Chapter 6.6, Problem 29E

(a).

To determine

To Calculate: The hydrostatic pressure at the bottom of an aquarium.

Answer to Problem 29E

The pressure at the bottom of the aquarium is $187.5lb/f{t}^2$ .

Explanation of Solution

Given Information: The aquarium is fully filled with water.

• The height of the aquarium is $3ft$ .

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: Assume the density of water is Ï�g=62.5 lb/ft³

Put the values of $\rho ,g,d$ in the formula for $P$ to get

  $P=\rho gd=62.5\times 3=187.5$

Therefore, the pressure at the bottom of the aquarium is $187.5lb/f{t}^2$ .

(b).

To determine

To Calculate: The hydrostatic force on the bottom of an aquarium.

Answer to Problem 29E

The hydrostatic force on the bottom surface of the aquarium is $1875lb$ .

Explanation of Solution

Given Information:

• The width of the aquarium is $2ft$ .
• The length of the aquarium is $5ft$ .

Concept Used:

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

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

  $A=length\times width=2\times 5=10f{t}^2$

From the previous sub-part, the hydrostatic pressure at the bottom is $187.5lb/f{t}^2$

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

  $F=PA=187.5\times 10=1875$

Therefore, the hydrostatic force on the bottom surface of the aquarium is $1875lb$ .

(c).

To determine

To Calculate: The hydrostatic force on the lateral face of an aquarium.

Answer to Problem 29E

The total force on one end of the aquarium is $562.5lb$ .

Explanation of Solution

Given Information:

• The height of the aquarium is $3ft$ .
• The width of the aquarium is $2ft$ .
• The length of the aquarium is $5ft$ .

Concept Used:

• The hydrostatic pressure of a liquid of density $\rho$ at a depth $d$ from the surface of the liquid is $P=\rho gd$ .
• 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 water is

  $P_x=\rho gx=62.5x$

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

  $dA=width\times thickness=wdx=2dx$

Therefore, the force on this thin strip of area is

  $dF=P_{x}dA=62.5x\cdot 2dx$

  $=125xdx$

Now the total force on this lateral surface is the integral of $dF$ from $x=0$ (surface of the liquid) to $x=3$ (the lowest point of the liquid). So

  $F={{\displaystyle \int }}_0^{3}dF={{\displaystyle \int }}_0^{3}125xdx$

  $={125\frac{x^2}{2}]}_0^3$

  $=125\times \frac{3^2-0^2}{2}=562.5$

Therefore, the total force on one end of the aquarium is $562.5lb$ .