Chapter 3, Problem 81P

From Prob. 3-80, knowing that the density of the suspension depends on liquid depth and changes linearly from 800 kg m to 900 kg/m³ in the vertical direction, determine the resultant force acting on the gate ABC, and its line of action.

To determine

Resultant force acting on gate and its line of action.

Answer to Problem 81P

Resultant force acting perpendicular to the gate is $144.129455\text{kN}$.

Explanation of Solution

Given:

Density of liquid varies from $800\text{kg}/{\text{m}}^{\text{3}}$ to $900\text{kg}/{\text{m}}^{\text{3}}$ and the gate is parabolic.

Draw the diagram for the cross-section of the gate.

  

Figure (1)

Draw the diagram for the variation of density with variation of depth.

  

Figure (2)

Write the expression for the curve.

  $x=\sqrt{\frac{y}{2}}$   ...... (I)

Here, horizontal axis is denoted by the x and vertical axis is denoted by the y.

Write the expression for the density as a function of depth.

  $\rho =\left(\frac{{\rho }_2-{\rho }_1}{h_2-h_1}\right)h+{\rho }_1$   ...... (II)

Here, density is ${\rho }_2$ at height of $h_2$ and density is ${\rho }_1$ at height of $h_1$.

The density at a particular height is $\rho$.

Write the expression for the area of the elemental strip.

  $dA=2xdy$   ...... (III)

Here, area is $dA$.

Substitute $\sqrt{\frac{y}{2}}$ for $x$ in Equation (III).

  $dA=\sqrt{2}\sqrt{y}dy$   ...... (IV)

Write the expression for the resultant force.

  $dF=\rho ghdA$   ...... (V)

Here, the resultant force is $F$.

Write the expression for the vertical depth of the centre of gravity of gate from the free surface.

  $h=5-y\sin 60°$   ...... (VI)

Here, the vertical depth of the centre of gravity of gate from the free surface is $h$.

Calculation:

Substitute $800\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_1$, $900\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_2$ in Equation (II).

  $\begin{array}{c}\rho =\left(\frac{900\text{kg}/{\text{m}}^{\text{3}}-800\text{kg}/{\text{m}}^{\text{3}}}{5\text{m}-0\text{m}}\right)h+800\text{kg}/{\text{m}}^{\text{3}}\\ =\left(\frac{100}{5}\text{kg}/{\text{m}}^2\right)h+800\text{kg}/{\text{m}}^{\text{3}}\\ =\left(20\text{kg}/{\text{m}}^4\right)h+800\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Substitute $\left(20\text{kg}/{\text{m}}^4\right)h+800\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and Equation (IV) in Equation (V).

  $dF=\left[\left(20\text{kg}/{\text{m}}^4\right)h+800\text{kg}/{\text{m}}^{\text{3}}\right]gh\left(2xdy\right)$

Substitute Equation (VI) in above Equation.

  $\begin{array}{c}dF=\left[\left(20\text{kg}/{\text{m}}^4\right)\left(5\text{m}-y\sin 60°\right)+800\text{kg}/{\text{m}}^{\text{3}}\right]g\left(5\text{m}-y\sin 60°\right)\left(2xdy\right)\\ =\left[900\text{kg}/{\text{m}}^{\text{3}}-\left(20\text{kg}/{\text{m}}^4\right)y\right]g\left(5\text{m}-y\sin 60°\right)2xdy\\ =\left[\begin{array}{l}4500\text{kg}/{\text{m}}^2-\left(900\text{kg}/{\text{m}}^{\text{3}}\right)y\sin 60°\\ -\left(100\text{kg}/{\text{m}}^{\text{3}}\right)y\sin 60°+20\text{kg}/{\text{m}}^4\left(y^2{\left(\sin 60\right)}^2\right)\end{array}\right]g\sqrt{2}\sqrt{y}dy\\ =\left[\left[\begin{array}{l}\left(4500\text{kg}/{\text{m}}^2\right)\sqrt{y}-1000\text{kg}/{\text{m}}^{\text{3}}{y}^{3/2}\sin 60°\\ +\left(20\text{kg}/{\text{m}}^4\right)y^{5/2}{\sin }^{2}60°\end{array}\right]g\sqrt{2}\right]\end{array}$

Integrate both sides of the above Equation.

  $\begin{array}{c}F=\left[\sqrt{2}g{\displaystyle \underset{0}{\overset{3}{\int }}\left[\begin{array}{l}\left(4500\text{kg}/{\text{m}}^2\right)\sqrt{y}-1000\text{kg}/{\text{m}}^{\text{3}}{y}^{3/2}\sin 60°\\ +\left(20\text{kg}/{\text{m}}^4\right)y^{5/2}{\sin }^{2}60°\end{array}\right]}\right]\\ =\left[\sqrt{2}g\left[\begin{array}{l}\left(4500\text{kg}/{\text{m}}^2\right)\frac{2}{3}{y}^{3/2}-1000\text{kg}/{\text{m}}^{\text{3}}\left(\frac{5}{2}{y}^{5/2}\right)\sin 60°\\ +\left(20\text{kg}/{\text{m}}^4\right)\left(\frac{7}{2}{y}^{7/2}\right){\sin }^{2}60°\end{array}\right]\right]\\ =\sqrt{2}g\left[\begin{array}{l}\left\{3000\text{kg}/{\text{m}}^2\times \left(3^{3/2}\text{m}\right)\right\}-\left\{200\sqrt{3}\text{kg}/{\text{m}}^{\text{3}}\times 3^{5/2}\text{m}\right\}\\ +\left\{\frac{30\text{kg}/{\text{m}}^4}{7}\times \left(3^{7/2}\right)\text{m}\right\}\end{array}\right]\end{array}$

Substitute $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in above Equation.

  $\begin{array}{c}F=1.414\times 9.81\text{m}/{\text{s}}^{\text{2}}\left[\left(15588.4572-5400+200.423022\right)\text{kg}\right]\\ =144129.455\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\\ =144129.455\text{N}\end{array}$

  $\begin{array}{c}F=144129.455\text{N}\left(\frac{1\text{kN}}{1000\text{N}}\right)\\ =144.129455\text{kN}\end{array}$

Conclusion:

Resultant force acting on gate is $144.129455\text{kN}$