Chapter 10, Problem 29P

(a)

To determine

To Find: The buoyanct force on the chamber based on given information.

Answer to Problem 29P

The buoyant force is $\text{7}{\text{.4×10}}^{\text{5}}\text{N}$ .

Explanation of Solution

Given info:

Density of seawater, ${\rho }_w=1025{\text{kg/m}}^{\text{3}}$

Diameter, $d=5.20\text{m}$

So, radius, $r=2.\text{6}\text{m}$

Mass of the chamber, $m=7440\text{0}\text{kg}$

Formula used:

Buoyant force

  $B={\rho }_{w}gV$

Where, ${\rho }_w$ is the density of seawater, Volume of the chamber is V and g is the acceleration due to gravity.

Calculation:

Volume of the chamber is

  $\begin{array}{l}V=\frac{4}{3}\pi r^3\\ V=\frac{4}{3}\times 3.14\times {\left(2.6\right)}^3\end{array}$

So, the buoyant force is

  $\begin{array}{l}B=1025\times 9.8\times \left[\frac{4}{3}\times 3.14\times {\left(2.6\right)}^3\right]\\ B=7.4\times {10}^5\text{N}\end{array}$

Conclusion:

The buoyant force is $\text{7}{\text{.4×10}}^{\text{5}}\text{N}$ .

(b)

To determine

To Find: The tension in the cable on the basis of given information.

Answer to Problem 29P

The tension in the cable is $\text{10040}\text{N}$ .

Explanation of Solution

Given info:

Density of seawater, ${\rho }_w=1025{\text{kg/m}}^{\text{3}}$

Diameter, $d=5.20\text{m}$

So, radius, $r=2.\text{6}\text{m}$

Mass of the chamber, $m=7440\text{0}\text{kg}$

Formula used:

Tension in the cable, after simplification

  $F=\left(\frac{1}{6}\pi d^3{\rho }_w-m\right)g$

Here, ${\rho }_w$ is the density of seawater, $d$ is the diameter and $m$ is the mass of the chamber.

Calculation:

According to the second law of motion,

  $ma=B-F-mg$

But at rest, acceleration is zero. So, the above equation becomes

  $\begin{array}{l}0=B-F-mg\\ F=B-mg\end{array}$

Now use the value of buoyant force from the previous part result,

  $F=\left(\frac{1}{6}\pi d^3{\rho }_w-m\right)g$

  $\begin{array}{l}F=\left(\frac{1}{6}\times 3.14\times {\left(5.2\right)}^3\times 1025-74400\right)\left(9.8\right)\\ F=10040\mathrm{N}\end{array}$

Conclusion:

The tension in the cable is $10040\mathrm{N}$ .