Solving Torque Equalibrium Problem for Beam and Boy

[prophet05]

Problem: A 40kg, 5.0m long beam is supported by, but not attached to, the two posts in the figure. A 20kg boy starts walking along the beam. How close can he get to the right end of the beam without it falling over?
http://student.ucr.edu/~meliz003/phy.JPG

My answer:
\tau_net = (F_1*x_F_1)-(F_Beam*X_Beam)-(F_Boy*X_Boy)
Since, \tau_net = 0

0 = (F_1*x_F_1)-(F_Beam*X_Beam)-(F_Boy*X_Boy)
0 = (40*3)-(40*2)-(20*X_boy)

Comes out to X_Boy = 2?
So the boy has to step at the very edge of the beam for it to start falling over?

 

[Doc Al]

My answer:
\tau_net = (F_1*x_F_1)-(F_Beam*X_Beam)-(F_Boy*X_Boy)
Since, \tau_net = 0

0 = (F_1*x_F_1)-(F_Beam*X_Beam)-(F_Boy*X_Boy)
0 = (40*3)-(40*2)-(20*X_boy)

Looks like you are treating the beam as have two parts, which is OK. But:
(1) 40 kg is the mass of the entire beam, not each piece;
(2) The weight of an object acts through its center of mass.
You need to redo those torque calculations.

Even better is to treat the beam as one piece--that way you only have two torques to worry about. Where is its center of mass with respect to the pivot?