Chapter 3, Problem 83P

To determine

(a)

The magnitude of resultant force acting on surface $AB$.

The location of line of action of resultant force.

Answer to Problem 83P

The magnitude of resultant force acting on surface $AB$ is $699\text{kN}$.

The location of line of action of resultant force is $4.86\text{m}$.

Explanation of Solution

Given information:

The length of the tank is $2.5\text{m}$, the width of the tank is $8.1\text{m}$, and the height of the tank is $6\text{m}$ and the specific gravity of oil is $0.88$.

Below figure represent the schematic and force acting on the tank.

        Figure-(1)

Write the expression of pressure acting on the side $AB$.

   $P_{AB}=\rho g\left(s+\frac{h}{2}\right)$     ...... (I)

Here, the density of the fluid is $\rho$, the acceleration due to gravity is $g$, height of the water in the column is $s$ and the height of the tank is $h$.

Write the expression of force acting on the side $AB$.

   $F_{R,AB}=P_{AB}\times A$     ...... (II)

Here, the frontal area is $A$.

Write the expression of frontal area of the tank.

   $A=hw$     ...... (III)

Here, the width of the tank is $w$.

Write the expression of location of resultant force acting on the surface $AB$.

   $y_P=y_C+\frac{I_{xx,C}}{y_{C}A}$     ...... (IV)

Here, the distance from centroid is $y_C$ and the moment of inertia about the centroid is $I_{xx,C}$.

Write the expression of moment of inertia about the centroid.

   $$

   $I_{xx,C}=\frac{w{h}^3}{12}$

Write the expression of distance from the centroid of the surface.

   $y_C=\left(s+\frac{h}{2}\right)$

Substitute $\left(s+\frac{h}{2}\right)$ for $y_C$ and $\frac{w{h}^3}{12}$ for $I_{xx,C}$ in Equation (IV).

   $y_P=\left(s+\frac{h}{2}\right)+\frac{\left(\frac{w{h}^3}{12}\right)}{\left(s+\frac{h}{2}\right)A}$     ...... (V)

Calculation:

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $3.5\text{m}$ for $s$ and $2.5\text{m}$ for $h$ in Equation (I).

   $\begin{array}{c}{P}_{AB}=\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\times \left(\left(3.5\text{m}\right)+\frac{\left(2.5\text{m}\right)}{2}\right)\\ =\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\times \left(4.75\text{m}\right)\\ =\left(46597.5\text{kg}/\text{m}\cdot {\text{s}}^2\right)\times \left(\frac{1\text{Pa}}{1\text{kg}/\text{m}\cdot {\text{s}}^2}\right)\times \left(\frac{1\text{kPa}}{1000\text{Pa}}\right)\\ =46.6\text{kPa}\end{array}$

Substitute $2.5\text{m}$ for $h$ and $6\text{m}$ for $w$ in equation (III).

   $\begin{array}{c}A=\left(2.5\text{m}\right)\times \left(6\text{m}\right)\\ =15{\text{m}}^2\end{array}$

Substitute $15{\text{m}}^2$ for $A$ and $46.6\text{kPa}$ for $P_{AB}$ in equation (II).

   $\begin{array}{c}{F}_{R,AB}=\left(46.6\text{kPa}\right)\times \left(15{\text{m}}^2\right)\\ =\left(46.6\text{kPa}\right)\times \left(15{\text{m}}^2\right)\times \left(\frac{1000\text{Pa}}{1\text{kPa}}\right)\times \left(\frac{1\text{kg}/\text{m}\cdot {\text{s}}^2}{1\text{Pa}}\right)\\ =\left(699000\text{kg}\cdot \text{m}/{\text{s}}^2\right)\times \left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^2}\right)\times \left(\frac{1\text{kN}}{1000\text{N}}\right)\\ =699\text{kN}\end{array}$

Substitute $3.5\text{m}$ for $s$, $2.5\text{m}$ for $h$, $6\text{m}$ for $w$ and $15{\text{m}}^2$ for $A$ in Equation (V).

   $\begin{array}{c}{y}_P=\left(3.5\text{m}+\frac{2.5\text{m}}{2}\right)+\frac{\left(\frac{\left(6\text{m}\right)\times {\left(2.5\text{m}\right)}^3}{12}\right)}{\left(3.5\text{m}+\frac{2.5\text{m}}{2}\right)\left(15{\text{m}}^2\right)}\\ =\left(4.75\text{m}\right)+\frac{7.81{\text{m}}^{\text{4}}}{71.25{\text{m}}^{\text{3}}}\\ =\left(4.75\text{m}\right)+\left(0.1096\text{m}\right)\\ =4.86\text{m}\end{array}$

Conclusion:

The magnitude of resultant force acting on surface $AB$ is $699\text{kN}$.

The location of line of action of resultant force is $4.86\text{m}$.

To determine

(b)

The pressure acting on surface $BD$.

Weight of the oil in the tank.

Answer to Problem 83P

The pressure acting on surface $BD$ is $58.86\text{kPa}$.

Weight of the oil in the tank is $1067.01\text{kN}$.

Explanation of Solution

The below figure represent the pressure acting on the side $BD$.

        Figure-(2)

Write the expression of pressure acting on the bottom surface $BD$.

   $P_{BD}=\rho g\left(s+h\right)$     ...... (VI)

Write the expression of force acting on the side $AB$.

   $F_{R,BD}=P_{BD}\times A_{BD}$     ...... (VII)

Here, the frontal area is $A$.

Write the expression of bottom surface area of the tank.

   $A_{BD}=lw$     ...... (VIII)

Here, the length of the tank is $l$.

Substitute $lw$ for $A_{BD}$ in Equation (VII).

   $F_{R,BD}=P_{BD}\times lw$     ...... (IX)

Write the expression of weight of the tank.

   $W={\rho }_{oil}\left(V_t+V_c\right)g$     ...... (X)

Here, the density of the oil is ${\rho }_{oil}$ the volume occupied in tank is $V_t$ and the volume occupied in column is $V_c$.

Write the expression of volume occupied in the tank.

   $V_t=lhw$

Write the expression of volume occupied in the column.

   $V_c=l_{c}sw$

Here, the length of column is $l_c$.

Write the expression of density of the oil.

   ${\rho }_{oil}=S_g\times \rho$

Here, the specific gravity of oil is $S_g$ and the density of the water is $\rho$.

Substitute $l_{c}sw$ for $V_c$, $lhw$ for $V_t$ and $S_g\times \rho$ for ${\rho }_{oil}$ in Equation (X).

   $W=\left(S_g\times \rho \right)\left(lhw+l_{c}sw\right)g$     ....... (XI)

Calculation:

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $3.5\text{m}$ for $s$ and $2.5\text{m}$ for $h$ in Equation (VI).

   $\begin{array}{c}{P}_{BD}=\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\times \left(3.5\text{m}+2.5\text{m}\right)\\ =\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\times \left(6\text{m}\right)\\ =\left(58860\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\right)\times \left(\frac{1\text{Pa}}{1\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}}\right)\times \left(\frac{1\text{kPa}}{1000\text{Pa}}\right)\\ =58.86\text{kPa}\end{array}$

Substitute $58.86\text{kPa}$ for $P_{BD}$, $8.1\text{m}$ for $l$ and $6\text{m}$ for $w$ in Equation (IX).

   $\begin{array}{c}{F}_{R,BD}=\left(58.86\text{kPa}\right)\times \left(8.1\text{m}\right)\times \left(6\text{m}\right)\\ =\left(58.86\text{kPa}\right)\times \left(8.1\text{m}\right)\times \left(6\text{m}\right)\times \left(\frac{1000\text{Pa}}{1\text{kPa}}\right)\times \left(\frac{1\text{kg}/\text{m}\cdot {\text{s}}^2}{1\text{Pa}}\right)\\ =\left(2860596\text{kg}\cdot \text{m}/{\text{s}}^2\right)\times \left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^2}\right)\times \left(\frac{1\text{kN}}{1000\text{N}}\right)\\ =2860.6\text{kN}\end{array}$

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$, $0.88$ for $S_g$, $8.1\text{m}$ for $l$, $2.5\text{m}$ for $h$, $6\text{m}$ for $w$, $0.1\text{m}$ for $l_c$, $3.5\text{m}$ for $s$ and $6\text{m}$ for $w$ in Equation (XI).

   $\begin{array}{c}W=\left[\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(0.88\right)\left\{\begin{array}{l}\left(8.1\text{m}\right)\times \left(2.5\text{m}\right)\times \left(6\text{m}\right)\\ +\left(0.1\text{m}\right)\times \left(3.5\text{m}\right)\times \left(6\text{m}\right)\end{array}\right\}\times \left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\right]\\ =\left(8632.8\text{kg}/{\text{m}}^2\cdot {\text{s}}^{\text{2}}\right)\times \left\{\left(121.5{\text{m}}^{\text{3}}\right)+\left(2.1{\text{m}}^{\text{3}}\right)\right\}\\ =\left(8632.8\text{kg}/{\text{m}}^2\cdot {\text{s}}^{\text{2}}\right)\times \left(123.6{\text{m}}^{\text{3}}\right)\times \left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^2}\right)\times \left(\frac{1\text{kN}}{1000\text{N}}\right)\\ =1067.01\text{kN}\end{array}$

Conclusion:

The pressure acting on surface $BD$ is $58.86\text{kPa}$.

Weight of the oil in the tank is $1067.01\text{kN}$.