Data: Radius=1.5 m, Width=8 m

To find: Resultant force ($F_{A}$)=?, Location=$\theta$

a)Horizontal component ($F_{X}$)

$F_{X}=w_{w}.A.\bar{h}$

Here, $\bar{h} =\cfrac{1.5}{2}=0.75 m$

A=$1.5 \times 8 =12 m^{2}$

$w_{w}=9810 N/m^{2}$

∴$F_{X}=9810 \times 12 \times 0.75 = 88.29 \times 10^{3}$ N

b) Vertical component ($F_{Y}$)

$F_{Y}=w_{w} \times \text{cls area of ABC} \times width of door = w_{w} \times \cfrac{\pi}{4} \times 1.5^{2} \times 8 = 138.685 \times 10^{3} N$

Now, Resultant force is = R = $\sqrt{F_{X}^{2}+F_{Y}^{2}} = \sqrt{(88.29 \times 10^{3})^{2}+(138.685 \times 10^{3})^{2}} = 164.68 \times 10^{3}$ N

$\theta = tan^{-1} \left( \cfrac{F_{Y}}{F_{X}} \right) = 57.51^{\circ}$