Article about Archimedes principle

A ship with a huge mass does not sink, while a stone that has a small size can sink. Why is that? The answer is straightforward if you understand the concept of buoyancy and Archimedes’ principle.

In everyday life, we will find that objects that are inserted into a fluid, like a rock, have a smaller weight than when objects are not in the liquid. You may find it difficult to lift a stone from the ground, but the same stone is effortlessly raised from the bottom of the seawater. This is due to the buoyant force. Buoyancy occurs due to differences in fluid pressure at different depths. The fluid pressure increases with depth, the thicker the fluid, the greater the pressure of the fluid. When an object is inserted into the fluid, there will be a difference in pressure between the fluid at the top of the object and fluid at the bottom of the object. Fluid located at the bottom of the object has a pressure higher than the fluid at the top of the object.

In the figure, you can see an object floating in the water. The fluid at the bottom of the object has a pressure higher than the fluid located at the top of the object. This is because the fluid under the object has a depth more significant than the fluid above the object (h_{2} > h_{1}).

The amount of fluid pressure at a depth of h_{2} is:

The amount of fluid pressure at a depth of h_{1} is:

F_{2} = the force applied by the fluid at the bottom of the object, F_{1} = the force applied by the fluid on the top of the object, A = the surface area of the object

The difference between F_{2} and F_{1} is the total force given by the fluid on the object, which we know as the buoyancy force. The amount of buoyancy is:

F _{buoyancy} = F_{2} − F_{1}

F _{buoyancy} = (ρ g h_{2 }A) − (ρ g h_{1} A)

F _{buoyancy} = ρ g A (h_{2} − h_{1})

F _{buoyancy} = ρ F g A h

F_{ buoyancy} = ρ F g V

ρ_{F} = fluid density, g = gravitational acceleration, V = volume of objects in the fluid

So, we can write the equation that states the amount of buoyancy (F buoyancy) above:

F _{buoyancy} = ρ_{F }g V → m = ρ V

F _{buoyancy} = m_{F} g

F _{buoyancy} = w_{F}

m_{F} g = w_{F }= the weight of the fluid which has the same volume as the volume of the dipped object.

Based on the above equation, we can say that the buoyant force is equal to the weight of the fluid being displaced, the volume of fluid displaced is similar to the volume of the object dipped in the fluid.

If the object is inserted into the fluid, floating, where the part of the object that is dipped is only a part,

then the volume of fluid that is displaced = the volume of the part of the object dipped in the fluid. No matter what the object is and how its shape, all will experience the same thing. This is the work result of Archimedes (287-212 BC), known as the Archimedes’ Principle.

The Archimedes’s principle states that:

**When an object is entirely or partially dipped in a liquid, the liquid will give an upward force (buoyant force) to the object, where the amount of upward force (buoyant force) equals the weight of the fluid being displaced.**

The story of Archimedes

Archimedes, who lived between 287-212 BC, was commissioned by King Hieron II to investigate whether the crown made for the King was made of pure gold or not. To find out whether the crown is made of pure gold or the crown contains other metals, Archimedes was at first confused. The problem is, the shape of the crown is irregular and cannot be destroyed first so that it can be determined whether the crown is made of pure gold or not.

The idea of determining whether a crown is made of pure gold or not is to first determine the weight of the crown and then compare it to the specific gravity of gold. If the crown is made of pure gold, then the specific gravity of the crown = the specific gravity of gold.

The specific gravity of an object is the ratio between the weight of the object in the air and the weight of water that has the same volume as the volume of objects. Mathematically written:

How to determine the weight of water that has the same volume as the volume of objects?

According to Archimedes, the weight of water that has the same volume as the volume of an object = the amount of the buoyancy force when the object sinks (the whole part of the object is dipped in water). This is the same as the weight of objects lost when weighed in water. Therefore :

To determine the specific gravity of the crown, the crown is first weighed in the air (the weight of the crown in the air). Then the crown is inserted into the water and then weighed again to obtain the Weight Lost of the Crown. So:

After the specific gravity of the crown is obtained, then it is compared with the specific gravity of gold. Gold specific gravity = 19.3. If the specific gravity of the crown = specific gravity of gold, the crown is made of pure gold. But if the crown is not made of pure gold, then the specific gravity of the crown is not the same as the specific gravity of gold.

**Why does the ship not sink?**

If the density of an object is smaller than the density of water, then the object will float. Conversely, if the density of an object is greater than the density of water, then the object will sink. Most ships are made of iron and steel. Density of iron and steel = 7.8 x 10^{3} kg/m^{3} while the density of water = 1.00 x 10^{3} kg/m^{3}. It appears that the density of iron and steel is greater than the density of water. In this case, the specific gravity of iron and steel = 7.8. The ship should be a sink. Why doesn’t the ship sink? The total density of the ship is smaller than the density of water or seawater.

Example problem 1:

A stone with a mass of 40 kg is at the bottom of a pond. If the volume of stone = 0.2 m3, what is the minimum force needed to lift the stone?

__Known:__

Mass of stone (m) = 40 kg

Volume of stone (V) = 0.02 m^{3}

Density of water = 1000 kg/m^{3}

Acceleration of gravity (g) = 10 m/s^{2}

__Wanted:__ F minimum

__Solution:__

F _{buoyancy} = w_{F}

F _{buoyancy} = m_{F} g → m = pV

F _{buoyancy} = ρ_{F }g V

F _{buoyancy} = (1000 kg/m^{3})(10 m/s^{2})(0.02 m^{3})

F _{buoyancy} = 200 kg m/s^{2}

F _{buoyancy} = 200 N

weight of stone (w) = m g

weight of stone = (40 kg)(10 m/s^{2})

weight of stone = 400 kg m/s^{2}

weight of stone = 400 N

The minimum force needed to lift the stone:

weight of stone – buoyancy force = 400 N – 200 N = 200 N

Example problem 2:

Weight of object in air = 5000 kg m/s^{2 }and weight of object in water = 4000 kg m/s^{2}. If density of object = 2000 kg/m^{3 }, what is the mass and volume of the object? g = 10 m/s^{2}

Solution

Acceleration of gravity (g) = 10 m/s^{2}

Density of object = 2000 kg/m^{3}

Density of water = 1000 kg/m^{3}

Weight of object in air = 5000 kg m/s^{2}

Weight of object in water = 4000 kg m/s^{2}

Buoyancy force (F buoyancy) = weight of object in air – weight of object in water

F buoyancy = 5000 kg m/s^{2} – 4000 kg m/s^{2}

F buoyancy = 1000 kg m/s^{2}

F buoyancy = weight of water displaced

F buoyancy = (mass of water)(g)

F buoyancy = (volume of water displaced)(density of water)(g)

Volume of water that displaced = volume of object in water

Volume of object = 0.1 m^{3}

Mass of object = ?

ρ = m / V

m = ρ V

m = (2000 kg / m^{3})(0.1 m^{3})

m = 200 kg

Mass of object = 200 kg

Example problem 3:

What volume of helium is needed if a balloon has to lift 500 kg of load?

Solution:

Density of helium = 0.1786 kg/m^{3}

Density of air = 1.293 kg/m^{3}

Buoyancy force = weight of air that displaced = weight of object + weight of helium

Buoyancy force = weight of object + weight of helium

Buoyancy force = (mass of load)(g) + (mass of helium)(g)

Buoyancy force = (mass of load + mass of helium)g —- equation 1

Buoyancy force = weight of air that displaced

Buoyancy force = (mass of air that displaced)(g) —- equation 2

We combine the equation 1 and the equation 2:

(mass of load + mass of helium)(g) = (mass of air that displaced)(g)

mass of load + mass of helium = mass of air that displaced

500 kg + (ρ helium)(V helium) = (ρ air)(V air)

500 kg = (ρ air)(V air) – (ρ helium)(V helium)

Volume of air that displaced (V air) = Volume of helium in balloon (V helium)

500 kg = (ρ air – ρ helium)(V)

This is the minimum volume of helium needed to lift weights on the earth’s surface. For the balloon to float higher, the volume of helium needs to be added. The volume of helium has to be increased because the density of air decreases with height.