Yay! It works. I can't believe it was something so simple. Thanks a lot for your time! :smile:.
On a side note, my teacher said that we could do it both ways, but engineers prefer taking the perpendicular distance, but physicists prefer taking the perpendicular force. I suppose it's a matter...
OK, let me try again! :smile:
Torque is \Sigma\tau=F*d
Where:
F is the perpendicular force acting on the beam;
d is the distance from the pivot point.
The mass of the beam applies a downwards force on the beam of mg, i.e.: 10*9.8=98 N.
However, for the torque, one requires the...
Ooh. OK, now I got the angular velocity and acceleration! Thanks a lot :smile:
I am still having problems with the first part. (the tension in the cable) Any idea what's wrong?
Angular motion + Inertia [SOLVED]
Homework Statement
A uniform beam of length, L, and mass, M, is freely pivoted at one end about an attachment point in a wall. The other end is supported by a horizontal cable also attached to the wall, so that the beam makes an angle phi with the...