Moment of a force

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Homework Statement


Find the moment of a force about Point A

- I have attached a diagram of the problem as well

EDIT It's actually 62 degrees.. I drew it incorrectly ^^
[PLAIN]http://img823.imageshack.us/img823/1315/diagramp.png [Broken]

Homework Equations


Moment = Force * perpendicular distance to point


The Attempt at a Solution


what I've done is the following

Moment of C = Fx * distance
Moment of C = 22cos62 * 950
Moment of C = 9811.96kN
_____________________________
Moment of C = Fy * distance
Moment of C = 22sin62 * 0
Moment of C = 0kN



I would just like someone to revise my work, and see if I have any errors :S I believe I'm doing it right, but I don't feel confident. Any help would be appreciated thanks.

I also feel as if the answer is:
Moment of C = 22 * 950, but that's just me :D

My final question would be... how would I present the final answer in a proper manner?

like... The horizontal moment of C is... and the vertical moment of C is...? or should there be just 1 solid answer? thanks a lot!
 
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Answers and Replies

  • #2
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ok nvm I think I got it.. perpendicular distance... meaning I should extend the force in it's line of action and then the perpendicular distance from the point C to the line of action * the force ^^ thanks for the views though everyone appreciate it :D
 
  • #3
PhanthomJay
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Homework Statement


Find the moment of a force about Point A

- I have attached a diagram of the problem as well

EDIT It's actually 62 degrees.. I drew it incorrectly ^^
[PLAIN]http://img823.imageshack.us/img823/1315/diagramp.png [Broken]

Homework Equations


Moment = Force * perpendicular distance to point


The Attempt at a Solution


what I've done is the following

Moment of C = Fx * distance
Moment of C = 22cos62 * 950
Moment of C = 9811.96kN
_____________________________
Moment of C = Fy * distance
Moment of C = 22sin62 * 0
Moment of C = 0kN



I would just like someone to revise my work, and see if I have any errors :S I believe I'm doing it right, but I don't feel confident. Any help would be appreciated thanks.
You are finding moment of the force about C, but the problem is aking for the moment of the force about A.
I also feel as if the answer is:
Moment of C = 22 * 950, but that's just me :D
and that is wrong...you are not using the perpendicular distance..
.

My final question would be... how would I present the final answer in a proper manner?

like... The horizontal moment of C is... and the vertical moment of C is...? or should there be just 1 solid answer? thanks a lot!
Don't forget it is the moment about A you are looking for....there is just one solid answer....ccw momnts can be comsidered positive and cw moments as negative...then add the two moments of the component forces together to get the resulting moment about A, is it clockwise (-) or counterclockwise (+)?
 
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  • #4
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oh wait, so I can just add the two resulting components (x and y) about C together to get the total moment about C?

I just finished getting C... as 18453.6kN CCW or (+)

and just got B as well ... as 65804.74kN CW or (-)

for these two, I'm actually usuing M = F * perpendicular distance from the line of action to the point :D feels very nice understanding it a bit better haha
 
  • #5
PhanthomJay
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Please show how you are calculating these numbers about C and B..what happened to the moment about A, which is what your problem is asking ?
 
  • #6
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oh no! WHOOPS! totally my mistake. the question was supposed to be about point C haha.

B and A i did extra i guess... having it revised wouldn't be bad either ^^
my work is:

Mc = F * d
Mc = 22 * (950sin62)
Mc = 18453.6kN ; CCW
====================
Mb = 22 * (6371.28sin28)
Mb = 65804.97kN ; CW
====================
Ma = 22 * (8871.28sin28)
Ma = 91625.9kN ; CW
 
  • #7
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thanks again phanthomJay for the quick response :)
 
  • #8
PhanthomJay
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You had the moment about C correct in your first post...what happened since then?
 
  • #9
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wait... it was correct? but it had 2 answers :( 0 and 9811.96kN it was also in component form O_O.... whereas my second solution only had 1 answer
 
  • #10
PhanthomJay
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wait... it was correct? but it had 2 answers :( 0 and 9811.96kN it was also in component form O_O.... whereas my second solution only had 1 answer
The first answer was more or less correct, you have a moment of 9811 kN-cm ccw from the x component of the force, and 0 kN-cm from the y component of the force.....add them up algebraically to get 9811 kN-cm ccw, or + 9811 kN-cm. Note the units..I am assuming the force is given in kN and the distances in cm.....9811 kN-cm = 98.11 kN-m..or round it to 98 kn-m. Continue.....
 
  • #11
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hmm.. so if I consider doing the solution that way with components, what if I end up with a value not equal to 0 for the x and y component...

lets just use an example of... Mx = 5000kn-mm and My = 4500kn-mm

how would I present the final answer? just like
Mx = 5000kn-mm and My = 4500kn-mm

or would I have to like...

M = root( 5000^2 + 4500^2)

M = 1 value only? i dunno :S
 
  • #12
PhanthomJay
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hmm.. so if I consider doing the solution that way with components, what if I end up with a value not equal to 0 for the x and y component...

lets just use an example of... Mx = 5000kn-mm and My = 4500kn-mm

how would I present the final answer? just like
Mx = 5000kn-mm and My = 4500kn-mm

or would I have to like...

M = root( 5000^2 + 4500^2)

M = 1 value only? i dunno :S
Don't confuse subscripts. This is a problem where the forces lie in a 2-D x-y plane, and the moments (Mz) of these forces are about an axis through the point in the z direction coming into and out of the plane into the 3rd dimension perpendicular to the plane. Both components of the force produce moments about the z axis, so you just add up the results, no square roots involved. For a 3D system of forces, then you have in general moments about all axes, and you've got to do that square root stuff...or else just leave it in terms of Mx, My. and Mz.
 
  • #13
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mmmmm okay thanks! the last piece of vital information was very helpful for me. thanks again!
 
  • #14
PhanthomJay
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mmmmm okay thanks! the last piece of vital information was very helpful for me. thanks again!
Ok, just remember for forces in the x-y plane only, then [tex]M_z = F_x(y) + F_y(x)[/tex], watch signage...and in general for 3D problems,
[tex] M_z = F_x(y) + F_y(x)[/tex],

[tex] M_x = F_z(y) + F_y(z)[/tex], and

[tex] M_y = F_z(x) + F_x(z)[/tex]

Watch plus and minus signs...ccw is plus, cw is minus.
 

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