Question

A $\mathbf{8}\mathbf{.60}\mathbf{kg}$sphere of radius $\mathbf{6}\mathbf{.22}\mathbf{cm}$is at a depth of $\mathbf{2}\mathbf{.22}\mathbf{km}$in seawater that has an average density of $\mathbf{1025}\mathbf{kg}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$.What are the (a) gauge pressure, (b) total pressure, and (c) corresponding total force compressing the sphere’s surface? What are (d) the magnitude of the buoyant force on the sphere and (e) the magnitude of the sphere’s acceleration if it is free to move? Take atmospheric pressure to be $\mathbf{1}\mathbf{.01}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{Pa}$.

Short answer

1. The gauge pressure acting on the sphere is$2.23\times {10}^7\mathrm{Pa}$

2. The total pressure acting on the sphere is$2.24\times {10}^{\mathrm{\prime }}\mathrm{Pa}$

3. The total force compressing the sphere’s surface is $1.09\times {10}^6\mathrm{N}$.

4. The magnitude of the buoyant force on the sphere is$9.7\times {10}^4\mathrm{N}$

5. The magnitude of the sphere’s acceleration if it is free to move is$8.62\mathrm{m}/{\mathrm{s}}^2$

Step-by-step solution

Listing the given quantities

• The mass of the sphere is$m=8.6\mathrm{kg}$.

• The radius of the sphere is $R=6.22\mathrm{cm}=0.0622\mathrm{m}$

• The depth at which the sphere is in the seawater is,$h=2.22\mathrm{km}=2220\mathrm{m}$

• The density of the seawater is p =$1025\mathrm{kg}/{\mathrm{m}}^3$.

Understanding the concept of pressure

A fluid is a substance that can flow; it conforms to the boundaries of its container because it cannot withstand shearing stress. It can, however, exert a force perpendicular to its surface. If the force is uniform over a flat area, then the force is described in terms of pressure p as,

$\mathit{p}\mathbf{=}\frac{\mathbf{F}}{\mathbf{A}}$

We can find the gauge pressure, total pressure, and buoyant force acting on the sphere from the density of seawater and the depth at which the sphere is submerged in it. Then we can find the acceleration of the sphere using Newton’s second law.

Formulae:

1. $p=\rho gh$

2.$\begin{array}{r}p=p+\rho gh\end{array}$

3.$F_b=\rho Vg$

4.$F_{n\theta t}=ma$

(a) Calculations of weight of the water on the top of the submarine

The gauge pressure acting on the sphere from seawater is

$\begin{array}{r}p=\rho gh\\ =1025\mathrm{kg}/{\mathrm{m}}^3\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(2220\mathrm{m}\right)\\ =22.3\times {10}^6\mathrm{Pa}\\ =2.23\times {10}^7\mathrm{Pa}\end{array}$

Therefore, the gauge pressure acting on the sphere is$2.23\times {10}^7\mathrm{Pa}$

(b) Calculations of total pressure acting on the sphere from seawater

The total pressure acting on the sphere from seawater is

$\begin{array}{r}P=P_0+\rho gh\\ =1.01\times {10}^5\mathrm{Pa}+2.23\times {10}^7\mathrm{Pa}\\ =2.24\times {10}^7\mathrm{Pa}\end{array}$

Therefore, the total pressure acting on the sphere is$2.23\times {10}^7\mathrm{Pa}$

(c) Calculations of the total force compressing the sphere’s surface

The total force compressing the sphere’s surface is

$p=\frac{F}{A}$

So,

$\begin{array}{r}F=PA\\ F=P\left(4\pi R^2\right)\\ =2.24\times {10}^7\mathrm{Pa}\left(4\times 3.142\times (0.0622\mathrm{m}{)}^2\right)\\ =1.089\times {10}^6\mathrm{N}\\ \approx 1.09\times {10}^6\mathrm{N}\end{array}$

Therefore, the total force compressing the sphere’s surface is $1.09\times {10}^6\mathrm{N}$

(d) Calculations of the magnitude of the buoyant force on the sphere

The magnitude of the buoyant force on the sphere is

$\begin{array}{r}{F}_b=\rho Vg\\ =\rho \left(\frac{4}{3}\pi R^3\right)g\\ =\left(1025\mathrm{kg}/{\mathrm{m}}^3\right)\left(\frac{4}{3}\pi (0.0622\mathrm{m}{)}^3\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\\ =10.1\mathrm{N}\end{array}$

Therefore, the magnitude of the buoyant force on the sphere is$10.1\mathrm{N}$

(e) Calculations of the magnitude of the sphere’s acceleration if it is free to move

The forces acting on the sphere if it is free to move with acceleration a are weight in the downward direction and buoyancy force in the upward direction. Then Newton's second law gives

$\begin{array}{r}{F}_{\text{not}}=ma\\ F_b-mgma\\ a=\frac{F_b-mg}{m}\\ =\frac{10.1\mathrm{N}-8.6\mathrm{kg}\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}{8.6\mathrm{kg}}\\ =-8.62\mathrm{m}/{\mathrm{s}}^2\end{array}$

Therefore, the magnitude of the sphere’s acceleration if it is free to move is$8.62\mathrm{m}/{\mathrm{s}}^2$