Chapter 2, Problem 2.95P

To determine

(a)

The specific gravity of the body if $h=0$.

Answer to Problem 2.95P

$SG=\frac{3}{2}$

Explanation of Solution

Given information:

The width of uniform body is b. It’s in static equilibrium, when pivoted about point O.

Assume that the horizontal force acting on the panel is ${\text{F}}_{\text{H }}$ and can be given as below.

${\text{F}}_{\text{H }}=\gamma A{h}^1$

In above equation,

$\gamma$ - Specific weight of water

$A$ - Area which the horizontal force is acting on

$h^1$ - Distance between the top water level to the center of gravity of the panel (DG)

Assume that the vertical force acting on the panel is ${\text{F}}_{\text{v }}$ and can be given as below:

${\text{F}}_{\text{v }}=\gamma {(Vol)}_{above}OC$

$=\gamma Ah$

In above equation,

$\gamma$ - Specific weight of water

$A$ - Area above the panel

$h$ - Width of the panel

The below equation is used to find the line of action of horizontal force ${\text{F}}_{\text{H }}$

$y=-\frac{I}{A{h}^1}$

In above equation,

$y$ - Distance from the center of gravity (G) to the line of action.

$I$ - Inertia of the plane which the horizontal force is acting on.

$A$ - Area which the horizontal force is acting on

$h^1$ - Distance between the top water level to the center of gravity of the panel (DG)

Calculation:

To find horizontal force:

According to the above-mentioned explanation:

$\begin{array}{l}{\text{F}}_{\text{H }}=\gamma A{h}^1\\ =\gamma (Rb)\left(\frac{R}{2}\right)\\ =\gamma \left(\frac{R^{2}b}{2}\right)\end{array}$

Therefore, the horizontal component of the hydrostatic force = $\gamma \left(\frac{R^{2}b}{2}\right)$

To find the line of action,

$y=-\frac{I}{A{h}^1}$

In this case, the inertia of the rectangular plane is equal to,

$I=\left(\frac{1}{12}b{h}^3\right)$

Therefore,

$\begin{array}{l}y=-\frac{I}{A{h}^1}\\ =-\frac{\left(\frac{1}{12}\left(b\right){\left(R\right)}^3\right)}{\left(Rb\right)\left(\frac{R}{2}\right)}\\ =-\left(\frac{R}{6}\right)\end{array}$

The horizontal force is acting at a distance $\left(\frac{R}{2}\right)-\left(\frac{R}{6}\right)=\left(\frac{R}{3}\right)$ from point O.

To find vertical force,

According to the above-mentioned explanation:

${\text{F}}_{\text{v }}=\gamma {(Vol)}_{above}BC$

$=\gamma Ah$

$\begin{array}{l}=\gamma h(AreaofOCD)\\ =\gamma b\left(\frac{\pi {\left(R\right)}^2}{4}\right)\\ =\left(\frac{\pi R^2\gamma b}{4}\right)\end{array}$

The vertical force acts at a distance equals to $\frac{4R}{3\pi }$ from left of point O.

According to above diagram:

The body to be in static equilibrium, the moment about point O is equal to zero.

${\displaystyle \sum M_O=0}$

$F_H\left(\frac{R}{3}\right)+F_V\left(\frac{4R}{3\pi }\right)-W_A\left(\frac{R}{2}\right)-W_B\left(\frac{4R}{3\pi }\right)=0$

$\gamma \left(\frac{R^{2}b}{2}\right)\left(\frac{R}{3}\right)+\left(\frac{\pi R^2\gamma b}{4}\right)\left(\frac{4R}{3\pi }\right)-{\gamma }_s\left(Rhb\right)\left(\frac{R}{2}\right)-{\gamma }_s\left(\frac{\pi R^{2}b}{4}\right)\left(\frac{4R}{3\pi }\right)=0$

$\gamma \left(\frac{R^{3}b}{6}\right)+\gamma \left(\frac{R^{3}b}{3}\right)-{\gamma }_s\left(\frac{R^{2}bh}{2}\right)-{\gamma }_s\left(\frac{R^{3}b}{3}\right)=0$

$\gamma \left(\frac{R^{3}b}{2}\right)={\gamma }_s{R}^{2}b\left(\frac{h}{2}+\frac{R}{3}\right)$

$\left(\frac{{\gamma }_s}{\gamma }\right)=\left(\frac{\left(\frac{R}{2}\right)}{\left(\frac{h}{2}+\frac{R}{3}\right)}\right)$

$\left(\frac{{\gamma }_s}{\gamma }\right)={\left(\frac{h}{R}+\frac{2}{3}\right)}^{-1}$

$\begin{array}{l}forh=0,\\ SG=\frac{3}{2}\end{array}$

Conclusion:

The specific gravity of uniform body, when $h=0$ is equal to $SG=\frac{3}{2}$.

To determine

(b)

The specific gravity of the body if $h=R$.

Answer to Problem 2.95P

$SG=\frac{3}{5}$

Explanation of Solution

Given information:

The width of uniform body is b. It’s in static equilibrium when pivoted about point O.

Assume that the horizontal force acting on the panel is ${\text{F}}_{\text{H }}$ and can be given as below:

${\text{F}}_{\text{H }}=\gamma A{h}^1$

In above equation,

$\gamma$ - Specific weight of water

$A$ - Area which the horizontal force is acting on

$h^1$ - Distance between the top water level to the center of gravity of the panel (DG)

Assume that the vertical force acting on the panel is ${\text{F}}_{\text{v }}$ and can be given as below:

${\text{F}}_{\text{v }}=\gamma {(Vol)}_{above}OC$

$=\gamma Ah$

In above equation,

$\gamma$ - Specific weight of water

$A$ - Area above the panel

$h$ - Width of the panel

The below equation is used to find the line of action of horizontal force ${\text{F}}_{\text{H }}$ :

$y=-\frac{I}{A{h}^1}$

In above equation:

$y$ - Distance from the center of gravity (G) to the line of action.

$I$ - Inertia of the plane which the horizontal force is acting on.

$A$ - Area which the horizontal force is acting on

$h^1$ - Distance between the top water level to the center of gravity of the panel (DG)

Calculation:

To find horizontal force,

According to the above-mentioned explanation:

$\begin{array}{l}{\text{F}}_{\text{H }}=\gamma A{h}^1\\ =\gamma (Rb)\left(\frac{R}{2}\right)\\ =\gamma \left(\frac{R^{2}b}{2}\right)\end{array}$

Therefore, the horizontal component of the hydrostatic force = $\gamma \left(\frac{R^{2}b}{2}\right)$

To find the line of action:

$y=-\frac{I}{A{h}^1}$

In this case, the inertia of the rectangular plane is equal to:

$I=\left(\frac{1}{12}b{h}^3\right)$

Therefore,

$\begin{array}{l}y=-\frac{I}{A{h}^1}\\ =-\frac{\left(\frac{1}{12}\left(b\right){\left(R\right)}^3\right)}{\left(Rb\right)\left(\frac{R}{2}\right)}\\ =-\left(\frac{R}{6}\right)\end{array}$

The horizontal force is acting at a distance $\left(\frac{R}{2}\right)-\left(\frac{R}{6}\right)=\left(\frac{R}{3}\right)$ from point O.

To find vertical force,

According to the above-mentioned explanation:

${\text{F}}_{\text{v }}=\gamma {(Vol)}_{above}BC$

$=\gamma Ah$

$\begin{array}{l}=\gamma h(AreaofOCD)\\ =\gamma b\left(\frac{\pi {\left(R\right)}^2}{4}\right)\\ =\left(\frac{\pi R^2\gamma b}{4}\right)\end{array}$

The vertical force acts at a distance equals to $\frac{4R}{3\pi }$ from left of point O.

According to above diagram:

The body to be in static equilibrium, the moment about point O is equal to zero.

${\displaystyle \sum M_O=0}$

$F_H\left(\frac{R}{3}\right)+F_V\left(\frac{4R}{3\pi }\right)-W_A\left(\frac{R}{2}\right)-W_B\left(\frac{4R}{3\pi }\right)=0$

$\gamma \left(\frac{R^{2}b}{2}\right)\left(\frac{R}{3}\right)+\left(\frac{\pi R^2\gamma b}{4}\right)\left(\frac{4R}{3\pi }\right)-{\gamma }_s\left(Rhb\right)\left(\frac{R}{2}\right)-{\gamma }_s\left(\frac{\pi R^{2}b}{4}\right)\left(\frac{4R}{3\pi }\right)=0$

$\gamma \left(\frac{R^{3}b}{6}\right)+\gamma \left(\frac{R^{3}b}{3}\right)-{\gamma }_s\left(\frac{R^{2}bh}{2}\right)-{\gamma }_s\left(\frac{R^{3}b}{3}\right)=0$

$\gamma \left(\frac{R^{3}b}{2}\right)={\gamma }_s{R}^{2}b\left(\frac{h}{2}+\frac{R}{3}\right)$

$\left(\frac{{\gamma }_s}{\gamma }\right)=\left(\frac{\left(\frac{R}{2}\right)}{\left(\frac{h}{2}+\frac{R}{3}\right)}\right)$

$\left(\frac{{\gamma }_s}{\gamma }\right)={\left(\frac{h}{R}+\frac{2}{3}\right)}^{-1}$

$\begin{array}{l}forh=R,\\ SG=\frac{3}{5}\end{array}$

Conclusion:

The specific gravity of uniform body, when $h=R$ is equal to $SG=\frac{3}{5}$.