the net torque would be zero.
In that case, no problem. It's equivalent to reversing the sign convention for clockwise versus counterclockwise--doesn't really matter.
If the problem were not static and I consistently used this definition my final result would have a negative symbol?
Depends on what you are asked to find. If you are actually calculating the torque vector, then your result would be in the opposite direction to the actual torque.

For simple problems that are restricted to 2 dimensions (rotation in a single plane), you may not even need the torque vector. Clockwise versus counter-clockwise is often good enough.