Centre of mass for a triangle with 3 different lengths Watch

jobo3
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Hi,

I've got M2 tomorrow and the only thing I can't understand in my notes is how to find this. There's some reference to (1/3) but my diagrams crap.

Any help?

Thanks a lot

Edit: it's a solid triangle
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Kolya
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Don't we need to know if we are working with a wire-frame or a lamina?
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jobo3
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(Original post by Kolya)
Don't we need to know if we are working with a wire-frame or a lamina?
Indeed, just realised I forgot that detail, it's a lamina
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emily10
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(Original post by jobo3)
Hi,

I've got M2 tomorrow and the only thing I can't understand in my notes is how to find this. There's some reference to (1/3) but my diagrams crap.

Any help?

Thanks a lot

Edit: it's a solid triangle
I’ve just done a question like this on the ocr jan 2007 paper and it took me a while to figure it out! It helps to draw lines from each vertex to the opposite side and the centre of mass lies where all the lines meet.
It’s hard to explain without reference to a particular question but the x and y values can be deduced from the smaller triangles made by drawing the lines. Say the letter l is one of the lines drawn from one vertex, through trigonometry of right angled triangle, you can see that one side (for example the x value on the horizontal) is equal to lcos(theta). You don't actually need to calculate lcos(theta). Instead to find the actual length, you multiply both sides by 2/3 to get the x value
I hope that makes sense!

Oh and by the way, the 2/3 come from the formula that says that the centre of mass for a triangular lamina is 2/3 along the median from the vertex
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DFranklin
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Centre of mass for a triangle with corners a, b, c is just (a+b+c)/3.

If a geometric solution suits you better, given a triangle ABC, find the midpoint of AB and join it to C. Then find the midpoint of BC and join it to A. Where the two lines meet is the centre of mass.
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jobo3
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(Original post by DFranklin)
Centre of mass for a triangle with corners a, b, c is just (a+b+c)/3.

If a geometric solution suits you better, given a triangle ABC, find the midpoint of AB and join it to C. Then find the midpoint of BC and join it to A. Where the two lines meet is the centre of mass.

Right so if I've got a triangle with sides of 4, 6, and 3, could you run through finding the centre of mass for that using your a+b+c / 3 formula.

Thanks a lot
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DFranklin
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That's not the right formula for the scenario you describe. I'd use the geometric approach. Can you post an actual exam question?
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TheRandomer
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bump this.
How about this from OCR June 04 paper...

How could we find the COM distance from the left of the triangle, or the bottom of the triangle, please?


Edit: I didn't really realise that the OP wanted one with 3 different lengths, but I'd like help on this one all the same!
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TheRandomer
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buuuuuump!
Exam is in 1hr 15!!!
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DFranklin
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Taking the position of A as the origin (0,0), we have B = (6,0) and C=(0,6).

Then (A+B+C)/3 = (6,6) /3 = (2,2).

This is the position on the center of mass. The distance from bottom and left edges is therefore 2.

Incidentally, if you think about it, you'll see a problem with the OP's question. Whatever the answer is, it's not at all obvious how you can express it without knowing the coordinates of A, B and C.
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Icy_Mikki
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(Original post by TheRandomer)
bump this.
How about this from OCR June 04 paper...

How could we find the COM distance from the left of the triangle, or the bottom of the triangle, please?


Edit: I didn't really realise that the OP wanted one with 3 different lengths, but I'd like help on this one all the same!
The formula (A+B+C)/3 uses the vector coordinates of the points on the triangle. So A = (0,0), B = (0,6) and C = (6,0)

Centre of Mass = [A+B+C]/3 [(0,0) + (0,6) + (6,0)] / 3
x coordinate = (0 + 0 + 6) /3 = 2
y coordinate = (0 + 6 + 0) /3 = 2 as well.

So COM = (2,2)
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