#### Answer

$$(15 b+30 a) h^{2} \text { lb } $$

#### Work Step by Step

Place the origin along the top edge of the trough with the positive $y$ -axis pointing downward. The width of the front side of the trough varies linearly from $b$ when $y=0$ to $a$ when $y=h ;$ thus, the width of the front side of the trough at depth $y$ feet is given by
$$
b+\frac{a-b}{h} y
$$
Then
\begin{aligned}
F&=w \int_{a}^{b} y f(y) d y\\
&=90 \int_{0}^{h} y\left(b+\frac{a-b}{h} y\right) d y \\
&=90 b \int_{0}^{h} y d y+\frac{90(a-b)}{h} \int_{0}^{h} y^{2} d y \\
&=90 b\left[\frac{y^{2}}{2}\right]_{0}^{h}+\frac{90(a-b)}{h}\left[\frac{y^{3}}{3}\right]_{0}^{h} \\
&=45 b h^{2}+30 \frac{(a-b)}{h} h^{3} \\
&=(45 b+30(a-b)) h^{2} \\
&=(45 b+30 a-30 b) h^{2} \\
&= (15 b+30 a) h^{2} \text { lb }
\end{aligned}