Chapter 15, Problem 17PQ

The dolphin tank at an amusement park is rectangular in shape with a length of 40.0 m, a width of 15.0 m, and a depth of 7.50 m. The tank is filled to the brim to provide maximum splash during dolphin shows. What is the total amount of force exerted by the water on a. the bottom of the tank, b. the longer wall of the tank, and c. the shorter wall of the tank?

To determine

The total amount of force exerted by the water on the bottom of the tank.

Answer to Problem 17PQ

The total amount of force exerted by the water on the bottom of the tank is $4.53\times {10}^7\text{N}$ .

Explanation of Solution

Write the expression for the hydrostatic pressure on the bottom due to water.

  $P_b=\rho gy$                                                                                                                   (I)

Here, $P_b$ is the pressure at the bottom of tank due to water, $\rho$ is the density of water, $g$ is the acceleration due to gravity and $y$ is the depth of water.

Write the expression for the force acting on the bottom of tank.

  $F_b=P_{b}A$                                                                                                                   (II)

Here, $F_b$ is the force acting on the bottom of the tank and $A$ is the area of the tank.

Write the expression for the area of the tank.

  $A=lw$                                                                                                                    (III)

Here, $l$ is the length of the tank and $w$ is the width of the tank.

Substitute equation (III) in (II).

  `$F_b=P_{b}lw$                                                                                                              (IV)

Conclusion:

The length of the tank is $40.0m$ , the width of the tank is $15.0\text{m}$ , height of the tank is $7.50\text{m}$ , density of water is $1025.18\text{kg}/{\text{m}}^3$ and acceleration due to gravity is $9.81\text{m}/{\text{s}}^2$ .

Substitute $1025.18\text{kg}/{\text{m}}^3$ for $\rho$ , $9.81\text{m}/{\text{s}}^2$ for $g$ and $7.50\text{m}$ for $y$ in equation (I) to get $P_b$ .

  $\begin{array}{c}{P}_b=\left(1025.18\text{kg}/{\text{m}}^3\right)\left(9.81\text{m}/{\text{s}}^2\right)\left(7.50\text{m}\right)\\ =75427.62\text{Pa}\end{array}$

Substitute $75427.62\text{Pa}$ for $P_b$ , $40.0\text{m}$ for $l$ and $15.0\text{m}$ for $w$ in equation (IV) to get $F_b$ .

  $\begin{array}{c}{F}_b=\left(75427.62\text{Pa}\right)\left(40.0\text{m}\right)\left(15.0\text{m}\right)\\ =4.53\times {10}^7\text{N}\end{array}$

Therefore, the total amount of force exerted by the water on the bottom of the tank is $4.53\times {10}^7\text{N}$ .

To determine

The total amount of force exerted by the water on the longer wall of the tank.

Answer to Problem 17PQ

The total amount of force exerted by the water on the longer wall of the tank is $1.13\times {10}^7\text{N}$ .

Explanation of Solution

Consider strip of height $dz$ at a height $z$ and having length $L$ .

Write the expression for the force on the strip.

  $dF=PdA$                                                                                                                (V)

Here, $dF$ is the force acting on the strip, $P$ is the pressure and $dA$ is the area of the strip.

Write the expression for the pressure at height $z$ .

  $P=\rho gz$                                                                                                                  (VI)

Write the area of the strip.

  $dA=Ldz$                                                                                                              (VII)

Put equations (VI) and (VII) in equation (V) to get force acting the strip.

  $dF=\rho gzLdz$                                                                                                       (VIII)

The above equation is written in terms of differential height. Therefore, integrating above equation with limit from 0 to $h$ will give total force acting on wall of greater length.

Integrate equation (VIII) to get total force acting on the wall of greater length.

  $\begin{array}{c}F={\displaystyle {\int }_0^h\rho gzLdz}\\ =\rho gL{\displaystyle {\int }_0^{h}zdz}\\ =\rho gL{\left(\frac{z^2}{2}\right)}_0^h\\ =\frac{1}{2}\rho gL{h}^2\end{array}$                                                                                                      (IX)

Conclusion:

Substitute $1025.18\text{kg}/{\text{m}}^3$ for $\rho$ , $9.81\text{m}/{\text{s}}^2$ for $g$ , $40.0\text{m}$ for $L$ and $7.50\text{m}$ for $h$ in equation (IX) to get $F$ .

  $\begin{array}{c}F=\frac{1}{2}\left(1025.18\text{kg}/{\text{m}}^3\right)\left(9.81\text{m}/{\text{s}}^2\right)\left(40.0\text{m}\right){\left(7.50\text{m}\right)}^2\\ =1.13\times {10}^7\text{N}\end{array}$

Therefore, the total amount of force exerted by the water on the longer wall of the tank is $1.13\times {10}^7\text{N}$ .

To determine

The total amount of force exerted by the water on the shorter wall of the tank.

Answer to Problem 17PQ

The total amount of force exerted by the water on the shorter wall of the tank is $4.24\times {10}^6\text{N}$ .

Explanation of Solution

The total force acting on the shorter side of the tank can be obtained from the equation (IX) with length $L$ should replace by width.

Write the expression for the total force acting on the shorter side of the tank.

  $F_w=\frac{1}{2}\rho gw{h}^2$                                                                                                         (X)

Here, $F_w$ is the force acting on the shorter length.

Conclusion:

Substitute $1025.18\text{kg}/{\text{m}}^3$ for $\rho$ , $9.81\text{m}/{\text{s}}^2$ for $g$ , $15.0\text{m}$ for $w$ and $7.50\text{m}$ for $h$ in equation (X) to get $F_w$ .

  $\begin{array}{c}{F}_w=\frac{1}{2}\left(1025.18\text{kg}/{\text{m}}^3\right)\left(9.81\text{m}/{\text{s}}^2\right)\left(15.0\right){\left(7.50\text{m}\right)}^2\\ =4.24\times {10}^6\text{N}\end{array}$

Therefore, the total amount of force exerted by the water on the shorter wall of the tank is $4.24\times {10}^6\text{N}$ .