Chapter 3, Problem 68EP

A long, solid cylinder of radius 2 ft hinged at point A is used as an automatic gate, as shown in Fig. P3-79E. When the water level reaches 12 ft, the cylindrical gate opens by turning about the hinge at point A. Determine (a) the hydrostatic force acting on the cylinder and its line of action when the gate opens and (b) the weight of the cylinder per ft length of the cylinder.

To determine

(a)

The hydrostatic force when gate opens.

Answer to Problem 68EP

The hydrostatic force when gate opens is $1373.90\text{lbf}$.

Explanation of Solution

Given:

Water level is $12\text{ft}$.

Radius of the cylinder is $2\text{ft}$.

Draw a free body diagram acting on the cylinder as shown in the below figure.

  

Figure (1)

Write the expression for the horizontal force on the vertical surface.

  $F_H=\rho g{h}_{c}A$   ...... (I)

Here, the horizontal force on vertical surface is $F_H$, density of fluid is $\rho$, acceleration due to gravity is $g$, distance from surface to the centre of gravity is $h_c$ and area is $A$.

Write the expression for the area.

  $A=Rl$   ...... (II)

Here, radius is $R$ and length is $l$.

Write the expression for the distance from surface to the centre of gravity.

  $h_c=y+\frac{R}{2}$   ...... (III)

Substitute the Equation (II) and Equation (III) in Equation (I).

  $F_H=\rho g\left(y+\frac{R}{2}\right)\left(Rl\right)$   ...... (IV)

Calculation:

Substitute $62.4\text{lbm}/{\text{ft}}^3$ for $\rho$, $32.174\text{ft}/{\text{s}}^{\text{2}}$ for $g$, $10\text{ft}$ for $y$, $2\text{ft}$ for $R$ and $1\text{ft}$ for $l$ in Equation (IV).

  $\begin{array}{c}{F}_H=\left(62.4\text{lbm}/{\text{ft}}^3\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left[10\text{ft}+\frac{2\text{ft}}{2}\right]\left(2\text{ft}\times 1\text{ft}\right)\\ =2009.28\text{lbm}/{\text{ft}}^2\cdot {\text{s}}^2\left(22{\text{ft}}^3\right)\\ =44204.16\text{lbm}\cdot \text{ft}/{\text{s}}^{\text{2}}\left(\frac{1\text{lbf}}{32.174\text{lbm}\cdot \text{ft}/{\text{s}}^{\text{2}}}\right)\\ =1373.90\text{lbf}\end{array}$

Conclusion:

Hydrostatic force is $1373.90\text{lbf}$.

To determine

(b)

Weight of the cylinder per unit length.

Answer to Problem 68EP

Weight of the cylinder per unit length is $1443.75\text{lbf}$.

Explanation of Solution

Write the expression for the vertical force on the projected horizontal surface.

  $F_y=\rho g{h}_{bottom}\left(Rl\right)$   ...... (V)

Here, height of the liquid column from free surface of water till the bottom of the gate is $h_{bottom}$, vertical force on the projected horizontal surface is $F_y$.

The figure below shows the diagram for the volume enclosed per unit width.

  

Figure-(2)

Write the expression for the volume of element.

  $V=\left[R^2-\left(\frac{\pi R^2}{4}\right)\right]l$   ...... (VI)

Here, radius is $R$ and length is $l$.

Write the expression for the weight of the element per unit width of the gate.

  $W=\rho Vg$   ...... (VII)

Here, weight is $W$ and mass is $m$.

Write the expression for the vertical force.

  $F_V=F_y-W$   ...... (VIII)

Here, vertical force is $F_V$.

Write the expression for the resultant force.

  $F_R=\sqrt{{\left(F_H\right)}^2+{\left(F_V\right)}^2}$   ...... (IX)

Here, resultant force is $F_R$.

Write the expression for the direction of hydrostatic force.

  $\theta ={\tan }^{-1}\left(\frac{F_V}{F_H}\right)$   ...... (X)

The below figure represents direction of hydrostatic force.

  

Figure-(3)

Write the expression for the moment about the hinged point.

  $F_R\left(R\sin \theta \right)-\left(W_{cylinder}\right)R=0$   ...... (XI)

Here, the weight of the cylinder is $W_{cylinder}$.

Calculation:

Substitute $62.4\text{lbm}/{\text{ft}}^3$ for $\rho$, $32.174\text{ft}/{\text{s}}^{\text{2}}$ for $g$, $12\text{ft}$ for $h_{bottom}$, $2\text{ft}$ for $R$ and $1\text{ft}$ for $l$ in Equation (V).

  $\begin{array}{c}{F}_y=\left(62.4\text{lbm}/{\text{ft}}^3\times 32.174\text{ft}/{\text{s}}^{\text{2}}\right)\left(12\text{ft}\right)\left(2\text{ft}\times 1\text{ft}\right)\\ =48183.78\text{lbf}\cdot \text{ft}/{\text{s}}^{\text{2}}\left(\frac{1\text{lbf}}{32.1742\frac{\text{lbm}\cdot \text{ft}}{{\text{s}}^{\text{2}}}}\right)\\ =1497.6\text{lbf}\end{array}$

Substitute $2\text{ft}$ for the $R$ in Equation (VI).

  $\begin{array}{c}V=\left[{\left(2\text{ft}\right)}^2-\left(\frac{\pi {\left(2\text{ft}\right)}^2}{4}\right)\right]\left(1\text{ft}\right)\\ =\left(4{\text{ft}}^2-3.14{\text{ft}}^2\right)\times 1\text{ft}\\ =0.854{\text{ft}}^3\end{array}$

Substitute $62.4\text{lbm}/{\text{ft}}^3$ for $\rho$, $32.174\text{ft}/{\text{s}}^{\text{2}}$ for $g$ and $0.854{\text{ft}}^3$ for $V$ in Equation (VII).

  $\begin{array}{c}W=\left(62.4\text{lbm}/{\text{ft}}^3\right)\left(32.174\text{ft}/{\text{s}}^{\text{2}}\right)\left(0.854{\text{ft}}^3\right)\\ =\left(1714.53\text{lbf}\cdot \text{ft}/{\text{s}}^{\text{2}}\right)\left(\frac{1\text{lbf}}{32.174\frac{\text{lbf}\cdot \text{ft}}{{\text{s}}^{\text{2}}}}\right)\\ =53.54\text{lbf}\end{array}$

Substitute $1497.6\text{lbf}$ for $F_y$, $53.54\text{lbf}$ for $W$ in Equation (VIII).

  $\begin{array}{c}{F}_V=1497.6\text{lbf}-53.54\text{lbf}\\ =1444.06\text{lbf}\end{array}$

Substitute $1373.90\text{lbf}$ for $F_H$ and $1444.06\text{lbf}$ for $F_V$ in Equation (IX).

  $\begin{array}{c}{F}_R=\sqrt{{\left(1373.90\text{lbf}\right)}^2+{\left(1444.06\text{lbf}\right)}^2}\\ =\sqrt{3737262.24{\text{lbf}}^2}\\ =1993.2\text{lbf}\end{array}$

Substitute $1373.90\text{lbf}$ for $F_H$ and $1444.06\text{lbf}$ for $F_V$ in Equation (X).

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{1444.06\text{lbf}}{1373.90\text{lbf}}\right)\\ ={\tan }^{-1}\left(1.051\right)\\ =46.42°\end{array}$

Substitute $1993.2\text{lbf}$ for $F_R$ and $46.42°$ for $\theta$ in Equation (XI).

  $\begin{array}{l}1993.2\text{lbf}\left(2\text{ft}\times \sin \left(46.42°\right)\right)-\left(W_{cylinder}\right)\left(2\text{ft}\right)=0\\ W_{cylinder}=\left(\frac{2887.798\text{lbf}\cdot \text{ft}}{2\text{ft}}\right)\\ W_{cylinder}=1443.75\text{lbf}\end{array}$

Conclusion:

The weight of the cylinder is $1443.75\text{lbf}$.