Buoyant Force and Archimede’s Principle

Buoyant Force

When something is submerged on water, it experiences an upward force by the water. This upward force causes things to float in water. We call this upward force as the buoyant force. For example, when a ball is pushed downward on water, it bulges out of the water, since the water helps to overcome the downward force on submerged objects. Another example is when carrying heavy objects, such as rock, it seems to weigh much less underwater that at the surface of the water.

Buoyant force occurs because water exerts pressure into the submerged objects. As the depth increases, the buoyant force also increases. This is also because the force deep down the water is much higher than the pressure at the top of the water, thus, force is also greater under than at the top.


Archimede’s Principle

When an object is immersed on fluid, either as a whole or partially, a buoyant force is exerted on the object which is equal in magnitude to the gravitational force on the displaced fluid.

The buoyant force is expressed mathematically as

\(F_b = mg\)

Since \(m = \rho V\), the above equation can also be written as

\(F_b = \rho Vg\),

where

\(F_b\) is the buoyant force, in Newtons,

\(V\) is the volume displaced measured in cubic meters,

\(\rho\) is the density of the fluid, in kilograms per cubic meter,

\(g\) is the gravitational field strength, in newtons per kilogram.

When an object is submerged in water, it seems to weigh much less than its original weight. This apparent weight of the object can be determined by getting the difference between the object's weight (force of gravity, g) and the buoyant force. Mathematically, it can be written as

\(F_g\; \text{apparent} = F_g - F_b\).

In case the apparent mass is given, the apparent weight can also be solved using the equation

\(F_g\; \text{apparent} = m_ag\)

where ma is the apparent mass.


Example

A rock, submerged in water, has an apparent mass of 7.5 kg.

  1. What is the actual weight of the rock if its mass is 13.0 kg?
  2. Solve for the rock’s apparent weight.
  3. Calculate the buoyant force exerted on the rock.
  4. What is the mass and volume of the water?
  5. What is the volume of the water?
  6. What is the density of the rock?

Solution:

  1. To solve for the actual weight of the rock, we use the equation for gravitational force, F = mg, given its mass = 13.0 kg and g = 9.8 N/kg.

 

\(F_g = mg = (13.0\;kg)(9.8\;N/kg) = 127.4\;N\)

 

  1. The apparent weight, Fg(apparent), can also be solve using the above equation, but instead of the actual mass, we use the given apparent mass of the rock = 7.5 kg.

 

\(F_g\;\text{apparent} = m_ag = (7.5\;kg)(9.8\;N/kg) = 73.5\;N\)

 

  1. To solve for the buoyant force, we use the equation for apparent weight: \(F_g\;\text{apparent} = F_g \;– F_b\).

 

\(F_g\;\text{apparent} = F_g \;– F_b \implies F_b = F_g - F_g\;\text{apparent}\)

 

       Substituting the values for weight and apparent weight, we have

\(F_b = 127.4\;N \;– \;73.5\;N = 53.9\;N\)

 

  1. According to Archimede’s principle, the buoyant force is equal in magnitude to the force of gravity on the displaced fluid, this follows that, Fb = 53.9 N is also the weight of the water displaced by the submerged rock. Thus, we can now solve for the mass of the water using the equation \(F_{gw} = m_wg\), \(F_{gw}\) and \(m_w\) and are the weight and the mass of the water, respectively.

 

\(F_gw = m_wg \implies m_w = {F_{gw} \over g}\)

        Thus, the mass of the water is

\(m_w = {53.9\;N \over 9.8\;N/kg} = 5.5\;kg\).

 

  1. Using the equation for density of water, \(\rho_w = {m_w \over V_w}\), we can solve for the volume.

 

\(\rho_w = {m_w \over V_w} \implies V_w = {m_w \over \rho_w}\)

     

Substituting the known values, we get the volume of the water as,

\(V_w = {5.5\;kg \over 1.0 \times 10^3\;kg/m^3} = 5.5 \times 10^{-3}\;m^3\).

 

  1. Since, the rock is submerged in water, this also implies that the volume of the water is also the volume of the rock. Thus, to get the density of the rock, we use the value of the water’s volume and the mass of the rock.

\(\rho_r = {m_r \over V_r} = {13.0\;kg \over 5.5 \times 10^{-3}\;m^3} = 2.36 \times 10^3\;kg/m^3\).