Problem 2B-09**: Center of mass of a rectangular triangle

Find the center of mass of a uniform thin board of rectangular triangle ABC as shown in Fig. 1. The side AB has a length a and the side BC has a length b.

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An image of ΔABC is folded along the line AB and the point C is relabeled as point D. ΔABC and ΔABD are congruent and are mirror image of each other. The mass of ΔABC and the mass of ΔABD are equal, and both of them are denoted as m. Triangle ACD is an equilateral triangle with a mass 2m. The center of mass of ΔACD is denoted as point P, and the position vector AP as R. According to Problem 2B-08, point P lies on line AB, and the length of AP is (2/3)·AB = 2a/3. Defining a unit length vector j that originates from A, lies on line AB and points toward point B, we can write

                R = (2a/3)·j.                                 (1)

We denote the center of mass of ΔABC as Q, and the position vector AQ as r. The center of mass of ΔABD is denoted as Q' and the position vector AQ' as r'. Eq. (5) of Topic 2B-02 says that

                2mR = mr + mr'     →     r + r' = 2R.                                 (2)

Position vectors r and r' can be written as

                r = rj·j + r ,                                 (3A)


                r' = r'j·j + r' ,                                 (3B)

where r and r' are perpendicular to the position vector j. Substituting Eq. (1) and Eqs. (3A) and (3B) into Eq. (2), we get

                (rj + r'jj + (r + r') = 2R = 2(2a/3)·j.                                 (4)

Since position vectors r and r' are mirror images of each other with the line AB serving the role of the mirror, we must have

                r + r' = 0 ,


                rj = r'j.

Thus Eq. (4) becomes

                rj = r'j = 2a/3.

In other words the center of mass point of ΔABC, Q, lies on the line that passes through point P and is parallel to the line BC.

Next, we fold an image of ΔABC along the line BC and relabel point A as point E as shown in Fig. 3. Triangle CAE is an equilateral triangle and its center of mass point, S, lies on line BC. According to Problem 2B-08, CS = (2/3)·BC = 2b/3. By similar arguments as for the case of Fig. 2, the center of mass point of ΔABC lies on a line that passes through point S and is parallel to the line AB.

In summary, to find the center of mass of the rectangular triangle ABC, we first draw a point P on line AB such that AP = 2a/3 as shown in Fig. 4. Then draw a line PP' that is parallel to the side BC. Next, we draw a point S on line BC such that CS = 2b/3. Then draw a line SS' that is parallel to the side AB. The crossing point between line PP' and line SS' is the center of mass point, Q, of the rectangular triangle ABC.