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m2 moments

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.

Can someone explain it please :frown:
m2mom1.png12.6KB
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.

Can someone explain it please :frown:


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?

m2mom1.png
Reply 2
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Katiee224
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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?

m2mom1.png


ah ok, but where does the distance 'a' come from in 'mg x asin(theta)'?
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.
(edited 7 years ago)
Reply 4
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Katiee224
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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 :colondollar:
(edited 7 years ago)
m2mom1.png115.9KB
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 :colondollar:


The force F acts horizontally. The perpendicular distance of it's line of action is 2a from D. Hence F2a.

With your labelling.
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.
Reply 6
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Katiee224
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Original post by ghostwalker
The force F acts horizontally. The perpendicular distance of it's line of action is 2a from D. Hence F2a.

With your labelling.
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?
(edited 7 years ago)
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.
(edited 7 years ago)
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Katiee224
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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 :biggrin:
Original post by Katiee224
ohhhhh I understand it now, thanks you've been very helpful :biggrin:


Cool, and you're welcome.

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