Find the center of mass of the triangle with vertices `(a,0), (0,b), (0,c) (a>0, b>c)`

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ishpiro | College Teacher | (Level 1) Educator

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In order to evaluate the integral for `y_c` , we need

`(f(x))^2 = (b-b/ax)^2 = b^2 - 2b^2/ax + b^2/a^2 x^2`

Similarly

`(g(x))^2 = (c - c/ax)^2 = c^2 - 2c^2/ax + c^2/a^2x^2`

Then

`(f(x))^2 - (g(x))^2 = (b^2 - c^2) -2(b^2 - c^2)/a x + (b^2 - c^2)/a^2 x^2`

The antiderivative of this function is

`(b^2 -c^2)x - (b^2 - c^2)/a x^2 + (b^2 - c^2)/a^2 x^3/3`

Plugging this into integral for `y_c` and evaluating it at the limits a and 0, we get

`y_c =1/A int^a_0 1/2((f(x))^2 - (g(x))^2)dx = 1/2(b^2 - c^2)a(1 - 1 + 1/3)`

`` `y_c = 1/A1/6a(b^2 - c^2)` 

Since `A = 1/2a(b-c)` ,

`y_c = 1/3 (a(b^2 - c^2))/a(b-c) = 1/3 ((b-c)(b+c))/(b-c) = 1/3(b+c)`

Therefore the coordinates of the triangle with vertices (a, 0), (0, b) and (0, c) are

`x_c = a/3` and `y_c = 1/3(b+c)`

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ishpiro | College Teacher | (Level 1) Educator

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This is a graph of an example of such a triangle with a = 5, b = 4 and c = -3:

First, let's find the area of this triangle. It can be seen from the picture above that the height of this triangle is h = a dropped on the base formed by the vertices (0, b) and (0, c). Thus the base has the length of b - c and the area is `A = 1/2a(b-c)` .

The formulas for the coordinates of the center of mass are, since the values of x here are limited by 0 and a,

`x_c = 1/A int^a_0 x(f(x) - g(x))dx `

`y_c = 1/A int^a_0 1/2 ((f(x))^2-(g(x))^2)dx`

f(x) is the line connecting (0,b) and (a, 0). It has y-intercept b and the slope -b/a, so its equation is

`f(x) = b - b/ax`

g(x) is the line connecting (0, c) and (a, 0). It has y-intercept c and the slope -c/a, so its equation is

`g(x) = c- c/ax`

`f(x) - g(x) = b - c - (b-c)/a x`

Plugging this in the integral for x_c, get

`int^a_0 x(b-c - (b-c)/a x)dx = [(b-c)x^2/2 - (b-c)/a x^3/3]^a _0=(b-c)a^2(1/2 - 1/3)=1/6a^2(b-c)` 

Dividing this by A = 1/2 a(b-c), get x_c = a/3.

In order to evaluate the integral for y_c, we need

 

 

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