Chapter 8, Problem 8.105P

To determine

(a)

The hydrodynamic mass of the sphere.

Answer to Problem 8.105P

The required value of the mass is $2.78kg$.

Explanation of Solution

Given Information:

$\begin{array}{l}D=22cm\\ SG=2.7\\ a=12m/s^2\\ T={20}^{o}C\end{array}$

Formula used:

$m_h=\frac{2}{3}{\rho }_{water}\pi a^3$

Calculation:

To solve this problem for water at ${20}^{o}C$ we are taking the value of density as $\rho =998kg/m^3$.

The radius of the sphere is $a=11cm$.

We are using Eq. $\left(8.103\right)$ which gives the hydrodynamic mass and proportional to the water density;

$m_h=\frac{2}{3}{\rho }_{water}\pi a^3$ ;

Put the values in the above equation;

$\begin{array}{l}{m}_h=\frac{2}{3}{\rho }_{water}\pi a^3=\frac{2}{3}\left(998kg/m^3\right)\pi {\left(0.11m\right)}^3\\ \Rightarrow 2.78kg\end{array}$.

To determine

(b)

The value of the force being applied to the sphere.

Answer to Problem 8.105P

The required value of the force is $214N$.

Explanation of Solution

Given information:

$\begin{array}{l}D=22cm\\ SG=2.7\\ a=12m/s^2\\ T={20}^{o}C\end{array}$

Formula used:

$\left(\pi /6\right){\rho }_{sphere}\left(D^3\right)$

${\displaystyle \sum F}=\left(m_{sphere}+m_h\right)\frac{dV}{dt}$

Calculation:

Firstly we will calculate the mass of the sphere;

$\begin{array}{l}\left(\pi /6\right){\rho }_{sphere}\left(D^3\right)=\left(\pi /6\right)\left(2.7\times 998\right){\left(0.22\right)}^3\\ \Rightarrow 15.02kg\end{array}$

Now we will use Eq. $\left(8.101\right)$ to find the applied force;

${\displaystyle \sum F}=\left(m_{sphere}+m_h\right)\frac{dV}{dt}$ ;

Put the values in the above equation;

$\begin{array}{l}{\displaystyle \sum F}=\left(m_{sphere}+m_h\right)\frac{dV}{dt}=\left(15.02+2.78kg\right)\left(12m/s^2\right)\\ \Rightarrow 214N\end{array}$.