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# m2 moments watch

1. I'm confused as to how I am meant to take moments from D

I understand how to take moments from a point along the rod as you just do the force x perp dist along the rod. But for D I am so confused.

Attached Images

2. (Original post by Katiee224)
I'm confused as to how I am meant to take moments from D

I understand how to take moments from a point along the rod as you just do the force x perp dist along the rod. But for D I am so confused.

It's not "force x perp" as such, but rather "force x perpendicular distance of the line of action of the force".

See attached:

For the weight mg, the line of action is in red, and it's perpendicular distance from D is the top green line, which is the same length as the bottom green line.

Does that cover it?

3. (Original post by ghostwalker)
It's not "force x perp" as such, but rather "force x perpendicular distance of the line of action of the force".

See attached:

For the weight mg, the line of action is in red, and it's perpendicular distance from D is the top green line, which is the same length as the bottom green line.

Does that cover it?

ah ok, but where does the distance 'a' come from in 'mg x asin(theta)'?
4. (Original post by Katiee224)
ah ok, but where does the distance 'a' come from in 'mg x asin(theta)'?
I presume the rod is uniform and of length 2a, so refering to the lower triangle with angle theta, in my diagram, distance of CofM to B is length "a", and so perp. dist. is a sin(theta).

Edit: Corrected A to B.
5. (Original post by ghostwalker)
I presume the rod is uniform and of length 2a, so refering to the lower triangle with angle theta, in my diagram, distance of CofM to A is length "a", and so perp. dist. is a sin(theta).
is the length needed not the one I labelled in this diagram here?

I know I sound stupid right now haha but im not sure why tha'ts not it, as we did that same thing with the F force
Attached Images

6. (Original post by Katiee224)
is the length needed not the one I labelled in this diagram here?

I know I sound stupid right now haha but im not sure why tha'ts not it, as we did that same thing with the F force
The force F acts horizontally. The perpendicular distance of it's line of action is 2a from D. Hence F2a.

The distance from D to the mgsin(theta) isn't going to be "a".
I'm not sure what your mgsin(theta) is meant to represent. Are you trying to resolve mg into components? How do you arrive at it? This may be the root of your problem.
7. (Original post by ghostwalker)
The force F acts horizontally. The perpendicular distance of it's line of action is 2a from D. Hence F2a.

The distance from D to the mgsin(theta) isn't going to be "a".
I'm not sure what your mgsin(theta) is meant to represent. Are you trying to resolve mg into components? How do you arrive at it? This may be the root of your problem.
Are you saying the green line is length a? as that is the perp dist to the line of action of mg?
8. (Original post by Katiee224)
Are you saying the green line is length a? as that is the perp dist to the line of action of mg?
No.

The distance of the CofM from B, is length "a". I.e. half the length of the rod.
The length of the green line is therefore a sin(theta)

Edit: I said "from A" prevously, incorrectly, and that may have confused you. Sorry about that.
9. (Original post by ghostwalker)
No.

The distance of the CofM from B, is length "a". I.e. half the length of the rod.
The length of the green line is therefore a sin(theta)

Edit: I said "from A" prevously, incorrectly, and that may have confused you. Sorry about that.
ohhhhh I understand it now, thanks you've been very helpful
10. (Original post by Katiee224)
ohhhhh I understand it now, thanks you've been very helpful
Cool, and you're welcome.

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Updated: June 12, 2016
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