Question

A spherical vessel used for deep-sea exploration has a radius of$\mathbf{\text{1.5\hspace{0.17em}m}}$and a mass of role="math" localid="1663750322684" $\mathbf{1}\mathbf{.2}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{\text{\hspace{0.17em}kg}}$. To dive, the vessel takes on mass in the form of seawater. Determine the mass the vessel must take on if it is to descend at a constant speed ofrole="math" localid="1663750448075" $\mathbf{1}\mathbf{.2}\raisebox{1ex}{${\displaystyle \mathbf{\text{m}}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{\text{s}}$}\right.$, when the resistive force on it is $\mathbf{1100}\mathbf{\text{\hspace{0.17em}N}}$in the upward direction. The density of seawater is equal to $\mathbf{1.03}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\raisebox{1ex}{$\mathbf{\text{kg}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{\text{m}}$}\right.$.

Short answer

The vessel must take on $2666.14\text{\hspace{0.17em}kg}$of water to descend at a constant speed.

Step-by-step solution

Identification of the given data:

The given data can be listed below as,

• Radius of the vessel, $r=1.5\text{\hspace{0.17em}m}$
• Mass of the vessel, $m=1.2\times {10}^4\text{\hspace{0.17em}kg}$
• Upward resistive force, $F_r=1100\text{\hspace{0.17em}N}$
• Density of sea water, $d_s=1.03\times {10}^3\raisebox{1ex}{${\displaystyle \text{kg}}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{3}}$}\right.$
• Required speed of the vessel,$v=1.2\raisebox{1ex}{${\displaystyle \text{m}}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$

Buoyancy and gravitational force:

The buoyant force from a liquid of density $\mathit{d}$when volume of the liquid displaced$\mathit{v}$is

$\mathit{B}\mathbf{=}\mathit{d}\mathit{g}\mathit{v}$ ..... (1)

Here, $\mathit{g}$is the acceleration due to gravity having value,

$\mathit{g}\mathbf{=}\mathbf{9}\mathbf{.8}\raisebox{1ex}{${\displaystyle \mathbf{\text{m}}}$}\!\left/ \!\raisebox{-1ex}{${\mathbf{\text{s}}}^{\mathbf{\text{2}}}$}\right.$

The gravitational force on a body of mass $\mathit{m}$is

$\mathit{F}\mathbf{=}\mathit{m}\mathit{g}$ ..... (2)

Determining the mass of water intake by the vessel:

Volume of the vessel is expressed as,

$V=\frac{4}{3}\pi r^3$

Substitute all the value in the above equation,

$\begin{array}{rcl}V& =& \frac{4}{3}\times 3.14\times {\left(1.5\text{\hspace{0.17em}m}\right)}^3\\ & =& 14.13{\text{\hspace{0.17em}m}}^3\end{array}$

This is also the volume of sea water the vessel displaces. Since the vessel is required to sink at a constant speed, the net force on it has to be zero. The downward force is the gravitational force due to the total mass of the vessel and the water intake $m_t$and the upward force is due to buoyancy and resistive force.

Thus from equations (1) and (2) you get,

$$\begin{gathered} m_{t}g=d_{s}Vg+F_r \\ {m}_t=d_{s}V+\frac{F_r}{g} \end{gathered}$$

Substitute all the value in the above equation,

$\begin{array}{rcl}{m}_t& =& 1.03\times {10}^3\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}\times 14.13{\text{\hspace{0.17em}m}}^3+\frac{1100\text{\hspace{0.17em}N}}{9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}}\\ & =& 14553.9\text{\hspace{0.17em}kg}+112.24\cdot \left(\frac{1\text{\hspace{0.17em}kg}\cdot \text{m}/{\text{s}}^{\text{2}}}{1\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}}\right)\\ & =& 14666.14\text{\hspace{0.17em}kg}\end{array}$

The mass of the water intake is thus

$\begin{array}{rcl}{m}_w& =& m_t-m\\ & =& 14666.14\text{\hspace{0.17em}kg}-12000\text{\hspace{0.17em}kg}\\ & =& 2666.14\text{\hspace{0.17em}kg}\end{array}$

Hence, the required mass is $2666.14\text{\hspace{0.17em}kg}$.