Find the Y component of the Center of Mass

AI Thread Summary
The discussion focuses on finding the Y component of the center of mass for a triangular shape. The user attempts to calculate this using the integral formula but struggles with defining the area of horizontal strips due to varying dimensions as y increases. They initially arrive at (2/3)B but believe the correct answer should be (1/3)B, suspecting an error in their area definition. Participants suggest simplifying the mass density assumption and finding the equation of the hypotenuse to better define the area element. The conversation emphasizes clarifying the relationship between x and y to proceed with the calculations correctly.
johnnyboy53
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Homework Statement


Find the Y component of the Center of Mass

http://img607.imageshack.us/img607/7122/83370796.png


Homework Equations


∫(r*dm)/M


The Attempt at a Solution


I keep coming up with (2/3)B but i know since that there is more mass near the origin axis, it should be 1/3B. I think it is my definition of area that is wrong. I am using horizontal strips with A * dy but i am unsure how to substitute A since it is a varying as y increases

∫(r*dm)/M
σ=dm/da
σ*da=dm

∫(r*σ*da)/M
da= area of small strip = dy*Mslope=
Y=Mslope*x=
Mslope=B/A

∫(y*σ*dy*y(A/B))/M
y(A/B)=X

(σA/BM )∫y^2dy=

σ= M/Area
Area = 1/2AB
σAB=2M

(AσB2/3M)=
(σAB/3M)*B =
(2MB/3M)=
(2/3)B

Which is wrong. I think the error is somewhere around the da part. I am a little bit unsure of what the area of each horizontal strip would be since the left side of the shape is a function of x, or is it a function of y?
 
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If you write out what you mean by A, B, Mslope etc. it might become apparent to you what's wrong. But if it doesn't, it will give the rest of us a chance to spot it.
 
oh wow i forgot the picture. Sorry!

The lines are the dy's
 
It looks like σ is meant to represent the mass density of the material comprising the triangle, and that it is assumed to be constant (triangle material is of uniform density). If that's the case, then you can ignore it (or assume σ = 1 in whatever units you prefer) and just work with the areas.

If you can find an equation of the line for the hypotenuse of the triangle then you can determine x in terms of y (that is, x(y) ) for the line. With x(y) you can find the length of your area element for a given y (its thickness is dy and its length lies between x(y) and x = A).

So, begin by finding the equation of the line on which the hypotenuse "lives". It's a linear equation of the form y = m*x. Solve for x in terms of y and carry on...
 
Hypotenuse of the triangle would be:
y=Mslope*x
Mslope = B/A
y=(B/A)x
x=(A/B)y

is that right? That's what i did on my attempt
 
johnnyboy53 said:
Hypotenuse of the triangle would be:
y=Mslope*x
Mslope = B/A
y=(B/A)x
x=(A/B)y
OK. What do you have for the mass (or area) of the strip width dy at height y?
 
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