What is the effect of adding weight to a balanced beam?

AI Thread Summary
The discussion centers on calculating the force F2 required to balance a beam under two scenarios. Initially, with a negligible beam weight, the equation (10)(1.5) - F2(6.5) = 0 yields F2 = 2.31 N. In the second scenario, where the beam has a weight of 3 N acting at its center of mass, the calculation must account for this additional force. The weight's position, 4 meters from one end, and the pivot point at 1.5 meters are crucial for determining the new value of F2. Proper attention to the signs in the equations is emphasized for accurate results.
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A beam with negligible mass is balanced by two external forces at R1 = 1.5 meters and R2 = 6.5 meters. If F1 = 10 N, what is F2.

i got this:

(10)(1.5) + F2(6.5) = 0

so F2 = 2.31 N

but the second part i don't get:

If the beam actually has a weight of 3 N, what would F2 be now?
 
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cookie monsta said:
A beam with negligible mass is balanced by two external forces at R1 = 1.5 meters and R2 = 6.5 meters. If F1 = 10 N, what is F2.

i got this:

(10)(1.5) + F2(6.5) = 0

so F2 = 2.31 N

but the second part i don't get:

If the beam actually has a weight of 3 N, what would F2 be now?
Well, your first equation in Part1 should read
10(1.5) - F2(6.5) = 0
F2 = 2.31N
You've got to watch your plus and minus signs, one moment is clockwise, the other is counterclockwise.
For Part 2, the 3N weight of the beam acts at its c.m. (or c.g.), that is, since the beam is 8m long, at 4m from one end. Try it again to solve for F2, watching your plus and minus signs. Remmber also the pivot point is at 1.5m.
 
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