Moment of force from an axis - 3D problem

  • #1
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Homework Statement


An OAB pipe is embedded in O. A cable BC = 1000N and F = ( 0,-f , 0) act on B.
Question:
Determine the magnitude of the force F as the resultant moment about the axis OA is canceled.

Picture in attachment:

Homework Equations




The Attempt at a Solution


O=(0 0 0)
A=(1.5 0 0)
B=(1.5 -sin(40)*0.75 cos(40)*0.75)
C=(0 0.6 1.2)
VTBC= 1000*unit(C-B)
VF=(0 -f 0)
VO=(0 Oy Oz) ==translation forces at o
vMo=(0 momenty momentz) == rotational forces at o
translation equilibrium = VO + VTBC + VF = (0 0 0)
VFxrOB*uAO => crossP(b-o,vf)*unitV(a-o)= (0 0 0)
crossp function = cross product of two vectors.
unitV function = unit vector
F = ??

Calculating moment from an axis is confusing me a lot.
I thank you in advance for helping me.
 

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Answers and Replies

  • #2
haruspex
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VFxrOB
OB? It says:
the resultant moment about the axis OA is canceled.

VFxrOB*uAO => crossP(b-o,vf)*unitV(a-b)= (0 0 0)
Not sure why you have unitV(a-b) in there. The cross product produces a vector, and you need a vector result. What is the multiplication by unitV(a-b) doing?

Also, the cancellation should be between the torque of F about OA and the torque of the cable tension about OA. You don't seem to have that in your equation at all.
 
  • #3
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OB? It says:



Not sure why you have unitV(a-b) in there. The cross product produces a vector, and you need a vector result. What is the multiplication by unitV(a-b) doing?

Also, the cancellation should be between the torque of F about OA and the torque of the cable tension about OA. You don't seem to have that in your equation at all.
sorry, its unitv(a-o) and not a-b

I could simply do crossP(b-a,vf)*crossP(b-a,vtbc)?
that gives me a weird answer if I dont include unitv(a-o), that part is to use the axis ob
 
  • #4
haruspex
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its unitv(a-o) and not a-b
OK, that makes more sense. You are taking the dot product with that to extract the torque component in the AO direction.
I could simply do crossP(b-a,vf)*crossP(b-a,vtbc)
You don't cancel two vectors by multiplying them together.
crossP(b-a,vf) is the torque of F about OB. Do as you suggested, taking the dot product with unitv(a-o).
Now do all the same with the cable tension as you did with F.
 
  • #5
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OK, that makes more sense. You are taking the dot product with that to extract the torque component in the AO direction.

You don't cancel two vectors by multiplying them together.
crossP(b-a,vf) is the torque of F about OB. Do as you suggested, taking the dot product with unitv(a-o).
Now do all the same with the cable tension as you did with F.
Yes, this is what I have now: dotP(crossP(b-o,vf)+crossP(b-a,vtbc),unitV(a-o))= 0
f= -585N
 
  • #7
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Is there a typo in there?
yes :(
I am sleepy.
dotP(crossP(b-a,vf)+crossP(b-a,vtbc),unitV(a-o))= 0
I hope no more typo now
 
  • #8
haruspex
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yes :(
I am sleepy.
dotP(crossP(b-a,vf)+crossP(b-a,vtbc),unitV(a-o))= 0
I hope no more typo now
Yes, that looks right.
 
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  • #9
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Yes, that looks right.
Question 2) With the strength found in a) , determine the components along y and z according to the time which is O.

solve(crossp(b-a,vf)+crossp(b-a,vtbc)+crossp(o-a,vmo))= (0 0 0)
momx = 0N
momy= -246.9N
momz = 294.26N

Is that correct?
 
  • #10
haruspex
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according to the time which is O.
I have no idea what that means. Is this a translation?
crossp(o-a,vmo)
What's this term? The cross product of a distance vector and a moment? What does that produce?
 
  • #11
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I have no idea what that means. Is this a translation?

What's this term? The cross product of a distance vector and a moment? What does that produce?
Basically I have to find the moments at point O from point A(I decided that point)
 
  • #12
haruspex
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Basically I have to find the moments at point O from point A(I decided that point)
OK, but moments of what about O? F? F+cable tension?
(And I don't understand what you mean by "from point A")
 
  • #13
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OK, but moments of what about O? F? F+cable tension?
(And I don't understand what you mean by "from point A")
I am not really sure, can I just do it from point o?
and do this
solve(crossp(b-o,vf)+crossp(b-o,vtbc)+vmo= (0 0 0))
That make more sense I think now.
 
  • #14
haruspex
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I am not really sure, can I just do it from point o?
and do this
solve(crossp(b-o,vf)+crossp(b-o,vtbc)+vmo= (0 0 0))
That make more sense I think now.
That makes more sense, yes. It assumes you want the total moment, not just that from F, but which is wanted is not clear to me from the question.
One error though - you are asked for the moment, not for what moment you would have to add to get a net zero moment.
 
  • #15
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That makes more sense, yes. It assumes you want the total moment, not just that from F, but which is wanted is not clear to me from the question.
One error though - you are asked for the moment, not for what moment you would have to add to get a net zero moment.
solve(crossp(b-o,vf)+crossp(b-o,vtbc)= (0 0 0))
= 0 -921.91 -773.575
 
  • #17
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?
sorry, I did not put that in my calculator actually.
(crossp(b-o,vf)+crossp(b-o,vtbc) right here.
 
  • #18
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?
Last question:
Determine the three components of the reaction force in O.

I would simply do this ? (vtbc+vf)
and this gives me (-768 -268 320)

Also thank you very much for helping it really do help me a lot.
 
  • #19
haruspex
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Last question:
Determine the three components of the reaction force in O.

I would simply do this ? (vtbc+vf)
and this gives me (-768 -268 320)

Also thank you very much for helping it really do help me a lot.
Yes, except that this time you are looking for a force that gives a net zero. It's the reaction force you are asked for, i.e. the force exerted on the pipe by its wall support.
 
  • #20
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Yes, except that this time you are looking for a force that gives a net zero. It's the reaction force you are asked for, i.e. the force exerted on the pipe by its wall support.
do I have to include vo in my equation?
(vtbc+vf+vo)=(0 0 0) ?
 
  • #21
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do I have to include vo in my equation?
(vtbc+vf+vo)=(0 0 0) ?
That'll work.
 
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  • #22
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That'll work.
just wondering stupid question, when I do 3d calculation with vectors, crossp, unitv my calculator has to be in rad or degree?
 
  • #23
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just wondering stupid question, when I do 3d calculation with vectors, crossp, unitv my calculator has to be in rad or degree?
AFAIK, the rad/degree setting only affects trig functions.
 
  • #24
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AFAIK, the rad/degree setting only affects trig functions.
weird I get different answers for this:
(vtbc+vf+vo)=(0 0 0) ?

my component y change from -1377.24 to 268.803 not sure which one is good lol.
 

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