Chapter 11, Problem 9P

(a)

To determine

The hydrostatic force acting on the cylinder and its line of action when the gate opens.

Explanation of Solution

Given:

Radius of the solid cylinder is 2 ft.

Height of the water level 15 ft.

Calculation:

Determine the horizontal force on vertical surface.

  $\begin{array}{c}{F}_H=F_x\\ =P_{ave}A\\ =\rho g{h}_{C}A\\ =\rho g\left(s+R/2\right)A\end{array}$

  $\begin{array}{c}{F}_H=\left(62.4\text{lbm}/{\text{ft}}^3\right)\left(32.2\text{ft}/{\text{s}}^2\right)\left(13\text{ ft}+\frac{2\text{ ft}}{2}\right)\left(2\text{ ft}\times \text{1 ft}\right)\left(\frac{1\text{ lbf}}{32.2\text{lbm}\cdot \text{ft}/{\text{s}}^2}\right)\\ =1747\text{ lbf}\end{array}$

Determine the vertical force on horizontal surface.

  $\begin{array}{c}{F}_y=P_{\text{avg}}A\\ =\rho g{h}_{C}A\\ =\rho g{h}_{\text{bottom}}A\end{array}$

  $\begin{array}{c}{F}_y=\left(62.4\text{lbm}/{\text{ft}}^3\right)\left(32.2\text{ft}/{\text{s}}^2\right)\left(15\text{ ft}\right)\left(2\text{ ft}\times \text{1 ft}\right)\left(\frac{1\text{ lbf}}{32.2\text{lbm}\cdot \text{ft}/{\text{s}}^2}\right)\\ =1872\text{ lbf}\end{array}$

Determine the weight of fluid block per ft length.

  $\begin{array}{c}W=mg\\ =\rho gV\\ =\rho g\left(R^2-\frac{\pi }{4}{R}^2\right)\left(1\text{ ft}\right)\\ =\rho g{R}^2\left(1-\frac{\pi }{4}\right)\left(1\text{ ft}\right)\end{array}$

  $\begin{array}{c}W=\left(62.4\text{lbm}/{\text{ft}}^3\right)\left(32.2\text{ft}/{\text{s}}^2\right){\left(2\text{ ft}\right)}^2\left(1-\frac{\pi }{4}\right)\left(1\text{ ft}\right)\left(\frac{1\text{ lbf}}{32.2\text{lbm}\cdot \text{ft}/{\text{s}}^2}\right)\\ =54\text{ lbf}\end{array}$

Determine the net upward vertical force.

  $\begin{array}{c}{F}_V=F_y-W\\ =1872\text{ lbf}-54\text{ lbf}\\ =1818\text{ lbf}\end{array}$

Determine the magnitude of the hydrostatic force acting on the cylinder.

  $\begin{array}{c}{F}_R=\sqrt{F_H^2+F_V^2}\\ =\sqrt{{\left(1747\text{ lbf}\right)}^2+{\left(1818\text{ lbf}\right)}^2}\\ =2521\text{ lbf}\end{array}$

Thus, the magnitude of the hydrostatic force acting on the cylinder is $\underset{\_}{2521\text{ lbf}}$.

Determine the direction of the hydrostatic force acting on the cylinder.

  $\begin{array}{l}\tan \theta =\frac{F_V}{F_H}\\ \tan \theta =\frac{18181\text{bf}}{17471\text{bf}}\\ \tan \theta =1.041\\ \theta =46.1°\end{array}$

Thus, the line of action passes through the center of the cylinder making an angle $\underset{\_}{46.1°}$ upwards from the horizontal.

(b)

To determine

The weight of the cylinder per ft length of the cylinder.

Explanation of Solution

Determine the weight of the cylinder per ft length.

  $\begin{array}{l}{F}_{R}R\sin \theta -W_{cyl}R=0\\ W_{sl}=F_R\sin \theta \\ W_{sl}=\left(2521\text{ lbf}\right)\sin 46.1°\\ W_{sl}=1817\text{ lbf}\end{array}$

Thus, the weight of the cylinder per ft length is $\underset{\_}{1817\text{ lbf}}$.