Problem 2B-08**: Center of mass of an equilateral triangle

Find the center of mass of a uniform thin equilateral triangular board. The equilateral triangle has a height h, and two equal sides span an angle of 2θ as shown in Fig. 1

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Solution:

The equilateral triangle is cut into N parallel strips each of that has a width of Δh = h / N. Each strip is then approximated as a rectangular strip as shown in Fig. 2. The center of mass of each rectangular strip is at its midpoint as discussed in Problem 2B-04. Let us denote the center of mass of the k-th rectangular strip as Ck and the points that the bottom line of the strip crosses two sides of the equilateral triangle as Bk and Dk respectively. The line that passes through point A and point CN must cross BD perpendicularly. Every Ck, from k = 1 to k = N - 1, all lie on this line. Let the cross point of line ACN and line BkDk be denoted as Gk. In this configuration of Fig. 2, we have

∠GkABk = θ,

and

AGk = k·Δh = kh / N.

Thus

BkGk = AGk·tanθ = (h / N)·(tanθ)·k.

The area of the k-th rectangular strip, ΔSk, is

ΔSk = 2BkGk·Δh = 2(h/N)2·(tanθ)·k .

We denote the mass density of the board, that is, the mass of unit area of the uniform material that makes up the board as ρ (kg/m2). The mass of the k-th strip, Δmk, is

Δmk = ρ·ΔSk = 2ρ·(h/N)2·(tanθ)·k .                                 (1)

The distance ACk is

ACk = AGk - (Δh / 2) = (kh / N) - (h / 2N) = (h / N)·(k - 1/2).

The position vector ACk is denoted as Rk. We have

| Rk| = ACk = (h / N)·(k - 1/2) .

Since every Rk, (k = 1, 2, · · ·, N) originates from point A, lies on the center line AG, and pointing toward point G, it is convenient to define a unit length vector j that originates from point A, lies on line AG and points toward point G. We can now write

Rk = |Rkj = (h/N)·(k - 1/2)·j.                                 (2)

Combining Eq. (1) and Eq. (2), we have

Δmk·Rk = 2ρ(h/N)3·(tanθ)·k·(k - 1/2)·j .                                 (3)

Therefore,

Δm1·R1 + Δm2·R2 + · · · + ΔmN·RN = 2ρ(h/N)3·(tanθ)·j· {1·(1 - 1/2) + 2·(2 - 1/2) + · · · + N·(N - 1/2)} .

Since

1·(1 - 1/2) + 2·(2 - 1/2) + · · · + N·(N - 1/2) = {12 + 22 + · · · + N2} - (1/2)·(1 + 2 + · · · + N)
= {(N3/3) + (N2/2) + (N/6)} - (1/2)·(1/2)·(N2 + N) = (N3/3) + (N2/4) - (N/12) ,

we have

Δm1·R1 + Δm2·R2 + · · · + ΔmN·RN = 2ρh3·(tanθ)·[{(1/3) + (1/4N) - (1/12N2)}·j .                                 (4)

The mass of N rectangular strips as a whole, denoted as M(N), can be calculated by repeatedly using Eq. (1) as

M(N) = Δm1 + Δm2 + · · · + ΔmN = 2ρ(h/N)2·(tanθ)·(1 + 2 + · · · + N)
= 2ρ(h/N)2·(tanθ)·(N/2)·(N + 1) = ρh2·(tanθ)·(1 + 1/N) .                                 (5)

Using Eq. (4) and Eq. (5), the position vector, R(N), of the center of mass of the N-strip system can be calculated from Eq. (5) of Topic 2B-02 with point A as the reference point as

R(N) = [{2ρh3 ·(tanθ)} / {ρh2 ·(tanθ)}] ·[{(1/3) + (1/4N) - (1/12N2)} / {1 + (1/N)}]·j
= 2h·[{(1/3) + (1/4N) - (1/12N2)} / { 1 + (1/N)}]·j .

As N → ∞, the N-strip system approaches the actual equilateral triangular board and R(N) approaches the position vector of the center of mass of the actual board, R. Thus

R = limN → ∞R(N) = (2/3)·h·j .