# Quick help need with finding the MAGNITUDE of the TORQUE of this wrench

I tried using Frsin θ

(270.756)(46 cm)sin (68) but I got 11547.86

And got it wrong...

And then I tried doing.

101.426982807+251.040591727 = 352.4 N*m but I got it wrong.

^ I don't remember how I did that but it was saved in my notes.

Help?

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Doc Al
Mentor
I tried using Frsin θ

(270.756)(46 cm)sin (68) but I got 11547.86
You have the wrong F. You used the result from part a, which is a torque.

You have the wrong F. You used the result from part a, which is a torque.
(60)(9.81) = 588.6

588.6(46)sin(68) = 25104.0592??

Doc Al
Mentor
(60)(9.81) = 588.6
Right.
588.6(46)sin(68) = 25104.0592??
Check your units. (How did you do part a?)

Right.

Check your units. (How did you do part a?)
I think I changed r to meters

so...

588.6(0.46)sin (68) = 251.04 N*m

And direction would be counterclockwise, right?

Doc Al
Mentor
I think I changed r to meters

so...

588.6(0.46)sin (68) = 251.04 N*m
Right.

And direction would be counterclockwise, right?
No. Which way would the wrench tend to turn? Just look at the diagram and use your common experience.

Right.

No. Which way would the wrench tend to turn? Just look at the diagram and use your common experience.
clockwise

thanks!