Finding torque from a wrench at an angle

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


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

clockwise


thanks!