Moment of uniformly distributed load

[chetzread]

Homework Statement

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

Homework Equations

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) ??

 

[collinsmark]

Homework Statement

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

Homework Equations

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) ??

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).

 

[chetzread]

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).

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 ?

 

[collinsmark]

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 ?

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.]

 

[chetzread]

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.]

i can't 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)

 

[collinsmark]

i can't 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)

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?

 

[David Lewis]

w = force per meter

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