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

[cookie monsta]

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?

 

[PhanthomJay]

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.

 

[m2003]

Since the beam now has a weight of 3 N, it will contribute to the forces acting on the beam. The new equation would be:

(10)(1.5) + F2(6.5) + 3 = 0

Solving for F2, we get F2 = -0.69 N. This means that the force at R2 would have to be in the opposite direction of the original calculation, at 0.69 N, in order to maintain static equilibrium with the added weight of 3 N.

 

[kyle8921]

When the beam has a weight of 3 N, the equation for static equilibrium becomes:

(10)(1.5) + F2(6.5) = 3

Solving for F2, we get:

F2 = 0.46 N

This means that in order to balance the beam with a weight of 3 N, F2 would need to be 0.46 N. This also shows the importance of considering the weight of the beam in addition to the external forces in order to achieve static equilibrium.