How do I calculate the torque of a crank?

Question

So, I have a crank that has a lever attached to a fulcrum, which rotates about an axis. But I also have a handle attached perpendicular to this lever, and the effort force is applied to the handle 90° downward.
I am trying to calculate torque = RFsin(theta)
So, my question is: what is the radius in this instance. Is it the distance from the axis of rotation to the edge of the lever, or is it the distance from the axis of rotation to the edge of the handle where the force is applied? See image for more clarity.
If it were the latter distance, then trigonometry would need to be used to find the diagonal.
Thanks in anticipation of your help.

kamran · Accepted Answer

The torque is $tau= RF  ,$ no other term or sin needed if you apply force trying to rotate the lever.
The torque of a force about an axis is its perpendicular projection to a radial passing through the center or axis we want to measure the torque for. And no matter how long the  round handle is as long as it is perpendicular to the handle it doesn't affect the torque.
Let's say your diagonal would be 45 or 60 degrees it does not make a difference. Of course, the handle where you grab and turn acts as a cantilever and has bending stress, but it is orthogonal to the force so it soes not matter.

How do I calculate the torque of a crank?

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