Chapter 19, Problem 14A

When the buoyant force on a submerged object is more than the weight of the object, how do the densities of the object and water compare?

To determine

To Compare: The density of water and the object when the buoyant force on the submerged object is more than the weight of the object.

Answer to Problem 14A

The density of water is greater than density of the object

Explanation of Solution

Introduction:

Archimedes principle states that upward buoyancy force acts on an immersed object in a fluid is of same magnitude as to the weight of the fluid which the object displaces.

The weight of the object is,

  $F_{\text{obe}}=\rho gV$

Here, $\rho$ is the density of the object, $g$ is the acceleration due to gravity, and $V$ is the volume of the object.

The buoyancy force acting on the object is,

$F_{\text{buoyancy}}={\rho }_{\text{w}}g{V}_{\text{w}}$

Here, ${\rho }_w$ is the density of the water and $V_w$ is the volume of the water.

When the object is submerged in water then it displaces the volume of water equal to its own volume. Thus, the volume of displaced water is equal to volume of the object.

$V=V_w$

The buoyancy force on a submerged object is more than the weight of the object.

  $F_{\text{buoyancy}}>F_{\text{object}}$

  ${\rho }_{\text{w}}g{V}_{\text{w}}>\rho gV$

Substitute $V$ for $V_w$

  $\begin{array}{c}{\rho }_{\text{w}}gV>\rho gV\\ {\rho }_{\text{w}}>\rho \end{array}$

Conclusion:

Therefore, the density of water is greater than density of the object.