Chapter 10, Problem 24P

Solution

(a)

To Find: The tension in the crane’s cable when the hull is submerged in water.

The tension in the crane’s cable is $1.5\times {10}^5\text{N}$ .

Given information:

Mass of the steel hull, $m=18000\mathrm{kg}$

Density of the steel, ${\rho }_s=8050\mathrm{kg}/m^3$

Formula Used:

• The tension force acts in the reverse direction to that of the weight force during the pulling of an object by the string.

  $T=mg$

• When an object is submerged into the fluid, then a force is exerted by the fluid to the object, this force is known as the buoyancy force.

This force is written as,

  $F_B={\rho }_{F}Vg$

Where, ${\rho }_F$ is the fluid density, $F_B$ is the buoyancy force, $V$ is the volume of the fluid and $g$ is the acceleration due to gravity.

Calculation:

Now substitute the values in above formula, then the result is

  $\begin{array}{l}T+F_B=mg\\ \Rightarrow T=mg-F_B\\ =m_{hull}g-{\rho }_{water}{V}_{sub}g\\ =m_{hull}g-{\rho }_{water}\frac{m_{hull}}{{\rho }_{hull}}g\\ =m_{hull}g\left(1-\frac{{\rho }_{water}}{{\rho }_{hull}}\right)\\ =18000kg\times \left(9.8m/s^2\right)\left(1-\frac{1.0\times 10^3\text{kg}/m^3}{7.8\times 10^3\text{kg/}{m}^3}\right)\\ T=1.5\times 10^{5}N\end{array}$

Conclusion:

The tension in the crane’s cable is $1.5\times {10}^5\text{N}$ .

(b)

To Find: The tension in the crane’s cable when the hull is completely out of the water.

  $1.8\times {10}^5\text{N}$

Given information:

Mass of the steel hull, $m=18000\mathrm{kg}$

Formula Used:

• The tension force acts in the reverse direction to that of the weight force during the pulling of an object by the string.

  $T=mg$

Calculation:

Now substitute the values in above formula, then the result is

  $\begin{array}{l}T=mg=18000\times 9.8\\ =176400N\\ =1.8\times 10^{5}N\end{array}$

Conclusion:

The tension when the hull is completely out of the water is $1.8\times {10}^5\text{N}.$ .