Archimedes’ Law: Definition, History, Examples of Problems and Their Application

Understanding Archimedes’ Law – Have you ever seen the Titanic movie? The film is enough to get the world’s attention because it tells the true story of the sinking of the Titanic. Titanic is known as a luxury ship. His voyage at that time was highly anticipated. The sinking of the ship occurred after the ship hit an iceberg and a leak occurred.

The ship, which was originally sailing calmly in the ocean, then collided so that this was a little friction, causing Titanic’s hull plates to bend inward in a number of places on the starboard side and tore five of the sixteen watertight compartments. Over the next two and a half hours, the boat slowly filled with water and sank.

We will witness the appearance of passengers and floating objects as well as passengers and objects that sink in the film. Why do some things sink? Why are some objects floating? And why are there objects floating?

Well, this film is actually full of theories about Archimedes’ Law! This theory was initiated by a scientist named Archimedes from Greece. There is a unique story behind the discovery of Archimedes’ law, want to know?

History of Archimedes’ Law

So you see, Archimedes was a prominent scientist who came from Syracuse, Greece. This scientist probably lived in the era of 287 – 212 BC. As a major scientist at that time, Archimedes could master several fields at once, from Mathematics, Physics, Philosophy, Astronomy, and Engineering.

Archimedes lived in an era before paper and pencils, but he was able to formulate various laws regarding buoyancy, principles regarding levers, pulley systems, and much more which is discussed in the book STD 16: ARCHIMEDES.

Grandpa is really great, even though he was an ancient person who had limited information, he didn’t even have a cell phone. It’s all of us who are ready to study math blankly, do confused physics, peek at chemistry and faint suddenly.

Archimedes became known to people after he received an order from King Hiero II to prove the purity of the king’s crown gold. Either for fun or indeed the king wanted to know the greatness of Archimedes.

Being challenged with such a task, Archimedes was confused. But he thought hard to find a solution to this problem. When his confusion peaked because he had yet to find a way to prove the purity of gold, Archimedes chose instead to throw himself into the bath.

Yes, Archimedes was also a human, maybe he felt his brain was starting to heat up and had to cool it down by taking a bath.

As he stepped into the bathtub full of water, he noticed that some water had spilled onto the floor. Curious he tried again and again.

While the mind was wandering thinking about the king’s crown, an idea came down. And he shouted
“EUREKA!”

Unexpectedly from the incident he took a bath he got a theory which in the end was able to prove the purity of the king’s gold.
That’s why you have to bathe diligently, gaess, haha

Well, how did Archimedes prove⁰?. Armed with his hypothesis after bathing Archimedes then put the crown into the water. The change in water volume due to the entry of the crown is then used as the basis for determining the density. Then he divided the mass of the crown by the change in volume of water.

Well, the density that he got from the results of the experiment was then equated with the density of pure gold. So he put the crown and pure gold in equal volumes of water.
Uh, the water that was displaced by the gold and the crown was not what should have been the same, but the reality was different. Well, it was concluded that the gold that was in the king’s crown was not pure gold, but a mixture.

Now we discuss more in detail the law of Archimedes:

Understanding / Sounding of Archimedes’ Law

Archimedes’ law states that an object that is partially or completely immersed in a liquid will experience an upward force equal to the weight of the liquid it displaces.

So this Law explains the relationship between gravity and upward force on an object when it is put in water. As a result of the upward lift (buoyancy), of course, objects that are in the liquid will experience a reduction in weight. As a result, objects lifted in water will feel lighter than when lifted on land.

Well, Archimedes’ law implies 3 conditions

1.1. Sink

This situation occurs when the density of the liquid is less than the density of the object. For example iron or steel will sink if put in water because the density of iron is greater than the density of water.
When submerged, the weight of the object in the liquid is greater than the upward force exerted by the liquid.

See also  difference between soil and land

Pressure force of water < weight of object

1.2. Flying

This situation occurs when the density of the liquid is the same as the density of the object. The floating object is between the bottom of the vessel and the surface of the liquid. For example, if an egg is placed in water and then a little salt is added, it will float because the density of the two is the same.

Pressure force of water = weight of object

The same thing happens to fish? fish can equalize their weight with the weight of the water they displace so they can float. The trick is whether the fish will fill its swimming coffers with air.

1.3. Floating

This situation occurs when the density of the liquid is greater than the density of the object. For example, styrofoam or plastic will float if placed in water. An example of applying Archimedes’ law of floating bodies \

The force of water pressure > the weight of the object.

Archimedes’ Law Formula

So Archimedes’ Law will calculate the upward compressive force (Fa) affected by 3 things namely

2.1. density of fluid ( ρ)

The greater the density of the fluid, the greater the upward force generated, conversely,
the smaller the type of fluid, the smaller the resulting upward force

2.2. band volume (in)

The greater the volume of the object, the greater the upward force generated, conversely
the smaller the volume of the object lifted, the smaller the resulting upward force

2.3. object gravity (g)

The greater the object’s gravity, the greater the upward force generated, conversely,
the smaller the object’s gravity is lifted, the smaller the resulting upward force.

Formulas from Archimedes’ law

F = ρ . g . V

Where Fa = upward pressure in units of Newtons (N)
ρ = density in units of Kg/L
g = gravity in units of N/Kg
V= unit volume in m³

For floating conditions, the application of Archimedes’ law is as follows:
suppose there is wood. So there are 2 forces acting on the wood, namely the lifting force of water and the weight of the wood.

The magnitude of the lifting force is equal to the weight of the water pushed by the part of the wood below the water’s surface.

Fa = ρair . g. Vair

Where
Vair = volume of water pushed by wood (m³)
ρair = density of water in unit Kg/L

The weight of the wood is

w = ρkayu g Vtotal

w = weight of wood
V total = total volume of wood
ρwood = density of wood
Because wood is at rest on the surface of the water, there is a relationship

Lifting force = weight of wood
Fa = w
ρair.g. Vair= ρwood . g. Vtotal

Because g or the value of gravity on the right and left sides is the same, an equation will be obtained

ρair.Vair= ρkayu . Vtotal

Find various other physics formulas in the Physics Dictionary which summarizes various types of complicated and quite confusing physics formulas related to objects in nature, phenomena, and natural occurrences.

Example of Archimedes’ Law Problem

3.1. Example Question 1

A block is completely immersed in oil. If the volume of the block is 8 m³, determine the weight of the block in kerosene which has a density of 800 kg/m³ , what is the lifting force experienced by the block
?

Is known :

Volume of block : V = 8 m³
gravity : 10 N/ Kg
Density of oil ρ = 800 kg /m³
Wanted: lift force Fa?
Answer:
Fa = ρ . g. V
= 800 kg/m³ . 10 N/kg . 8 m³
= 64,000 N

3.2. Example Problem 2

A stone is completely immersed in oil. If the volume of the block is 10 cm³, determine the weight of the block in water which has a density of 8
1000 kg/m³ , what is the lifting force experienced by the block (gravity: 10 N/kg)

Solution

Is known :

Volume of block : V = 10 cm³
= 10______
1,000,000
= 1________
100,000
= 10-6 m³
gravity : g = 10 N/ kg
Density of water : ρ = 1000 kg /m³
Wanted: lift force Fa?
Answer:
Fa = ρ . g. V
= 1000 kg/m³ . 10 N/kg . 10-6 m³
= 103 . Kg/ m³ .10 N/kg. 10-6 m³
= 10-2 N
= 0.01 N

3.3. Example Problem 3

It is known that a block with a height of 30 cm has a density of 750 kg/m³ floating on a liquid with a density of 1,200 kg/m³ the height of the block that appears to the surface.

Solution :

Block height = 30 cm³
= 30 m³
1,000,000
= 3
100,000

= 3. 10⁵m³
Density of block : ρblock= 750 kg /m
Density of liquid : ρliquid= 1.200kg /m³
Overall volume of block = Vt
Volume of block submerged : V1
Wanted: height of block that appears on the surface

ρzatcairV1=  ρbenda.  Vtotal

1,200 V1 = 750 Vt
V1 = 750 Vt
1,200
V1 = 75 Vt
120 we divide by 15
V 1 = 5 Vt
8
If the whole beam is considered 1 part
Then the height of the beam that appears on the surface we can find
V2 = Vt – V1
= 1 – ⅝ V
= ⅜
Because total V = 30 cm
Then V2 = ⅜ .30 cm
= 11.25 cm
So V2 or the height that appears on the surface is 11.25 cm

See also  difference between graph and tree

3.4. Example Problem 4

A ball is completely submerged in water. If the density of water is 1,000 kg/m³ , and the lifting force experienced by the ball is 0.2 N,
what is the volume of the ball?
Resolution:

It is known
that lifting force : Fa = 0.2 N
gravity : 10 N/ Kg
Density of water : ρ = 1,000 kg /m³
What : volume : V = …?
Answer:
Fa = ρ . g. V
0.2 N = 1000 kg/m³ . 10 N/Kg . V
0.2 N = 10,000. V N. m³
0.2 N = 10⁴ . V. N/ m³
V = 0.2 N
10⁴ N/ m³
= 2.10³ m³
= 2,000 m³

3.5 Example Problem 5

The weight of a piece of metal in air is 0.63 N. When weighed in tarebinth oil, the metal loses weight by 0.0522 N. How much does it weigh when weighing gasoline? It is known that the density of gasoline is 700 kg/m³ and the density of tarebin oil is 870 kg/m³

For example:
Mass of turbantine : Fa1 = 0.0522 N
Mass of turban : ρ1= 870 kg /m3
Mass of gasoline : ρ1= 700 kg /m3
Fa1= ρ1 g V
0.0522 N = 870kg/m3 10 N/ Kg V
0.0522 N = 8700 N/ m3 V
V = 0.0522 N
8700 N
= 522.10-4 N
87.102 N/m3
= 6.10-6 m3

We find the lift force of gasoline
Fa2= ρ2. g. V
= 700kg/m³ . 10 N/Kg . 6.10-6 m³
= 7.10² kg/m³ .10 N/ Kg . 6.10-6 m³
= 42.10-3 N
= 0.042 N
So, the weight of metal in gasoline is the difference in the lift force on tarebinth oil and gasoline
W metal = 0.63 N – 0.042 N
= 0.588N

 3.6. Example Problem 6

A piece of ice having a density of 900 kg/m³ floats in sea water which has a density of 1030 kg/m³. If the volume of the protruding part of the ice in sea water is 0.65 m³. What is the total volume of ice?

Solution
Known:
Density of ice: ρes = 900 kg/ m³
Density of water: ρair = 1,030 kg/ m³
Prominent volume: V1 = 0.65 m³
For example, we denote the total volume of ice as Vtotal

ρair.Vair=  ρes .  Vtotal

1.030 kg/ m³ ( Vtotal-V1) m³= 900. Vtotal
1.030 kg/ m³ ( Vtotal – 0,65 m³) = 900 kg/ m³ .Vtotal
1.030 kg/ m³ Vtotal – (1.030.0,65 kg/m³ .m³) = 900 kg/ m³. Vtotal
1.030 kg/ m³ Vtotal – (6.695 kg / m³.m³) = 900 kg / m³ .Vtotal
(1.030  Vtotal – 900  Vtotal) kg/ m³= 669,5  kg /m³.m³
130 kg / m³. Vtotal = 669,5 kg / m³.m³
Vtotal = 669,5 kg/m³.m³
130 kg/ m³
= 5,15 m³

3.7. Example Problem 7

An object with a density of 1,200 kg/m³ floats in 1/5 of a liquid, the density of the object is…
Solution: If the density of the liquid is
known : ρ1= 1200 kg/ m³ Floating volume = V1 Total volume = V1 V1= ⅕ Vt Submerged volume = V2 V2= Vtotal – V1 = 1- ⅕ = ⅘ Vt Wanted: object density: ρobject

Answer:
ρ1.Vt = ρbend. V2

1200 kg/ m³. Vt = ρbenda.  ⅘ Vt
ρbenda=  1200 kg/ m³. Vt
⅘ Vt
= 960 kg / m³

Various other questions and discussions that can help Sinaumed’s to better understand Archimedes’ law can also be found in the Physics Smart Pocket Book: Collection of Formulas and Applications for SMA/MA Class 10,11,12 below.

Example of Application of Archimedes’ Law

Archimedes’ law is widely used in everyday life. Some examples of objects that use this law are

1. Submarine

Uniquely the submarine he can float, he can float and he is fine diving. The submarine will adjust its density in the water so that its position can dive, float or float on the surface of the water. The trick is to remove or add water to reduce or add mass

2. Ships

Isn’t it strange? Ships are usually made of steel or iron. How come they can stay afloat on the sea?

Why is that? Well, apparently this is because the lift force on the ship is proportional to the weight of the ship. How to? You see, ships have a hollow shape so that the volume of water displaced is greater and the upward lift is greater as well.

Well, the Titanic could have sunk, possibly because the function of the cavity to move water had been damaged so that the weight of the ship became greater than water.

3. Hot air balloon

It turns out that the Application of Archimedes’ Law also applies to gaseous objects, namely hot air balloons. Gas has almost the same properties as water.

The concept of a hot air balloon is that, in order to float in the air, a hot air balloon is filled with a gas that has a density smaller than the density of air in the atmosphere.

We have finished discussing Archimedes’ Law. Hopefully it can be understood and applied in everyday life.