Find the Magnitude of the torque about P if a 36-lb force is applied as shown.
I'm not too sure what i'm doing wrong so i'll list the things I do know
I need to use the Cross Product to determine torque.
T = abs(r x F)
From the picture, I decided to use sqrt(32) seeing as that is the hypotenuse.
So basically I do (sqrt32)(36lb)sin(150)
The resulting answer is supposed to be 197 ft-lb but I'm getting something not even close to it. Any help is appreciated.
September 6th 2008, 02:18 PM
Where does your 150 degrees come from?
September 6th 2008, 02:29 PM
I used the supplementary angle of 30 degrees. Should I be using 30 degrees? Either way the sin of both is 1/2 so I didn't think it mattered too much =/
September 6th 2008, 02:40 PM
Yes but where is the moment of torque, is it at P? If the vector to that point is measured from the place where the vector is root 32, then that vector is most definitely not at 150 degrees (or even 30 degrees) from the line of action. From where I can see, it's 45 + 30 degrees.
September 6th 2008, 02:43 PM
DOH!!! You're right. Yet again I missed an easy detail (Headbang). Thanks alot for helping me. I appreciate it!