Chapter 26.2, Problem 1aTH

1. Three objects are at rest in three beakers of water as shown.

1. Compare the mass, volume, and density of the objects to the mass, volume, and density of the displaced water. Explain your reasoning in each case.

To determine

The comparison of mass, volume and density of the object with the displaced water.

Explanation of Solution

Given:

Three objects are at rest in three beakers of water as shown:

FormulaUsed:

Use Archimedes principle which is expressed as follows:

  $F_b={\rho }_{water}g{V}_d$   ......... (1)

Here, the buoyant force is $F_b$ , the density of water is ${\rho }_{water}$ , acceleration due to gravity is $g$ and the volume of fluid displaced is $V_d$ .

Calculation:

For Object 1

Free body diagram of the 1^st object is shown below:

As the object floats on the top of the surface, the displaced volume must be less than the volume of the object. Thus, $V_d<V_1$ .

  

As the body is floating on the top surface, then the weight of the body is equal to the buoyant force. Thus, for equilibrium,

  $\begin{array}{l}{m}_{object}g=F_b\\ m_{object}g={\rho }_{water}g{V}_d\left(m_d={\rho }_{water}{V}_d\right)\\ m_{object}g=g{m}_d\\ m_{object}=m_d\end{array}$

Thus, the mass of the object is equal to the mass of the displaced water.

For density, apply summation of force in the vertical direction as follows:

  $\begin{array}{l}{m}_{object}g=F_b\\ {\rho }_{object}g{V}_{object}={\rho }_{water}g{V}_d\left(m_{object}={\rho }_{object}{V}_{object}\right)\\ {\rho }_{object}={\rho }_{water}\frac{V_d}{V_{object}}\end{array}$

As the $V_d$ is less than the $V_{object}$ , then $\frac{V_d}{V_{object}}$ is less than 1. Thus, the density of object must be less than the density of water.

For Object 2, Free body diagram of the 2nd object is shown below:

  

As the object floats inside the water, the displaced volume must be equal to the volume of the object. Thus, $V_d=V_2$ .

As the body is floating inside the water then the weight of the body is equal to the buoyant force. Thus, for equilibrium,

  $\begin{array}{l}{m}_{object}g=F_b\\ m_{object}g={\rho }_{water}g{V}_d\left(m_d={\rho }_{water}{V}_d\right)\\ m_{object}g=g{m}_d\\ m_{object}=m_d\end{array}$

Thus, the mass of the object is equal to the mass of the displaced water.

For density apply summation of force in the vertical direction as follows:

  $\begin{array}{l}{m}_{object}g=F_b\\ {\rho }_{object}g{V}_{object}={\rho }_{water}g{V}_d\left(m_{object}={\rho }_{object}{V}_{object}\right)\\ {\rho }_{object}={\rho }_{water}\frac{V_d}{V_{object}}\left(V_d=V_{object}\right)\\ {\rho }_{object}={\rho }_{water}\end{array}$

Thus, the density of the object must be equal to the density of water.

For Object 3,

Free body diagram of the 3rd object is shown below:

  

The displaced volume must be equal to the volume of the object as it is kept inside the tank completely. Thus, $V_d=V_2$ .

As the body is not floating inside the water, then the weight of the body is more than the buoyant force. Thus, for equilibrium,

  $\begin{array}{l}{m}_{object}g>F_b\\ m_{object}g>{\rho }_{water}g{V}_d\left(m_d={\rho }_{water}{V}_d\right)\\ m_{object}g>g{m}_d\\ m_{object}>m_d\end{array}$

Thus, the mass of the object is more than the mass of the displaced water.

For density, apply summation of force in the vertical direction as follows:

  $\begin{array}{l}{m}_{object}g>F_b\\ {\rho }_{object}g{V}_{object}>{\rho }_{water}g{V}_d\left(m_{object}={\rho }_{object}{V}_{object}\right)\\ {\rho }_{object}>{\rho }_{water}\frac{V_d}{V_{object}}\left(V_d=V_{object}\right)\\ {\rho }_{object}>{\rho }_{water}\end{array}$

Thus, the density of object must be more thanthe density of water.

Conclusion:

For object 1,

  $V_d<V_1$ , $m_{object}=m_d$ and ${\rho }_{object}<{\rho }_{water}$ .

For object 2,

  $V_d=V_2$ , $m_{object}=m_d$ and ${\rho }_{object}={\rho }_{water}$ .

For object 3,

  $V_d=V_2$ , $m_{object}>m_d$ and ${\rho }_{object}>{\rho }_{water}$ .