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# For the cellar of a new house a hole is dug in the ground, withvertical sides going down 2.40m. A concrete foundation wallis built all the way across

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For the cellar of a new house a hole is dug in the ground, withvertical sides going down 2.40m. A concrete foundation wallis built all the way across the 9.6m width of the excavation. This foundation wall is 0.183m away from the front of the cellarhole. During a rainstorm, drainage from the streetfills up the space in front of the concrete wall, but not thecellar behind the wall. The water does not soak into the clays oil. Find the force the water causes on the foundation wall. For comparison, the weight of the water is given by:
$$\displaystyle{2.40}{m}\cdot{9.60}{m}\cdot{0.183}{m}\cdot{1000}{k}\frac{{g}}{{m}^{{3}}}\cdot{9.8}\frac{{m}}{{s}^{{2}}}={41.3}{k}{N}$$

2021-04-21
When the hole is dugging in the initial stage the the verticalsides from 0m to 2.40m
So average height is $$\displaystyle{h}={\frac{{{0}+{2.40}{m}}}{{{2}}}}={1.20}{m}$$
We know the formula for the gauge pressure is
$$\displaystyle{P}_{{{g}{a}{u}\ge}}=\rho{g}{h}$$
$$\displaystyle={\left({10}^{{3}}{k}\frac{{g}}{{m}^{{3}}}\right)}{\left({9.8}\frac{{m}}{{s}^{{2}}}\right)}{\left({1.20}{m}\right)}$$
$$\displaystyle={11.76}\cdot{10}^{{3}}\frac{{N}}{{m}^{{2}}}$$
Now the formula for the force the watercauses on the foundation wall is
$$\displaystyle{F}={\left({P}_{{{g}{a}{u}\ge}}\right)}{A}$$
$$\displaystyle={\left({11.76}\cdot{10}^{{3}}\frac{{N}}{{m}^{{2}}}\right)}{\left({2.40}{m}\right)}{\left({9.60}{m}\right)}$$
$$\displaystyle={270.9}\cdot{10}^{{3}}{N}$$