Problem 2B-10**: Center of mass of a triangle

Find the center of mass of a uniform thin board of triangle ABC as shown in Fig. 1. The lengths of sides AB, BC and CA are c, a and b respectively.

Scroll down for solution

Solution:

In Fig. 2 a line is dropped from point A perpendicularly to line BC and intersects line BC at point D. The length BD is denoted as d, and the length of AD as h. We need to find d and h first.

Triangle ABD is a rectangular triangle, so we have

h2 = c2 - d2 .

Similarly from the rectangular triangle ADC we get

h2 = b2 - (a - d)2 .

Equating the right-hand sides of the above two equations, we get

c2 - d2 = b2 - (a - d)2 .

Canceling the term d2, we get

d = (c2 + a2 - b2) / 2a.                                 (1)

Thus

h2 = c2 - {(c2 + a2 - b2)2 / 4a2} .                                 (2)

The mass of ΔABC is denoted as M. The area of ΔADB, SL, is

SL = hd / 2,

and the area of ΔADC, SR

SR = (a - dh / 2.

The mass of ΔADB, ML, is

ML = M·SL / (SL + SR) = hd·M / {hd + (a - d)h} = (d/a) ·M ,

and the mass of ΔADC, MR, is

MR = M·SR / (SL + SR) = h(a - d)·M / {hd + (a - d)h} = {(a - d)/a}·M .

We choose a coordinate system such that point D is the origin, x-axis is at the direction of DC and y-axis is at the direction of DA as shown in Fig. 3. The center of mass of ΔABD is denoted as QL, and the center of mass of ΔACD denoted as QR. Problem 2B-09 says that

QR = (QRx, QRy) = ((a - d)/3, h/3),

and

QL = (QLx, QLy) = (-d/3, h/3).

The center of mass of ΔABC, Q = (Qx, Qy), can be written, according to Eq. (5) of Topic 2B-02, as

Qx = (MRQRx + MLQLx) / M
= {(a - d)2 - d2}/3a = (a - 2d)/3,

and

Qy = (MRQRy + MLQLy) / M
= h·{(d/a) + (a - d)/a} = h/3,

where d and h are given in Eq. (1) and Eq. (2) respectively.