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

  • #1
Number3Physics.png


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?
 

Answers and Replies

  • #2
Doc Al
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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.
 
  • #3
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??
 
  • #4
Doc Al
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(60)(9.81) = 588.6
Right.
588.6(46)sin(68) = 25104.0592??
Check your units. (How did you do part a?)
 
  • #5
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?
 
  • #6
Doc Al
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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.
 
  • #7
Right.


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


thanks!
 

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