## I am having a hard time defining theta in Torque=rFsin(theta)

Can someone dummify what theta represents in Torque = rFsin(theta)?

if my understanding of theta is correct for this problem theta would be 53 degrees.
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 Quote by dadiezel07 Can someone dummify what theta represents in Torque = rFsin(theta)?
In general, θ would be the angle between the position vector (r), which describes the point of application with respect to some axis, and the force vector (F).
 if my understanding of theta is correct for this problem theta would be 53 degrees.
No. Assuming you are trying to express the torque on the pulley, realize that the tensions are tangential to the pulley.

 Quote by Doc Al No. Assuming you are trying to express the torque on the pulley, realize that the tensions are tangential to the pulley.
When you say they are tangential to the pulley, can you explain a little furthur.

Thats another subject I can't quite wrap my head around is the definition on tangential, is it the word to describe "linear" equations?

I have been continuing the subject without complete understanding of everything this is my attempt at understanding everything because my book does a horrible job.

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## I am having a hard time defining theta in Torque=rFsin(theta)

 Quote by dadiezel07 When you say they are tangential to the pulley, can you explain a little furthur.
I mean that the line of action of the tension force (which is the line that the ropes make) is tangential to the circle that is the pulley. Which means that if you draw a radius to the point of application of the force, the force would be at 90° to the radius.
 so in the case of the pulley will the theta angle always be 90 degrees?

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 Quote by dadiezel07 so in the case of the pulley will the theta angle always be 90 degrees?
Yes.

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