The L-shaped tank shown in Figure is filled with water and is open at the top. (a) If $\mathit{d}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathit{m}$, what is the force due to the water on face A and (b) If $\mathit{d}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathit{m}$, what is the force due to the water on face B?

In the given figure, d=5 m

Using theformula of pressure in terms of density, gravitational acceleration, and depth, we can find the pressure. Using the value of pressure in the formula for the pressure in terms of force and unit area,we can find thetotal force exerted by the water on face Aand faceB.

Formulae:

Force applied on a body in terms of pressure,F=pA (i)

Pressure applied on a fluid surface,$p=\rho gh$ (ii)

As for face A, (in the given figure)h=2d

Hence, the pressure at face A using equation (ii) and the given values, we get

$\begin{array}{ll}p=\rho g\left(2d\right)& \\ =(1.0x1.0^3& \mathrm{kg}/{\mathrm{m}}^3\left)\right(9.8\mathrm{m}/{\mathrm{s}}^2\left)\right(2\times 5\mathrm{m})\\ =98\times {10}^3\mathrm{a}& \end{array}$

Therefore, using equation (i), the force on face A,

$$\begin{gathered} F_A=\left(98\times {10}^3\mathrm{Pa}\right)\left(5.0^2\mathrm{m}\right)\left(\because A={\left(d\right)}^2{\mathrm{m}}^2\right) \\ =2.45\times {10}^6{\mathrm{d}}^2\mathrm{N} \end{gathered}$$

Now, the atmospheric pressure using equation (i) is given by:

$$\begin{gathered} F_0=pA \\ =\left(1.0\times {10}^5\mathrm{Pa}\right){\left(5.0\mathrm{m}\right)}^2 \\ =2.5\times {10}^6\mathrm{N} \end{gathered}$$

Therefore, the total force on face A is given by:

$$\begin{gathered} F_A=F_0+F_A \\ =\left(2.5\times {10}^6\mathrm{N}\right) \\ =\left(4.95\times {10}^6\mathrm{N}\right) \end{gathered}$$

Therefore, the total force exerted by the water on face Ais$4.95\times {10}^6\mathrm{N}$

As for face B, (in the given figure)

$$\begin{gathered} h=2d+\frac{d}{2} \\ =\frac{5d}{2} \end{gathered}$$

Hence, the pressure at face B using equation (ii) and the given values, we get

$$\begin{gathered} p=\rho gh\left(\frac{5d}{2}\right) \\ =\left(1.0\times {10}^3\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(\frac{5\times 5{\mathrm{m}}^2}{2}\right) \\ 122.5\times {10}^3\mathrm{Pa} \end{gathered}$$

Therefore,

$$\begin{gathered} F_B=\left(122.5\times {10}^3\mathrm{Pa}\right){\left(5.0\mathrm{m}\right)}^2\left(\because A={\left(d\right)}^2{m}^2\right) \\ =3.1\times {10}^6\mathrm{N} \end{gathered}$$

Now, the atmospheric pressure is,

$\begin{array}{l}{F}_0=PA\\ =(1.0\times {10}^5\mathrm{Pa}){\left(5.0\mathrm{m}\right)}^2\\ =2.5\times {10}^6\mathrm{N}\end{array}$

Therefore, total force on face A is given by:

$\begin{array}{l}F{\text{'}}_B=F_0+F_B\\ \left(3.1\times {10}^6\mathrm{N}\right)\left(2.5\times {10}^6\mathrm{N}\right)\\ \left(5.6\times {10}^6\mathrm{N}\right)\end{array}$

Therefore, the total force exerted by the water on face Bis$5.6\times {10}^6\mathrm{N}$.