Find the Y component of the Center of Mass

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Homework Help Overview

The discussion revolves around finding the Y component of the Center of Mass for a triangular shape, with participants exploring the relevant equations and their application to the problem context.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the Y component using integration and expresses uncertainty about the definition of area and the substitution of variables. Other participants suggest clarifying definitions and equations, particularly regarding the hypotenuse of the triangle and the mass density.

Discussion Status

Participants are actively engaging with the problem, with some providing guidance on simplifying assumptions, such as treating mass density as constant. There is an ongoing exploration of the geometric relationships within the triangle, particularly how to express x in terms of y.

Contextual Notes

There is mention of a diagram that is crucial for understanding the problem, which has not been included in the discussion. The original poster also notes confusion regarding the area of horizontal strips as y increases.

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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