- #1

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i got this:

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

so F2 = 2.31 N

but the second part i dont get:

If the beam actually has a weight of 3 N, what would F2 be now???

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- #1

- 4

- 0

i got this:

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

so F2 = 2.31 N

but the second part i dont get:

If the beam actually has a weight of 3 N, what would F2 be now???

- #2

PhanthomJay

Science Advisor

Homework Helper

Gold Member

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Well, your first equation in Part1 should read

i got this:

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

so F2 = 2.31 N

but the second part i dont get:

If the beam actually has a weight of 3 N, what would F2 be now???

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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