Torque Equalibrium

  • Thread starter prophet05
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  • #1
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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 [Broken]

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?
 
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Answers and Replies

  • #2
Doc Al
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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?
 

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