Moment of force from an axis - 3D problem

In summary: One error though - you are asked for the moment, not for what moment you would have to add to get a net zero...Yes, that makes more sense. You are asking for the total moment, not just the moment from F. I can do that though.Yes, that makes more sense. You are asking for the total moment, not just the moment from F. I can do that though.
  • #1
masterchiefo
212
2

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.
 

Attachments

  • inglol.PNG
    inglol.PNG
    10.2 KB · Views: 418
Last edited:
Physics news on Phys.org
  • #2
masterchiefo said:
VFxrOB
OB? It says:
masterchiefo said:
the resultant moment about the axis OA is canceled.

masterchiefo said:
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
haruspex said:
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 don't include unitv(a-o), that part is to use the axis ob
 
  • #4
masterchiefo said:
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.
masterchiefo said:
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
haruspex said:
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
 
  • #6
masterchiefo said:
dotP(crossP(b-o,vf)+crossP(b-a,vtbc),unitV(a-o))= 0
Is there a typo in there?
 
  • #7
haruspex said:
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
masterchiefo said:
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.
 
  • Like
Likes masterchiefo
  • #9
haruspex said:
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
masterchiefo said:
according to the time which is O.
I have no idea what that means. Is this a translation?
masterchiefo said:
crossp(o-a,vmo)
What's this term? The cross product of a distance vector and a moment? What does that produce?
 
  • #11
haruspex said:
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
masterchiefo said:
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
haruspex said:
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
masterchiefo said:
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
haruspex said:
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
 
  • #16
masterchiefo said:
= (0 0 0))
?
 
  • Like
Likes masterchiefo
  • #17
haruspex said:
?
sorry, I did not put that in my calculator actually.
(crossp(b-o,vf)+crossp(b-o,vtbc) right here.
 
  • #18
haruspex said:
?
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
masterchiefo said:
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
haruspex said:
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
masterchiefo said:
do I have to include vo in my equation?
(vtbc+vf+vo)=(0 0 0) ?
That'll work.
 
  • Like
Likes masterchiefo
  • #22
haruspex said:
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
masterchiefo said:
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
haruspex said:
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.
 

1) What is a moment of force from an axis in a 3D problem?

The moment of force from an axis in a 3D problem is a measure of the rotational force applied to an object around a specific axis. It is also known as torque and is typically measured in units of Newton-meters (Nm) in the metric system or pound-feet (lb-ft) in the imperial system.

2) How do you calculate the moment of force from an axis in a 3D problem?

To calculate the moment of force from an axis in a 3D problem, you need to know the magnitude and direction of the force being applied, as well as the distance from the axis of rotation to the point where the force is being applied. The formula for calculating moment of force is M = F x d, where M is the moment of force, F is the force, and d is the distance from the axis of rotation.

3) What is the difference between a 2D and 3D moment of force from an axis?

The main difference between a 2D and 3D moment of force from an axis is the number of dimensions being considered. In a 2D problem, the force and distance are both in the same plane, while in a 3D problem, the force and distance may be in different planes, requiring the use of vector calculations.

4) How does the direction of the force affect the moment of force from an axis?

The direction of the force has a significant impact on the moment of force from an axis. If the force is applied perpendicular to the axis of rotation, it will have the maximum effect on the rotational force. However, if the force is applied at an angle, the moment of force will be reduced based on the angle of the force relative to the axis.

5) What is the significance of moment of force from an axis in real-world applications?

Moment of force from an axis is a crucial concept in many real-world applications, particularly in engineering and physics. It is used to understand and predict the rotational motion of objects, such as in the design of machines and structures. It is also essential in sports, such as in the analysis of a baseball pitch or a tennis serve, where rotational force is a critical component of the movement.

Similar threads

  • Introductory Physics Homework Help
Replies
16
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
973
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
6K
  • Introductory Physics Homework Help
Replies
16
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
11K
  • Introductory Physics Homework Help
Replies
11
Views
4K
  • Introductory Physics Homework Help
Replies
8
Views
6K
Back
Top