Chapter 11, Problem 42RQ

To determine

The required force to be applied at the center of gravity.

Explanation of Solution

Given:

Radius of the semicircular gate $\left(R\right)$ is 0.5 m.

Calculation:

Determine the force applied by glycerin.

  $\begin{array}{c}{F}_{Rg}=m_{cg}A\\ =1.26\times 9810\text{N}/{\text{m}}^3\times \frac{4\times 0.5\text{ m}}{3\times \pi }\times \frac{\pi {\left(0.5\text{ m}\right)}^2}{2}\\ =1030\text{ N}\end{array}$

Determine the gage pressure of air entrapped on the top of the oil surface.

  $\begin{array}{c}p=80\text{ kPa}-100\text{ kPa}\\ =-20\text{ kPa }(\text{gage})\end{array}$

Determine the imaginary reduction in the oil level.

  $\begin{array}{c}h=\frac{20000\text{N}/{\text{m}}^2}{0.91\times 9810\text{N}/{\text{m}}^3}\\ =2.24\text{ m}\end{array}$

Determine the imaginary oil level.

  $\begin{array}{c}H=4.74\text{ m}-2.24\text{ m}\\ =2.50\text{ m}\end{array}$

Determine the force applied by oil.

  $\begin{array}{c}{F}_{Ro}=\gamma h_{cg}A\\ =0.91\times 9810\text{N}/{\text{m}}^3\times \left(2.5\text{ m}+\frac{4\times 0.5\text{ m}}{3\times \pi }\right)\times \frac{\pi {\left(0.5\text{ m}\right)}^2}{2}\\ =9508\text{ N}\end{array}$

Determine the area.

  $\begin{array}{c}A=\frac{1}{2}\pi {\left(0.5\text{ m}\right)}^2\\ =0.39267{\text{ m}}^2\end{array}$

Determine the moment of inertia.

  $\begin{array}{c}{I}_{xc}=0.1098R^4\\ =0.1098\times {\left(0.5\text{ m}\right)}^4\\ =0.0068625{\text{ m}}^4\end{array}$

Determine the locations of ${\text{F}}_{\text{Rg}}{\text{ and F}}_{\text{Ro}}$.

  $\begin{array}{c}{y}_{cp-g}=y_{cg-g}+\frac{I_{xc}}{y_{cg-g}A}\\ =0.2122\text{ m}+\frac{0.0068625{\text{ m}}^4}{0.2122\text{ m}\times 0.39267{\text{ m}}^2}\\ =0.2945\text{ m}\end{array}$

  $\begin{array}{c}{y}_{cg-g}=\frac{4\times 0.5\text{ m}}{3\times \pi }\\ =0.2122\text{ m}\end{array}$

  $\begin{array}{c}{y}_{cg-o}=2.5\text{ m}+\frac{4\times 0.5\text{ m}}{3\times \pi }\\ =2.712\text{ m}\end{array}$

  $\begin{array}{c}{y}_{cp-o}=y_{cg-o}+\frac{I_{xc}}{y_{cg-o}A}\\ =2.712\text{ m}+\frac{0.0068625{\text{ m}}^4}{2.712\text{ m}\times 0.39267{\text{ m}}^2}\\ =2.784\text{ m}\end{array}$

Take moment about hinge.

  $\begin{array}{l}F\times y_{cg-g}+F_{Rg}\times y_{cp-g}-F_{Ro}\left(y_{cp-o}-2.5\right)=0\\ F\times 0.2122\text{ m}+1030\text{ N}\times 0.2945\text{ m}-9508\text{ N}\left(2.784\text{ m}-2.5\text{ m}\right)=0\\ F=11296\text{ N}\end{array}$

Thus, the required force to be applied at the center of gravity is $\underset{\_}{11296\text{ N}}$.