Finding Center of Mass of an Isosceles Triangle

[armolinasf]

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

 

[SammyS]

What is the equation of the line which passes through (0, b/2) and (a, 0)?

What is the slope of this line?

 

[armolinasf]

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...

 

[SammyS]

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.

 

[armolinasf]

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?

 

[SammyS]

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.

 

[HallsofIvy]

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.

 

[armolinasf]

I got this one. Thanks for all the help.