Chapter 3, Problem 93P

To determine

The force needed to keep the gate stationary.

Answer to Problem 93P

The force needed to keep the gate stationary is $126.98\text{kN}$.

Explanation of Solution

Given information:

Specific gravity of the oil is $1.5$, height of the water is $4\text{m}$, height of the oil is $3\text{m}$.

Below figure shows the free body diagram of the gate.

  Figure-(1)

Write the expression for the hydrostatic force.

   $F_H=\rho g\overline{h}A$    ...... (I)

Density is $\rho$, acceleration due to gravity is $g$, centre of gravity is $\overline{h}$ and area is $A$.

Write the expression for the distance from where the hydrostatic force is acting.

   $h=\frac{x}{3}$    ...... (II)

Here, height of the water is $x$.

Write the expression for the vertical force acting on gate.

   $F_V=\rho g\left(\frac{2y}{3}\right)\left(A\right)$    ...... (III)

Here, the height of the oil is $y$ and area of the gate is $A$.

Write the expression for the distance from where the vertical force is acting.

   $h^{\prime }=\frac{2\left(3y\right)}{2x}$    ...... (IV)

Write the expression for the horizontal force.

   $F_{H,oil}={\rho }_{oil}g\overline{h}A$    ...... (V)

Here, the density of oil is ${\rho }_{oil}$.

Write the expression for the distance from where the horizontal force is acting.

   $h^{″}=\frac{y}{3}$    ...... (VI)

Write the expression for the moment along B.

   $9F+F_V{h}^{\prime }+F_{H,oil}\times h^{″}+F_{H,water}h=0$    ...... (VII)

Write the expression for the centre of gravity.

   $\overline{h}=\frac{x}{2}$    ...... (VIII)

Write the expression for the area of the gate.

   $A=lw$    ...... (IX)

Here, the length of the gate is $l$ and the width is $w$.

Write the expression for the gravity of the density of oil.

   ${\rho }_{oil}=1.5{\rho }_{water}$    ...... (X)

Calculation:

Substitute $4\text{m}$ for $x$ in Equation (VIII).

   $\begin{array}{c}\overline{h}=\frac{4\text{m}}{2}\\ =2\text{m}\end{array}$

Substitute $2\text{m}$ for $l$ and $4\text{m}$ for $w$ in Equation (IX).

   $\begin{array}{c}A=2\text{m}\left(4\text{m}\right)\\ =8{\text{m}}^2\end{array}$

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{water}$ in Equation (X).

   $\begin{array}{c}{\rho }_{oil}=1.5\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\\ =1500\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{water}$ and $9.81{\text{m}/\text{s}}^2$ for $g$, $8{\text{m}}^2$ for $A$ and $2\text{m}$ for $\overline{h}$ in Equation (I).

   $\begin{array}{c}{F}_H=1000\text{kg}/{\text{m}}^{\text{3}}\left(9.81{\text{m}/\text{s}}^2\right)2\text{m}\left(8{\text{m}}^2\right)\\ =9810\text{kg}/{\text{m}}^{\text{3}}\left({\text{m}/\text{s}}^2\right)16{\text{m}}^3\\ =156960\text{N}\left(\frac{1\text{kN}}{1000\text{N}}\right)\\ =156.96\text{kN}\end{array}$

Substitute $1500\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{oil}$ and $9.81{\text{m}/\text{s}}^2$ for $g$, $8{\text{m}}^2$ for $A$ and $2\text{m}$ for $\overline{h}$ in Equation (III).

   $\begin{array}{c}{F}_V=1500\text{kg}/{\text{m}}^{\text{3}}\left(9.81{\text{m}/\text{s}}^2\right)\left(2\times 3\text{m}\right)\left(8{\text{m}}^2\right)\\ =14715\text{kg}/{\text{m}}^{\text{3}}\left({\text{m}/\text{s}}^2\right)48{\text{m}}^3\\ =706320\text{N}\left(\frac{1\text{kN}}{1000\text{N}}\right)\\ =706.3\text{kN}\end{array}$

Substitute $4\text{m}$ for $x$ and $2\text{m}$ for $y$ in Equation (IV).

   $\begin{array}{c}{h}^{\prime }=\frac{2\times 3\left(3\text{m}\right)}{2\left(4\text{m}\right)}\\ =\frac{18\text{m}}{8}\\ =2.25\text{m}\end{array}$

Substitute $3\text{m}$ for $y$ in Equation (VI).

   $\begin{array}{c}{h}^{″}=\frac{3\text{m}}{3}\\ =1\text{m}\end{array}$

Substitute $1500\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{oil}$, $9.81{\text{m}/\text{s}}^2$ for $g$, $6{\text{m}}^2$ for $A$ and $2\text{m}$ for $\overline{h}$ in Equation (V).

   $\begin{array}{c}{F}_{H,oil}=1500\text{kg}/{\text{m}}^{\text{3}}\left(9.81{\text{m}/\text{s}}^2\right)\left(1.5\text{m}\right)\left(6{\text{m}}^2\right)\\ =1500\times 9.81\times 12\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\\ =132435\text{N}\\ =132.435\text{kN}\end{array}$

Substitute $156.96\text{kN}$ for $F_H$, $706.3\text{kN}$ for $F_V$, $1\text{m}$ for $h^{″}$, $2.25\text{m}$ for $h^{\prime }$ and $2\text{m}$ for $\overline{h}$ in Equation (VII).

   $\begin{array}{l}9F+\left(706.3\text{kN}\right)\left(2.25\text{m}\right)-\\ \left[\left(132.435\text{kN}\right)\times \left(1\text{m}\right)+\left(156.96\text{kN}\right)\left(2\text{m}\right)\right]=0\\ F=\frac{\left[\begin{array}{l}\left(132.435\text{kN}\right)\times \left(1\text{m}\right)+\left(156.96\text{kN}\right)\left(2\text{m}\right)-\\ \left(706.3\text{kN}\right)\left(2.25\text{m}\right)\end{array}\right]}{9\text{m}}\\ F=\frac{-1142.82\text{kN}\cdot \text{m}}{9\text{m}}\\ F=-126.98\text{kN}\end{array}$

Conclusion:

The force needed to keep the gate stationary is $126.98\text{kN}$.