# 239 Resultant of lift on the wing of an airplane

**Problem 239**

The 16-ft wing of an airplane is subjected to a lift which varies from zero at the tip to 360 lb per ft at the fuselage according to w = 90x^{1/2} lb per ft where x is measured from the tip. Compute the resultant and its location from the wing tip.

**Solution 239**

$\displaystyle R = \int_0^{16} y \, dx = 90 \int_0^{16} x^{1/2} \, dx$

$R = 90 \left[ \dfrac{x^{3/2}}{3/2} \right]_0^{16} = 60 \left[ x^{3/2} \right]_0^{16}$

$R = 60(16^{3/2} - 0^{3/2})$

$R = 3840 \, \text{ lb}$ upward

$\displaystyle Rd = \int_0^{16} x(y \, dx) = 90 \int_0^{16} x(x^{1/2}\, dx)$

$\displaystyle 3840d = 90 \int_0^{16} x^{3/2}\, dx$

$3840d = 90 \left[ \dfrac{x^{5/2}}{5/2} \right]_0^{16} = 36 \left[ x^{5/2} \right]_0^{16}$

$3840d = 36(16^{5/2} - 0^{5/2})$

$3840d = 36\,864$

$d = 9.6 \, \text{ ft}$

Thus, R = 3840 lb upward at 9.6 ft from the tip of the wing. *answer*

**Another Solution**

$R = \text{ area of parabola}$

$R = \frac{2}{3}bh = \frac{2}{3}(16)(360)$

$R = 3840 \, \text{ lb}$ *okay!*

$\bar{x} = \text{ centroid of parabola}$

$x = \frac{3}{5}h = \frac{3}{5}(16)$

$x = 9.6 \, \text{ ft}$ *okay!*