Finding Center of Mass of an Isosceles Triangle

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Homework Help Overview

The problem involves finding the center of mass of an isosceles triangle situated on the xy-plane, with specific dimensions and a constant density. The original poster mentions the triangle's mass and density, and seeks clarification on the integration process related to its height.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the integration of small strips to find the center of mass, questioning the division of height by the base length. There are inquiries about the slope of the line connecting the triangle's vertices and its implications on the integration process.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the triangle's dimensions and the mathematical relationships involved. Some guidance has been offered regarding the equation of the line and the integration process, but no consensus has been reached on the final approach.

Contextual Notes

Participants note potential confusion regarding the triangle's height and base placement, as well as the implications of integrating over the triangle's dimensions. There is also mention of terminology, distinguishing between "center of mass" and "centroid."

armolinasf
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Homework Statement



an isosceles triangle with mass m and constant density is placed on an xy plane with base on the y axis. its height ranges from (0,-b/2) to (0,b/2) and its height is (a,0)

Find its center of mass.

I know that its density is equal 2m/ab.

The Attempt at a Solution



The solution says that the shape of the small strips which i am supposed to integrate from zero to a are given by 2*b/2a*(a-x)

What I don't understand is why the heght b/2 is divided by a? I would appreciate it if someone could explain this to me. Thanks
 
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What is the equation of the line which passes through (0, b/2) and (a, 0)?

What is the slope of this line?
 
It would be -b/2a. so that would be the slope of the height of the rectangles that were integrating but since half the triangle is above the axis and the other half is below wouldn't we need to multiply that height by 2 leaving us with -b/a plus this also introduces a negative which isn't in the solution...
 
armolinasf said:
It would be -b/2a. so that would be the slope of the height of the rectangles that were integrating but since half the triangle is above the axis and the other half is below wouldn't we need to multiply that height by 2 leaving us with -b/a plus this also introduces a negative which isn't in the solution...
Correct, the slope is -b/2a .

What's the equation of the line?



The height is multiplied by 2.
 
so the equation of the line becomes
b(a-1)/2a I have a feeling that the 1 should be x.

But doesn't this like give you the height at any point of the triangle so would would multiply this by 2 right?
 
Last edited:
armolinasf said:
so the equation of the line becomes
b(a-1)/2a I have a feeling that the 1 should be x.

But doesn't this like give you the height at any point of the triangle so would would multiply this by 2 right?
Yes.

y = (-b/(2a))·x + b/2  →  y = (-b/(2a))·x + b/(2a)·a  →  y = (b/(2a))(a-x)

Then as you said earlier, we need to multiply this by 2.
 
By the way, the "center of mass" of a geometric figure of constant density is more properly called the "centroid". And it happens that the centroid of a triangle is the point whose coordinates are the average of the coordinates of the three vertices.

"its height ranges from (0,-b/2) to (0,b/2) and its height is (a,0)"
Surely you didn't mean to say that. Since the triangle has its base on y-axis, Its base "ranges from (0, -b/2) to (0, b/2)". And it has a third vertex at (a, 0) so its height is the number, a.
 
I got this one. Thanks for all the help.
 

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