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# Help on Centre of Mass questions - M2 (Edexcel) watch

1. I am having problems with the CoM questions on most of the past papers...
i am stuck on how to calculate the centre of mass of all triangles!!

can someone plz explain it to me, or if you have a website that might help ... that would be great as well!!!

THX
2. (Original post by KingAS)
I am having problems with the CoM questions on most of the past papers...
i am stuck on how to calculate the centre of mass of all triangles!!

can someone plz explain it to me, or if you have a website that might help ... that would be great as well!!!

THX
Draw a dotted line from the midpoint of one edge to the vertex of the opposite side.
The COM is 1/3 of the distance from the end which touches the midpoint to the end which touches the vertex.
To find the exact coordinates, consider the horizontal and vertical components seperately.
For example, if a line goes 6 across and 3 down, then the COM is (1/3)(6) = 2 across and (1/3)(3) down = 1 down, from the position of the mid point of the edge.
3. (Original post by KingAS)
I am having problems with the CoM questions on most of the past papers...
i am stuck on how to calculate the centre of mass of all triangles!!

can someone plz explain it to me, or if you have a website that might help ... that would be great as well!!!

THX
Draw the triangle and join the mid-point of each side to the opposite vertex. The intersection of the 3 lines is the CoM.

Also

draw the triangle with one side on the x-axis, and draw the other 2 sides.

measure one-third up each of these 2 sides and draw a line joining the points. Ybar is on this line.

Aitch
4. (Original post by Aitch)
Draw the triangle and join the mid-point of each side to the opposite vertex. The intersection of the 3 lines is the CoM.

Also

draw the triangle with one side on the x-axis, and draw the other 2 sides.

measure one-third up each of these 2 sides and draw a line joining the points. Ybar is on this line.

Aitch
thx, yeah i understand that its 1/3 the distance from the median (from the midpoint end).

but is there any simple way to calculate this mathematicall... without having to draw the lines and measure it?

i remember my teacher mentioning something about similar triagles to calculate it?? possible?
5. (Original post by KingAS)
thx, yeah i understand that its 1/3 the distance from the median (from the midpoint end).

but is there any simple way to calculate this mathematicall... without having to draw the lines and measure it?
If you've got a question you're stuck on, put it up. It's easier to do an example, particularly if there's a certain type of question that yuou get stuck on.

Aitch
6. (Original post by Aitch)
If you've got a question you're stuck on, put it up. It's easier to do an example, particularly if there's a certain type of question that yuou get stuck on.

Aitch
alrighty, thanks... watch this space
7. (Original post by KingAS)
thx, yeah i understand that its 1/3 the distance from the median (from the midpoint end).

but is there any simple way to calculate this mathematicall... without having to draw the lines and measure it?

i remember my teacher mentioning something about similar triagles to calculate it?? possible?
You don't have to measure anything.
Perhaps this can help things. After you begin to understand it becomes intuitive.

http://img113.exs.cx/img113/7126/swscan000934xh.jpg
8. (Original post by KingAS)
thx, yeah i understand that its 1/3 the distance from the median (from the midpoint end).

but is there any simple way to calculate this mathematicall... without having to draw the lines and measure it?

i remember my teacher mentioning something about similar triagles to calculate it?? possible?
Yes, let's take an easy example

Triangle A(0,0) B(3,9) C(0,12)

Can you see that if you calculate or measure 1/3 along AB and CB you are actually creating similar triangles? Similarly for AC and BC?
9. (Original post by KingAS)
alrighty, thanks... watch this space
I think Gaz031 and I both have the Edexcel Heinemann M2 text. He can correct me if I'm wrong!! He may not have it with him of course...

Aitch
10. A way of looking at it.

Get the mid point of AB (X) , and midpoint of AC (Y). If you then draw the lines accordingly. The point at which they over lap, 'G', will be the CoM, given that ABC is a uniform lamina.

Also, as a result you have similar triangles all over the place - so loads of proportional values :-)

You also know that that the CoM of a uniform lamina is 2/3 the way down the line from a vertex to the midpoint of the opposite side of that vertex.

i.e. you know GC = (2/3)XC

So now, you should be able to work out any value, given the necessary values by using the idea of similar triangles/proportional lines.

Hope that helps.

MdSalih

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