$W = \gamma (14)(10)$
$W = 140\gamma$
$F = \gamma [ \, \frac{1}{4}\pi (4^2) \, ]$
$F = 4\pi \gamma$
$R = W - F = 140\gamma - 4\pi \gamma$
$R = (140 - 4\pi)\gamma$
$Rx = 7W - 9F$
$(140 - 4\pi)\gamma x = 7(140\gamma) - 9(4\pi \gamma)$
$(140 - 4\pi)x = 7(140) - 9(4\pi)$
$(140 - 4\pi)x = 980 - 36\pi$
$x = \dfrac{980 - 36\pi}{140 - 4\pi}$
$x = 6.8 \, \text{ in.}$
$Ry = 5W - 6F$
$(140 - 4\pi)\gamma y = 5(140\gamma) - 6(4\pi \gamma)$
$(140 - 4\pi)y = 5(140) - 6(4\pi)$
$(140 - 4\pi)y = 700 - 24\pi$
$y = \dfrac{700 - 24\pi}{140 - 4\pi}$
$y = 4.9 \, \text{ in.}$
Thus, the centroid is located at 6.8 in. to the right of left edge and 4.9 in. above the bottom edge. answer
