# Homework Help: Moment of uniformly distributed load

1. Jul 21, 2016

1. The problem statement, all variables and given/known data

at R2 , is the moment wrong ? it should be 6R1 -200(2)(1) =1800 , am i right ?
2. Relevant equations

3. The attempt at a solution
it's 200(2)(1) because the uniformly distributed load is from 2m away from R2 , so force should be 200(2) ??

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2. Jul 21, 2016

### collinsmark

Maybe I'm missing something, but it looks to me like the 200 N/m distributed load is centered right smack on R2. In other words, half of that load is to the left of R2, and half to the right: It's distributed evenly spanning 2 m on either side of R2. So the moment around R2 isn't affected by the distributed load (hence the 0 in the "200(4)(0)" term in the attached solution section).

Last edited: Jul 21, 2016
3. Jul 21, 2016

we know that force = wx , where w = force per meter , x = distance , so , do you mean , the x of distributed load start from 0 at R2 , and it is measured from R2 to the left and to the right ....If so , then , it should be 200(2)(1) . am i right ?

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4. Jul 21, 2016

### collinsmark

There still seems to be a misunderstanding somewhere.

The distributed load is depicted as 200 N/m, and it spans a total of 4 meters. So the total force of the distributed load is 200(4) = 800 N.

This 800 N force is centered directly on top of R2. In other words, it is 0 m from R2. So it's contribution to the total moment around R2 is 200(4)(0) = 0 N⋅m.

If you wanted to break up the distributed load into two forces, each centered 1 m away from R2 on either side, and each spanning 2 meters, you still get the same answer. In that case, one moment will act counterclockwise around R2 and the other clockwise around R2, so the terms will cancel.
200(2)(1) - 200(2)(1) = 0 N⋅m.

[Edited a pair of typos immediately after posting.]

5. Jul 21, 2016

i cant understand the 800N is at 2m away from R2 , (either at the left of R2 or right of the R2 ) , am i right ? how could 800N at the center of R2 ? (refer to the diagram)

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6. Jul 21, 2016

### collinsmark

If you want to balance a cube* with 4 meter sides (such that one of its planes is parallel with the ground) you would find the center of balance (i.e., center of gravity) to be in the middle, correct? If you place the cube on the ground (such that one of its planes is flush with the horizontal ground), the center of force of the cube on the ground is going to be in the middle of that 4 m span, not at one of its sides.

*(assume the cube has uniformly distributed mass)

Does that make sense?

Last edited: Jul 21, 2016
7. Jul 22, 2016

### David Lewis

You're mixing a physical quantity with a unit of measure.
You may have force divided by distance, or newtons per metre.