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1) The two sides of the wall measure [tex]l_1 = 10~ft=3.05~m[/tex] and [tex]l_2=20~ft=6.10~m[/tex]. Therefore, the area of the wall is
[tex]A=l_1 l_2=18.6~m^2[/tex]
2) The net pressure acting on the wall is given by the difference between the pressure acting from inside and from outside:
[tex]p_{net} = p_{in}-p_{out} = 30~inHg-26.3~inHg=3.7~inHg=12494~Pa [/tex]
3) The relationship between pressure, force F and area of the wall is
[tex]p_{net}= \frac{F}{A}
[/tex]
from which we found
[tex]F=p_{net}A = 232388~N=232.4~kN [/tex]
And given that [tex]1~N=0.23~lbf[/tex], we have
[tex]F=232288~N=53499~lbf[/tex]
