# 717 Symmetrical Arcs and a Line | Centroid of Composite Line

**Problem 717**

Locate the centroid of the bent wire shown in Fig. P-717. The wire is homogeneous and of uniform cross-section.

**Solution 717**

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$L_1 = 2r\alpha = 2(30)\left( 30^\circ \times \dfrac{\pi}{180^\circ} \right)$

$L_1 = 10\pi = 31.42 \, \text{ cm}$

$z = \dfrac{r \sin \alpha}{\alpha} = \dfrac{30 \sin 30^\circ}{30^\circ (\pi / 180^\circ)}$

$z = 28.65 \, \text{ cm}$

$y = z \sin \alpha = 28.65 \sin 30^\circ$

$y = 14.325 \, \text{ cm}$

$x = 10 + r - z \cos \alpha = 10 + 30 - 28.65 \cos 30^\circ$

$x = 15.19 \, \text{ cm}$

$L = 20 + 2L_1 = 20 + 2(31.42)$

$L = 82.84 \, \text{ cm}$

By symmetry

$\bar{x} = 0$ *answer*

$L \, \bar{y} = \Sigma ly$

$82.84\bar{y} = 31.42(14.325)(2) + 20(0)$

$\bar{y} = 10.87 \, \text{ cm}$ *answer*

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