Answer
(a) $1057330.4\space N$
(b) $237687.9\space lb$
Work Step by Step
Please see the attached image first.
(a) Let's take the x, and y in vertical & horizontal directions respectively.
The y component of the force acting on the roof $F_{y}= 2Fsin60^{\circ}-(1)$
We know that $F=PA-(2)$
(2)=>(1),
$F_{y}= 2PAsin60^{\circ}-(1)$ ; Let's plug known values into this equation.
$F_{y}= 2\times1\times10^{4}Pa\times(14.5\space m)(4.21\space m)\times\frac{\sqrt {3}}{2}=1057330.4\space N$
The x component of the force acting on the roof $F_{x}= Fcos60^{\circ}-Fcos60^{\circ}=0$
Therefore, the net force on the roof = $1057330.4\space N$ Vertically upward
(b) Let's multiply the above value by $(\frac{0.2248\space lb}{1\space N})$ to convert the units into lb.
$Net\space force=1057330.4\space N\times(\frac{0.2248\space lb}{1\space N})=237687.9\space lb$
