Chapter 3, Problem 78P

For a gate width of 2 m into the paper (Fig. P3-78), determine the force required to hold the gate K at its location.

To determine

The force required to hold the gate.

Answer to Problem 78P

The force required to hold the gate is $17.873\text{kN}$.

Explanation of Solution

Given:

In the following figure all the parameter is given.

  b

Figure-(1)

Write the expression for the density of fluid in the top portion.

  ${\rho }_1={\left(SG\right)}_1\times {\rho }_w$   ...... (I)

Here, the density of the fluid situated at top is ${\rho }_1$, specific gravity of the top fluid is ${\left(SG\right)}_1,$ and density of water is ${\rho }_w$.

Write the expression for the density of fluid in the lower portion.

  ${\rho }_2={\left(SG\right)}_2\times {\rho }_w$   ...... (II)

Here, the density of the fluid situated at lower portion is ${\rho }_2$ and the specific gravity of the lower fluid is ${\left(SG\right)}_2$.

Draw free body diagram for the portion AB.

  

Figure-(2)

In Figure-(2) hydrostatic are shown as $F_{H_1}$ and $F_{H_2}$, centers of pressure from the free surface are $y_{P_1}$ and $y_{P_2}$.

Write the expression for the hydrostatic force acting in top portion.

  $F_{H_1}={\rho }_{1}g\left(s+\frac{b_1}{2}\right)\left(a{b}_1\right)$   ...... (III)

Here, force is $F_{H_1}$, acceleration due to gravity is $g$ depth from point A is $s$, height of the portion is $b_1,$ and length of the portion is $a$.

Write the expression for the depth from surface at which the force $F_{H_1}$ is acting.

  $y_{p_1}=s+\frac{b_1}{2}+\frac{b_1^2}{12\left(s+\frac{b_1}{2}\right)}$   ...... (IV)

Write the expression for the hydrostatic force acting in lower portion.

  $F_{H_2}={\rho }_{2}g\left(\frac{b_2}{2}\right)\left(a{b}_2\right)$   ...... (V)

Here, force is $F_{H_2}$, height of the portion is $b_2$.

Write the expression for the depth from surface at which the force $F_{H_2}$ is acting.

  $y_{p_2}=0.5+s+\frac{b_2}{2}+\frac{b_2^2}{12\left(s+\frac{b_2}{2}\right)}$   ...... (VI)

The below figure shows the free body diagram of portion BC.

  

Figure (3)

Write the expression for force acting on the portion BC.

  $F_V=\left({\rho }_{1}g{h}_1+{\rho }_{2}g{h}_2\right)\left(al\right)$   ...... (VII)

Here, force is $F_V$, height of the top fluid is $h_1$, height of the lower fluid is $h_2$ and length of the BC is $l$.

The below figure shows the free body diagram for all portions.

  

Figure (4)

Write the expression for the distance of force $F_{H_1}$ from point A.

  $y_1=y_{p_1}-0.4$   ...... (VIII)

Here, the distance of force $F_{H_1}$ from point A is $y_1$.

Write the expression for the distance of force $F_{H_2}$ from point A.

  $y_2=y_{p_2}-0.4$   ...... (IX)

Here, the distance of force $F_{H_2}$ from point A is $y_2$.

Write the expression for the moment equilibrium condition about the point A.

  $F=\frac{F_{H_2}\times y_2+F_{H_1}\times y_1+F_V\times \left(\frac{l}{2}\right)}{l}$   ...... (X)

Calculation:

Substitute $0.86$ for ${\left(SG\right)}_1$ and $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_w$ in Equation (I).

  $\begin{array}{c}{\rho }_1=0.86\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\\ =860\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Substitute $1.23$ for ${\left(SG\right)}_2$ and $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_w$ in Equation (II).

  $\begin{array}{c}{\rho }_2=1.23\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\\ =1230\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Substitute $860\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_1$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $0.4$ for $s$, $0.1\text{m}$ for $b_1$ and $2\text{m}$ for $a$ in Equation (III).

  $\begin{array}{c}{F}_{H_1}=\left(860\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.4\text{m}+\frac{0.1\text{m}}{2}\right)\left(2\text{m}\times 0.1\text{m}\right)\\ =8436.6\text{kg}/{\text{m}}^2\cdot {\text{s}}^2\left(0.09{\text{m}}^{\text{3}}\right)\\ =759.294\text{N}\end{array}$

Substitute $0.4\text{m}$ for $s$, $0.1\text{m}$ for $b_1$ in Equation (IV).

  $\begin{array}{c}{y}_{p_1}=0.4\text{m}+\frac{0.1\text{m}}{2}+\frac{{\left(0.1\text{m}\right)}^2}{12\left(0.4\text{m}+\frac{0.1\text{m}}{2}\right)}\\ =0.45\text{m}+\frac{0.1{\text{m}}^2}{5.4\text{m}}\\ =0.45185\text{m}\end{array}$

Substitute $1230\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_2$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $0.4$ for $s$, $0.8\text{m}$ for $b_2$ and $2\text{m}$ for $a$ in Equation (V).

  $\begin{array}{c}{F}_{H_2}=1230\text{kg}/{\text{m}}^{\text{3}}\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{0.8\text{m}}{2}\right)\left(2\times 0.8\text{m}\right)\\ =12066.3\text{kg}/{\text{m}}^2\cdot {\text{s}}^2\left(0.64{\text{m}}^{\text{3}}\right)\\ =7722.432\text{N}\end{array}$

Substitute $0\text{m}$ for $s$, $0.8\text{m}$ for $b_2$ in Equation (VI).

  $\begin{array}{c}{y}_{p_2}=0.5+0+\frac{0.8\text{m}}{2}+\frac{{\left(0.8\text{m}\right)}^2}{12\left(0+\frac{0.8\text{m}}{2}\right)}\\ =0.5+0.4\text{m}+\frac{0.64{\text{m}}^2}{4.8\text{m}}\\ =1.033\text{m}\end{array}$

Substitute $860\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_1$, $1230\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_2$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $0.4$ for $l$, $2\text{m}$ for $a$, $0.8\text{m}$ for $h_2$ and $0.5\text{m}$ for $h_1$ in Equation (VII).

  $\begin{array}{c}{F}_V=\left[\left[\begin{array}{l}\left(860\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^{\text{2}}\times 0.5\text{m}\right)+\\ \left(1230\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^{\text{2}}\times 0.8\text{m}\right)\end{array}\right]\left(2\times 0.4\text{m}\right)\right]\\ =\left(8436.6\text{kg}/{\text{m}}^2\cdot {\text{s}}^{\text{2}}\left(0.5\text{m}\right)+12066.3\text{kg}/{\text{m}}^2\cdot {\text{s}}^{\text{2}}\left(0.8\text{m}\right)\right)\left(0.8\text{m}\right)\\ =11097.072\text{N}\end{array}$

Substitute $0.45185\text{m}$ for $y_{p_1}$ in Equation (VIII).

  $\begin{array}{c}{y}_1=0.45185\text{m}-0.4\text{m}\\ \text{=0}\text{.05185}\text{m}\end{array}$

Substitute $1.033\text{m}$ for $y_{p_2}$ in Equation (IX).

  $\begin{array}{c}{y}_2=1.0333\text{m}-0.4\text{m}\\ =0.6333\text{m}\end{array}$

Substitute $759.294\text{N}$ for $F_{H_1}$, $7722.432\text{N}$ for $F_{H_2}$, $0.05185\text{m}$ for $y_1$, $0.6333\text{m}$ for $y_2$ and $11097.072\text{N}$ for $F_V$ in Equation (X).

  $\begin{array}{c}F=\frac{\left[\begin{array}{r}\left(7722.432\text{N}\times 0.633\text{m}\right)+\left(759.294\text{N}\times 0.05185\text{m}\right)\\ +\left(11097.072\text{N}\times \left(\frac{0.4\text{m}}{2}\right)\right)\end{array}\right]}{0.4\text{m}}\\ =\frac{7149.3999\text{N}\cdot \text{m}}{0.4\text{m}}\\ =17873.5\text{N}\left(\frac{1\text{kN}}{1000\text{N}}\right)\\ =17.873\text{N}\end{array}$

Conclusion:

The force required is $17.873\text{kN}$.