Here are some pre-worked out problems for you to examine before trying some out for yourself.

Example 1:

A force of 5.0 N is applied at the end of a lever that has a length of 2.0 meters. If the force is applied directly perpendicular to the the lever, as shown in the above diagram, what is the magnitude of the torque acting the lever?

Solution:

This sample is a simple matter of plugging the values into the equation:
Torque = F * l
Torque = 5.0 N * 2.0 m
Torque = 10 N*m

Example 2:

If the same force as in example 1 is applied at an angle of 30 degrees at the end of the 2.0 meter lever, what will be the magnitude of the torque?

Solution:

First we must find the lever arm value using trig:

Sin30 = x/2.0
Sin30 (2.0) = x
1.0 m = x = l
Now we plug the value into the torque formula:
Torque = F * l
Torque = 5.0 N * 1.0 m
Torque = 5.0 N*m

Example 3:

What force is necessary to generate a 20.0 N*m torque at an angle of 50 degrees from along a 3.00 m rod?

Solution:

Solve for the lever arm value:

Sin50 = x/3.00
Sin50 (3) = x
2.30 m = x = l
Now plug the values into the formula:
Torque = F * l
20.0 N*m = F * 2.30 m
20.0 N*m/2.30 m = F
8.70 N = F

Example 4:

Suzie applies a force of 40.0 N at an angle of 60 degrees up from the horizontal to a wooden rod using a spring scale. If she generates a torque of 73.0 N*m, how long was the rod?

Solution:

First of all we create a trig equation for the value of the lever arm to use in our torque formula.
Let x = Rod Length
Let l = Lever Arm
Sin60 = l/x
Sin60 (x) = l
Now placing this into the formula, it should be easy to find the length of the rod.
Torque = F * l
73.0 N*m = 40.0 N * Sin60 (x)
x = 2.11 m

Now that you have seen some solved sample problems, you should head to the Torque Introduction Problem Page and try some on your own.