Calculate the torque that is produced by this force on a cylinder

[MatinSAR]

Homework Statement

Please take a look at the picture :

Relevant Equations

torque = r F sin(r, F)

Why it said that angle between r and F is 30?
I guess it should be 120 degrees... Am I wrong?

 

[Lnewqban]

As far as I can see, the length of the arm of the force respect to the center of rotation should be $rcos30$.

 

[kuruman]

Why it said that angle between r and F is 30?

It didn't say that. The angle between two vectors is obtained by putting them tail-to-tail. If you do that with r and F, you will get an angle of 120°. Then note that $rF\sin(120^{\circ})=rF\cancel{\sin}\cos(30^{\circ}).$

 

[Steve4Physics]

It didn't say that. The angle between two vectors is obtained by putting them tail-to-tail. If you do that with r and F, you will get an angle of 120°. Then note that $rF\sin(120^{\circ})=rF\sin(30^{\circ}).$

You probably meant to write $rF\sin(120^{\circ})=rF\cos(30^{\circ}).$

 

[kuruman]

You probably meant to write $rF\sin(120^{\circ})=rF\cos(30^{\circ}).$

Yes, of course. Good catch.

 

[MatinSAR]

It didn't say that. The angle between two vectors is obtained by putting them tail-to-tail. If you do that with r and F, you will get an angle of 120°. Then note that $rF\sin(120^{\circ})=rF\cancel{\sin}\cos(30^{\circ}).$

So the book is wrong since sin30 isn't equal to sin120 degrees... @Lnewqban
@kuruman
@Steve4Physics
Thanks for your help and time.