**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

Scroll down for
solution

**Solution:**

The equilateral triangle is cut into N parallel strips each of that has a width
of Δ*h* = *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 C_{k} and the points that the
bottom line of the strip crosses two sides of the equilateral triangle as
B_{k} and D_{k} respectively. The
line that passes through point A and point C_{N} must cross
BD perpendicularly. Every C_{k, 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 =
|Rk|·j =
(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 .
}