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# M2 centre of mass Watch

1. Need some help with 3ii.
What am I doing wrong?

Mass of rectangle = 16,000
Mass of semi circle = 800pi

Distance from AE of rectangle = 100
Distance from AE of semi circle = 200 + 160/3pi

(16,000 x 100) + 800pi(200+160/3pi) = (16000 + 800pi) Y bar

Y bar = 115.88cm
Answer is 117?
Attached Images

2. (Original post by Super199)
Need some help with 3ii.
What am I doing wrong?
Have you attached working?

It's possible I can't see anything due to my workplace's firewall (as imgur and other sites are blocked) but it's necessary to provide working in order for helpers to tell you where you went wrong!
3. (Original post by Kvothe the arcane)
Have you attached working?

It's possible I can't see anything due to my workplace's firewall (as imgur and other sites are blocked) but it's necessary to provide working in order for helpers to tell you where you went wrong!
Imgur wasn't working so I have tried to type it out.
4. (Original post by Super199)
Need some help with 3ii.
What am I doing wrong?

Mass of rectangle = 16,000
Mass of semi circle = 800pi
You've assumed that the area is proportional to the mass. This is true for a uniform lamina, but considering the door as a whole, the semicircle and rectangle are not necessarily of the same density, even though each is uniform in itself.

You need to use the weights rather than the areas.
5. (Original post by ghostwalker)
You've assumed that the area is proportional to the mass. This is true for a uniform lamina, but considering the door as a whole, the semicircle and rectangle are not necessarily of the same density, even though each is uniform in itself.

You need to use the weights rather than the areas.
I see. I'm a bit rusty with moments do you mind helping with 3i?

I tried taking moments about each point but I don't think I am doing it right.

M(G) = 40 * 60 = 140 * X
6. (Original post by Super199)
I see. I'm a bit rusty with moments do you mind helping with 3i?

I tried taking moments about each point but I don't think I am doing it right.

M(G) = 40 * 60 = 140 * X
You've missed the weight of the semicircular section. Else, looks fine.
7. (Original post by ghostwalker)
You've missed the weight of the semicircular section. Else, looks fine.
Cheers

Last one if you don't mind.

If I am taking moments about A do I need to know the distance of the dotted line?
Attached Images

8. (Original post by Super199)
Cheers

Last one if you don't mind.

If I am taking moments about A do I need to know the distance of the dotted line?
I believe you need only know the horizontal distance between A and where the dotted line touches AB as the weight is perpendicular to it.
9. (Original post by Super199)
Cheers

Last one if you don't mind.

If I am taking moments about A do I need to know the distance of the dotted line?
From A, yes, you do.

Note: The distance of the centre of mass of a uniform triangular lamina is 1/3 the perpendicular height from any side
10. (Original post by ghostwalker)
From A, yes, you do.

Note: The distance of the centre of mass of a uniform triangular lamina is 1/3 the perpendicular height from any side

hmm how would I take moments about A

12 * Tsin30 = 8 * 3 * 2g?
11. (Original post by Super199)
hmm how would I take moments about A

12 * Tsin30 = 8 * 3 * 2g?
Where's the "3" come from?
12. (Original post by ghostwalker)
Where's the "3" come from?
Isn't that the distance of the dotted line?
13. (Original post by Super199)
Isn't that the distance of the dotted line?
For moments, you want the perpendicular distance of the line of action of the force.

In this case, that distance from A is 8. The 3 doesn't come into it.
14. (Original post by ghostwalker)
For moments, you want the perpendicular distance of the line of action of the force.

In this case, that distance from A is 8. The 3 doesn't come into it.
Got it cheers

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