Question

Find the force acting on the wall of a house, measuring 10ft by 20ft, when the pressure inside the house is 30.0 inches of mercury, and the pressure outside is 26.3 inches of mercury. Express the answer in lbf and N.

Answer 1

Hello

1) The two sides of the wall measure
`l_1 = 10~ft=3.05~m` and
`l_2=20~ft=6.10~m`. Therefore, the area of the wall is

`A=l_1 l_2=18.6~m^2`

2) The net pressure acting on the wall is given by the difference between the pressure acting from inside and from outside:

`p_(net) = p_(in)-p_(out) = 30~inHg-26.3~inHg=3.7~inHg=12494~Pa`

3) The relationship between pressure, force F and area of the wall is

`p_(net)= (F)/(A)`
from which we found

`F=p_(net)A = 232388~N=232.4~kN`
And given that
`1~N=0.23~lbf`, we have

`F=232288~N=53499~lbf`