Answer
$$
F=\frac{200 \sqrt{3}}{3}(62.4) \approx 7205.33 \mathrm{lb}
$$
$$
F=\frac{200 \sqrt{3}}{3}(9800) \approx 1131606.53
$$
Work Step by Step
Since
$$
\frac{\Delta y}{\sin 60^{\circ}}=\frac{2\sqrt{3}}{3}\Delta y
$$
Then
\begin{aligned}
F=\rho g \int_{0}^{10} y f(y) d y &=\rho g \int_{0}^{10} \frac{3}{10} y \cdot \frac{2 \sqrt{3}}{3} y d y \\
&=\frac{\sqrt{3}}{5} \rho g \int_{0}^{10} y^{2} d y \\
&=\frac{\sqrt{3}}{5} \rho g\left[\frac{y^{3}}{3}\right]_{0}^{10} \\
&=\frac{\sqrt{3}}{5} \rho g\left[\frac{1000}{3}\right] \\
&=\frac{200 \sqrt{3}}{3} \rho g
\end{aligned}
Now if distances are in feet, then, $\rho g=w=62.4,$ and so,
$$
F=\frac{200 \sqrt{3}}{3}(62.4) \approx 7205.33 \mathrm{lb}
$$
If distances are in meters, then, $\rho g=9800,$ and so,
$$
F=\frac{200 \sqrt{3}}{3}(9800) \approx 1131606.53
$$
