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# How to take moment vertically? M2 Watch

1. In M1 you got a moment of 0 if the line of action passed through a point yet in M2 you can take moments about a point O to work out the center of mass of particles lined up vertically. How does one take moments about the x-axis too?
2. (Original post by Jackabc)
In M1 you got a moment of 0 if the line of action passed through a point yet in M2 you can take moments about a point O to work out the center of mass of particles lined up vertically. How does one take moments about the x-axis too?
I am having trouble understanding your question.

You always take moments about an axis, not about a point. When you take moments "about a point", you are really taking moments about an axis normal to the plane of the paper.

If you take moments about the x-axis, then you find the perpendicular distance from the axis to the line of action of the force.

I think the issue you may be having is that you think of the forces as acting in the x-y plane. We actually set it up so that the forces act in the direction of the z-axis. In other words, if you have a particle of weight W with Cartesian coordinates (0,k) and you take moments about the x-axis, then the moment of the force is Wk.
3. (Original post by Brister)
I am having trouble understanding your question.

You always take moments about an axis, not about a point. When you take moments "about a point", you are really taking moments about an axis normal to the plane of the paper.

If you take moments about the x-axis, then you find the perpendicular distance from the axis to the line of action of the force.

I think the issue you may be having is that you think of the forces as acting in the x-y plane. We actually set it up so that the forces act in the direction of the z-axis. In other words, if you have a particle of weight W with Cartesian coordinates (0,k) and you take moments about the x-axis, then the moment of the force is Wk.
What I have attached shows what I am having a problem with as how is it taking moments about the x-axis? Where is the perpendicular distance? And isn't the line of action, where gravity is a acting, is going through the axis so there is no moment. It's so confusing.

Posted from TSR Mobile
Attached Images

4. (Original post by Jackabc)
What I have attached shows what I am having a problem with as how is it taking moments about the x-axis? Where is the perpendicular distance? And isn't the line of action, where gravity is a acting, is going through the axis so there is no moment. It's so confusing.

Posted from TSR Mobile
See attachment.
Attached Images

5. (Original post by Brister)
See attachment.
I am still slightly confused. I see you rotated but in the initial diagram which is a plane but the force goes right through the x-axis.
6. (Original post by Jackabc)
I am still slightly confused. I see you rotated but in the initial diagram which is a plane but the force goes right through the x-axis.
No, it does not go through the axis. The force is perpendicular to the x-y plane. I did my best to make it look three-dimensional. Really, I did. The two pictures are exactly equivalent, just seen from a different view.

Imagine holding a piece of paper exactly parallel to the ground and let the plane of the paper be the x-y plane. Its weight acts perpendicular to the plane. If you hold the middle of an edge of the paper, it will rotate about that edge.

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Updated: April 6, 2013
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