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

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SUMMARY

The discussion centers on calculating the forces acting on a balanced beam subjected to external forces. Initially, with a negligible beam weight, the force F2 was determined to be 2.31 N using the equation (10)(1.5) + F2(6.5) = 0. However, when considering the beam's actual weight of 3 N, the calculation must account for the beam's center of mass located at 4 meters from one end, necessitating a revised equation: 10(1.5) - F2(6.5) - 3(4) = 0. This adjustment leads to a different value for F2, emphasizing the importance of sign conventions in torque calculations.

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cookie monsta
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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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