Question

For the cellar of a new house, a hole is dug in the ground, with the vertical sides going down 2.40 m. A concrete foundation wall is built all the way across the 9.60-m width of the excavation. This foundation wall is 0.183 m away from the front of the cellar hole. During a rainstorm, drainage from the street fills up the space in front of the concrete wall, but not the cellar behind the wall. The water does not soak into the clay soil. Find the force the water causes on the foundation wall. For comparison, the weight of the water is given by$2.40\text{m}\times 9.60\text{m}\times 0.183\text{m}\times 1000{\text{kg/m}}^{\text{3}}\times 9.8{\text{m/s}}^{\text{2}}=41.3\text{kN}$

Short answer

The force horizontally downward at the back of the hole is $F=2.71\times {10}^5\text{N}$

Step-by-step solution

Given Data

$$\begin{gathered} h=2.4m \\ b=9.6\text{m} \\ \text{T}\mathrm{he}\mathrm{weight}\mathrm{of}\mathrm{the}\mathrm{wate}\mathbf{r}\mathbf{}=41.3\text{kN} \end{gathered}$$

Concept

Theexcesspressuregenerallyknownasgaugepressureisgivenby:

$P_{gauge}=\rho gh$

The force formula is:

$F=P_{gauge}A$

Where,

role="math" localid="1663671952173" $F=$Force

$A=$Area

$P_{gauge}=$Pressure at gauge

Step 3: Find the force the water causes on the foundation wall.

The pressure of the water over air half way down is given by the formula:

$$\begin{gathered} P_{gauge}=\rho gh \\ {P}_{gauge}=1000{\text{kg/m}}^{\text{3}}\times 9.8{\text{m/s}}^{\text{2}}\times 120\text{m} \\ {P}_{gauge}=1.18\times {10}^4\text{pa} \end{gathered}$$

Theforceonthewallduetothewaterisgivenbyusingformula:

$$\begin{gathered} F=P_{gauge}\times A \\ F=1.18\times {10}^4\times 2.40\text{m}\times 9.60\text{m} \\ F=2.71\times {10}^5\text{N} \end{gathered}$$

Thisforceishorizontally downward at thebackofthehole.