Question

A hollow plastic sphere is held below the surface of a fresh-water lake by a cord anchored to the bottom of the lake. The sphere has a volume of 0.650 cubic meters and the tension in the cord is 900 N.

a) Calculate the buoyant force exerted by the water on the sphere.

b) what is the mass of the sphere?

c) the cord breaks and the sphere rises to the surface. When the sphere comes to rest, what fraction of its volume will be submerged?

Answer 1

Concepts and reason

The concept required to solve the given problem is buoyant force. Calculate the buoyant force by first computing mass in terms of density and volume and then taking the product of mass and acceleration due to gravity. Calculate the mass of the sphere with the help of equilibrium condition of force. Calculate the fraction of volume of the sphere submerged by taking the ratio of the volume of sphere submerged and volume of sphere and multiplying it by 100 .

Fundamentals

The upward force exerted by a fluid on an object when the object is placed in it is called buoyant force. Archimedes' Principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces and acts in the upward direction at the center of mass of the displaced fluid. The equilibrium condition for the force gives, \(\Sigma F=0\)

Here, \(F\) is force.

 

(a) The expression to calculate the buoyant force is, \(F_{B}=\rho_{\text {water}} V_{\text {sphere}} g\)

Substitute \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) for \(g, 1000 \mathrm{~kg} / \mathrm{m}^{2}\) for \(\rho_{w a t e r}\) and \(0.650 \mathrm{~m}^{3}\) for \(V_{s p h e r e}\) in the above equation.

$$ \begin{array}{c} F_{B}=\left(1000 \mathrm{~kg} / \mathrm{m}^{2}\right)\left(0.650 \mathrm{~m}^{3}\right)\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right) \\ =6370 \mathrm{~N} \end{array} $$

Part a The buoyant force exerted by water on the sphere is \(6370 \mathrm{~N}\).

The buoyant force is equal to the weight of fluid displaced by the body. The buoyant force exerted by water on the sphere will be, \(F_{B}=m g \ldots \ldots(1)\)

Also, the mass and density are related as, \(\rho=\frac{m}{V}\)

Here, \(m\) is the mass, \(\rho\) is the density and \(V\) is the volume. The volume of water displaced will be equal to the volume of the sphere since the sphere is completely submerged in water. Rearrange the above equation. \(m=\rho V\)

Substitute \(\rho V\) for \(m\) in equation \((1) .\)

$$ F_{B}=m g $$

\(=\rho_{\text {water}} V_{\text {waterdisplaced}} g\)

Here, \(\rho_{\text {water}}\) is the density of water, \(V_{\text {waterdisplaced}}\) is the volume of water displaced and \(g\) is acceleration due to gravity. But, \(V_{w a t e r d i s p l a c e d}=V_{\text {sphere}}\)

\(F_{B}=\rho_{\text {water}} V_{\text {sphere}} g\)

 

(b) The expression to calculate the mass of the sphere is, \(m=\frac{F B-T}{g}\)

Substitute \(6370 \mathrm{~N}\) for \(F_{B}, 900 \mathrm{~N}\) for \(T\) and \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) for \(g\) in the above equation.

$$ \begin{aligned} m &=\frac{6370 \mathrm{~N}-900 \mathrm{~N}}{9.8 \mathrm{~m} / \mathrm{s}^{2}} \\ &=558 \mathrm{~kg} \end{aligned} $$

Part b The mass of the sphere is \(558 \mathrm{~kg}\).

The net force acting on the hollow plastic sphere held below the surface of a fresh water lake is, \(F_{n e t}=F_{B}-T-m g\)

Here, \(F_{B}\) is the buoyant force, \(T\) is the tension, \(\mathrm{m}\) is the mass and \(\mathrm{g}\) is the gravitational force. But since the sphere is stationary, the net force acting on the sphere will be zero. \(F_{B}-T-m g=0\)

Rearrange the above equation. \(m=\frac{F B-T}{g}\)

 

(c) When the cord breaks the only forces acting on the sphere will be buoyant force and the weight of the sphere. The density of the sphere will be, \(\rho_{\text {sphere}}=\frac{M}{V_{\text {sphere}}}\)

Here, \(M\) is the mass of sphere and \(V_{\text {sphere}}\) is the volume of sphere. Substitute \(558 \mathrm{~kg}\) for \(M\) and \(0.650 \mathrm{~m}^{3}\) for \(V\) in the above equation.

$$ \begin{array}{l} \rho_{\text {sphere}}=\frac{558 \mathrm{~kg}}{0.650 \mathrm{~m}^{3}} \\ \quad=858 \mathrm{~kg} / \mathrm{m}^{3} \end{array} $$

The volume of the sphere submerged will be, \(V_{\text {submerged}}=V_{\text {sphere}}\left(\frac{\rho_{\text {sphere}}}{\text {\rhowater}}\right)\)

Substitute \(0.650 \mathrm{~m}^{3}\) for \(V_{\text {sphere}}, 858 \mathrm{~kg} / \mathrm{m}^{3}\) for \(\rho_{\text {sphere}}\) and \(1000 \mathrm{~kg} / \mathrm{m}^{3}\) for \(\rho_{\text {water}}\) in the above equation.

\(V_{\text {submerged}}=\left(0.650 \mathrm{~m}^{3}\right)\left(\frac{858 \mathrm{~kg} / \mathrm{m}^{3}}{1000 \mathrm{~kg} / \mathrm{m}^{3}}\right)\)

$$ =0.558 \mathrm{~m}^{3} $$

The fraction of volume submerged will be, \(\%=\frac{V_{\text {submerged}}}{V_{\text {sphere}}} \times 100\)

Substitute \(0.558 \mathrm{~kg} / \mathrm{m}^{3}\) for \(V_{\text {submerged}}\) and \(0.650 \mathrm{~kg} / \mathrm{m}^{3}\) for \(V_{\text {sphere}}\) in the above equation.

$$ \begin{array}{c} \%=\frac{0.558 \mathrm{~kg} / \mathrm{m}^{3}}{0.650 \mathrm{~kg} / \mathrm{m}^{3}} \times 100 \\ =85.8 \% \end{array} $$

Part c The fraction of volume submerged will be \(85.8 \%\).

The mass per unit volume is known as density. The units per density is \(\mathrm{kg} / \mathrm{m}^{3}\). As the density is inversely proportional to the volume is expressed as,

$$ \begin{array}{c} \rho \alpha \frac{1}{V} \\ \frac{\rho 1}{\rho 2}=\frac{V_{2}}{V_{1}} \\ \frac{\rho_{\text {sphere}}}{\rho_{\text {water}}}=\frac{V_{\text {water}}}{V_{\text {sphere}}} \end{array} $$

Rewrite the above expression as follows:

\(V_{\text {water}}(\)or\() V_{\text {submerged}}=V_{\text {sphere}}\left(\frac{\rho_{\text {sphere}}}{\rho_\text {water}}\right)\)

 

Part a 

The buoyant force exerted by water on the sphere is \(6370 \mathrm{~N}\).

Part b 

The mass of the sphere is \(558 \mathrm{~kg}\).

Part c

The fraction of volume submerged will be \(85.8 \%\).