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# M2: Centre of Mass - Uniform Square watch

1. Q: The diagram shows a thin uniform square of metal which has a corner folded over. calculate the distance of the centre of mass of ABCEF from the corner A.

Here the diagram, red is what I have added, and I've drawed the table alos. Ignore the the total mass is 72, it should be 56. I think

Don't you do the sqaure - triangle. Then pythrasos to find the disntance from A.

So I get

x: 240=56x

x=4 2/7

y: 208=56y

y= 3 5/7

then phy x^2+y^2 all sqaure rooted = 5.67

whereas I show get 5.42 ??
2. Think about whether the total mass of something changes when you fold over a corner.
3. (Original post by L-x)
Think about whether the total mass of something changes when you fold over a corner.
I'm confused... if we fold the corner over it's 2D, so the area of the triangle is 16 not 8 isn't it? What's the total mass I keeping getting different values 72, 58, 80 which is it???
4. If you took a piece of paper and folded a corner over, would the total mass of the paper have changed?

Be careful not to count the area under the folded part twice.
5. (Original post by L-x)
If you took a piece of paper and folded a corner over, would the total mass of the paper have changed?

Be careful not to count the area under the folded part twice.
No.

I'm soo confused now, so the mass of the sqaure is 64, and the total mass is 64, but what about the triangle?
6. I'm just trying to get you to think about what you mean by the "total", especially if you decide to "take away" one mass from another. The mass of the entire shape is obviously going to be 64. The mass of the bit you're folding is going to be 8. If you "take away" some mass (i.e. the bit you're folding) from the corner of the sheet, remember that the mass hasn't disappeared, it's been moved, so you'll need to "add it back in" again in it's new position.
7. (Original post by L-x)
I'm just trying to get you to think about what you mean by the "total", especially if you decide to "take away" one mass from another. The mass of the entire shape is obviously going to be 64. The mass of the bit you're folding is going to be 8. If you "take away" some mass (i.e. the bit you're folding) from the corner of the sheet, remember that the mass hasn't disappeared, it's been moved, so you'll need to "add it back in" again in it's new position.
I'm still not understanding this -- so now I'm dealing with 3 shapes

Mass 8 -8 64 Total M 84
X 2 2 4 x
Y 6 6 4 y
8. sort of.
If you start off with the [mass of the whole square times the distance of the CM of the square from A], then take away the [mass of the triangle times the distance of the CM of the triangle from A] you've found the moment of mass from A of a shape where you cut off the corner (and then you'd divide by the total mass of that new shape to find the distance of the CM)

However because we're folding the corner, not cutting it off, that's not good enough. What we have to do is as before, except we then add the [mass of the triangle time it's new distance from A] and we divide by the total mass of the sheet.

DUCY?

If it helps you to understand what's going on, try doing it without worrying about moving the corner first, so just pretend you've cut it off and then put it back in the same place, you should get the centre of mass of the square (obviously, because our imaginary glue we're sticking it back together with doesn't weigh anything, so nothing has changed)

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