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

[riseofphoenix]

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?

 

[Doc Al]

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.

 

[riseofphoenix]

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]

(60)(9.81) = 588.6

Right.

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

Check your units. (How did you do part a?)

 

[riseofphoenix]

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]

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.

 

[riseofphoenix]

Right.


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

clockwise


thanks!