Chapter 6.6, Problem 30E

(a).

To determine

To Calculate: The hydrostatic pressure at the bottom of a tank.

Answer to Problem 30E

The pressure at the bottom of the tank is $12054Pa$ .

Explanation of Solution

Given: The tank contains kerosene with density $820kg/m^3$ .

• The height of the tank is $2m$ .
• The depth of kerosene is $1.5m$ .

Concept Used: The hydrostatic pressure of a liquid of density $\rho$ at a depth $d$ from the surface of the liquid is $P=\rho \text{gd}$ .

Calculation: Put the values of $\text{ρ},g,d$ in the formula for $P$ to get

  $P=\rho gd=820\times 9.8\times 1.5=12054$

Therefore, the pressure at the bottom of the tank is $12054Pa$ .

(b).

To determine

To Calculate: The hydrostatic force on the bottom of a tank.

Answer to Problem 30E

The hydrostatic force on the bottom surface of the tank is $385728N$ .

Explanation of Solution

Given:

• The width of the tank is $4m$ .
• The length of the tank is $8m$ .

Concept Used:

The force on a surface due to pressure $P$ exertedis $F=PA$ , where $A$ is the area of the surface being considered.

Calculation: The area of the bottom surface of the tank is

  $A=length\times width=8\times 4=32 squaremeters$

From the previous sub-part, the hydrostatic pressure at the bottom of the tank is $12054Pa$

So, multiply the pressure and the area to get force as

  $F=PA=12054\times 32=385728$

Therefore, the hydrostatic force on the bottom surface of the tank is $385728N$

(c).

To determine

To Calculate: The hydrostatic force on the lateral face of a tank.

Answer to Problem 30E

The total force on one end of the tank is $36162N$ .

Explanation of Solution

Given: The tank contains kerosene with density $820kg/m^3$ .

• The height of the tank is $2m$ .
• The width of the tank is $4m$ .
• The length of the tank is $8m$ .
• The depth of kerosene is $1.5m$ .

Concept Used:

• The hydrostatic pressure of a liquid of density $\rho$ at a depth $d$ from the surface of the liquid is $P=\rho \text{gd}$ .
• The force on a surface due to pressure $P$ exerted is $F=PA$ , where $A$ is the area of the surface being considered.

The pressure in the liquid at any point that is at a depth $x$ from the surface of a liquid is same given by

  $P_x=\rho gx=820\times 9.8\times x=8036x$

The area of an infinitesimally thin strip of thickness $dx$ on one side of the tank with width $w$ at the depth $x$ is

  $dA=width\times thickness=wdx=4dx$

Therefore, the force on this shin strip of area is

  $dF=P_{x}dA=8036x\cdot 4dx$

  $=32144xdx$

Now the total force on this lateral surface is the integral of $dF$ from $x=0$ (surface of the liquid) to $x=1.5$ (the lowest point of the liquid). So

  $F={{\displaystyle \int }}_0^{1.5}dF={{\displaystyle \int }}_0^{1.5}32144xdx$

  $={32144\frac{x^2}{2}]}_0^{1.5}$

  $=32144\times \frac{{1.5}^2-0^2}{2}=36162$

Therefore, the total force on one end of the tank is $36162N$ .