Locate the centroid of the L-shaped area in Fig. 1.

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**A—Addition Method** Following the procedure outlined above, we divide the L-shaped area into two rectangles, as shown in Fig. 2.

The y and z axes are located along the outer edges of the area, with the origin at the lower-left corner. Since the composite area consists of only two areas, the composite centroid, C, lies between C_1 and C_2 on the line joining the two centroids, as illustrated in Fig. 2.

Area: From Eq. (C-4a),

\begin{array}{c}{{A=\sum_{i}A_{i},~~~Q_{z}=\overline{{{y}}}A=\sum_{i}\overline{{{y}}}_{i}A_{i},~~~Q_{y}=\overline{{{z}}}A=\sum_{i}\overline{{{z_i}}}}A_i}\end{array} (C-4)

A = A_1 + A_2 = (6t)(t) + (8t)(t) = 14t² (1)

Centroid: From Eqs. (C-4b) and (C-4c),

\bar{y}A = \bar{y}_1A_1 + \bar{y}_2A_2 = (t/2)(6t²) + 5t(8t²) = 43t²

\bar{y} = \frac{43t^3}{14t^2} = \frac{43}{14}t = 3.07t (2)

\bar{z}A = \bar{z}_1A_1 + \bar{z}_2A_2 = (3t)(6t²) + (t/2)(8t²) = 22t²

\bar{z} = \frac{22t^3}{14t^2}= \frac{22}{14}t = 1.57t (3)

**B—Subtraction Method** Sometimes (although not in this particular example) it is easier to solve composite-area problems by treating the area as the net area obtained by subtracting one or more areas from a larger area. Then, in Eqs. (C-4), the A_i’s of the removed areas are simply taken as negative areas. This method will now be applied to the L-shaped area in Fig. 1 by treating it as a larger rectangle from which a smaller rectangle is to be subtracted (Fig. 3). Area A_1 is the large rectangle PQRS; area A_2 is the smaller unshaded rectangle. The composite centroid, C, lies along the line joining the two centroids, C_1 and C_2, but it does not fall between them.

Area: From Eq. (C-4a),

A = A_1 + A_2 = (9t)(6t) + [-(8t)(5t)] = 14t² (4)

Centroid: From Eqs. (C-4b) and (C-4c),

\bar{y}A = \bar{y}_1A_1 + \bar{y}_2A_2 = (4.5t)(54t²) + [(5t)(-40t²)] = 43t²

\bar{y} = 43t³/14t² = 3.07t (5)

\bar{z}A = \bar{z}_1A_1 + \bar{z}_2A_2 = (3t)(54t²) + [(3.5t)(-40t²)] = 22t²

\bar{z} = 2t³/14t² = 1.57t (6)

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