How to Calculate Mass Using Torque?

[bballa99]

Torque question need help!

Homework Statement

Find the mass of the bar.
http://img23.imageshack.us/img23/8871/eeeeee.png
(My paint drawing sucks.)

A = 0.5 kg
B = 0.5 kg
L = 50cm
Angle = 76 degrees

Homework Equations

fd_\perp , Tcl = Tcc T_cl = T_cc (Torque clockwise = Torque counter-clockwise)

The Attempt at a Solution

Tcl = Tcc
m(9.8)(0.25 sin 76) = (0.5sin76)(9.8)

I have attemped the above, and ended up with mass = 1.49 kg which is impossible... (it was a plastic ruler).

I don't think I am missing anything given.

 

[bballa99]

Bump. Also the system is in eqilibrium, right?

 

[PhanthomJay]

What happened to the force and torque from the mass B? And you left out the mass of A or the lever arm of A to the pivot, I can't tell which one you forgot, because the numerical values are the same. Also, if I read your diagram correctly, the mass B seems to be attached to a rope around a pulley, with the other end of the rope teminating at the mid point of the beam. If that'scorrect, you need to know what angle the rope makes with the beam at its point of connection with the beam midpoint, and calculate the torque from that tensile force appropriately.

 

[bballa99]

Oh sorry. I forgot to add that in, I had it erased. The angle the rope makes with the beam is 30 degrees.

And I used:
F_ad = F_bd

 

[PhanthomJay]

The system is in equilibrium if it isn't moving after all loads are applied. When summing torques about the pivot, you must include the torques from all loads: from the bars weight, the weight of mass A, and the tension force applied at midpoint from the weight of mass B. Torques can be found from Torque=F(r)sin theta, where F is the force, r is the position vector measured along the bar from the point of application of the force to the pivot, and theta is the angle in between the force and position vector. Watch plus and minus signs.

 

[bballa99]

Thank you. I understand now.

Solved.

(How to mark it? I can't find it under thread tools)