V.5. Law of Archimedes.
π Experiment: The Archimede's Force of Water.
Required materials:
Balloon, thread.
Experiment description:
- Hold a balloon (plastic bag) filled with water on a string.
- What do you notice?
The rope is stretched.
- Immerse it in a bowl of water.
- What do you notice?
The rope is no longer stretched like in the air.
Experiment conclusion:
A vertical, bottom-up force acts on the balloon submerged in water, making the string no longer stretched.
The result of the hydrostatic pressure forces exerted by the liquid on the body immersed in it is called the Archimedean force, denoted FA, with a vertical direction, from the bottom up.
π Experiment: Determination of the Archimede's force using a dynamo-meter.
Required materials:
Water vessel, a hook weight, dynamometer.
Experiment description:
- Suspend a body from the dynamo-meter hook and measure its weight (G = 2 N).
- Immerse the body into a bowl of water, still held by the dynamo-meter hook, and read its instructions. Notice that the dynamometer indicates a lower force, called the apparent weight (Gap = 1,7 N).
Experiment conclusion:
The Archimede's force (FA) is equal to the difference between the weight of the body (determined in air, G) and the "apparent weight" of the body immersed in liquid (Gap), that is:
FA = G β Gap = m β’ g β map β’ g = (m β map) β’ g
FA = 2 N - 1,7 N = 0,3 N
π Experiment: Determination of the Archimede's force with the balance.
Required materials:
Vessel with water, Archimedes' cylinders, scales, marked masses, pipette.
Experiment description:
- Connect the cylinders of Archimedes to each other, with the solid one (metallic) under one plate of the scales and set up the scales in equilibrium by placing marked masses on the other plate.
- Insert only the full cylinder into a bowl of water and notice that the scales is unbalanced.
- Use a pipette to fill the empty cylinder and notice that the scales is balanced.
Experiment conclusion:
The Archimede's force acts upward on the cylinder full of water, which unbalances the scales.
When balancing the scales, the Archimede's force is equal to the weight of the volume of liquid introduced into the empty cylinder, which has the same volume as the full cylinder.
So the Archimede's force is equal to the weight of the volume of fluid displaced by the body.
The statement of the law of Archimedes:
A body immersed in a fluid is pushed by a force acting vertically, from the bottom up, called the Archimedean force (FA), equal to the weight of the volume of fluid displaced by the body.
According to Law of Archimedes, the force with which the fluid presses from the bottom up on the submerged body is equal to the weight of the volume of fluid displaced by the body, that is:
The body volume (Vc) = The volume of displaced fluid (Vf.dis. )
From the formula of the Archimedean force we observe that it depends directly proportionally, only by two factors:
1) The higher the density of the fluid is, the higher the FA is.
2) The higher the body volume (the volume of fluid displaced by the body) is, the higher the FA is.
π Experiment: The dependence of the Archimede's force on the density of the liquid.
π₯ Caution! This experiment is performed only by teachers!
π₯ Caution! Mercury is extremely toxic! Do not touch it and do not inhale its vapors!
Required materials:
Water vessel, mercury vessel, screw (nail)
Experiment description:
- Put a screw in a bowl of water.
- What do you notice?
The screw is immersed in water.
- Put the same screw in a mercury vessel.
- What do you notice?
The screw floats in mercury.
Experiment conclusion:
The screw floats in mercury because the mercury has a much higher density than water and the Archimedean force on the mercury side is higher than on the water side.
π Homework
1. What happens to a ship when it crosses the Danube into the Black Sea?
π Experiment: The dependence of the Archimede's force on the volume of the body.
Required materials:
Aluminum wire, water vessel, aluminum box having the same mass as the aluminum wire.
Experiment description
- Take the aluminum wire that has the same mass (hence the weight) as an aluminum box so that the box has a larger volume than the wire.
- Immerse them one by one in a bowl of water.
- What do you notice?
The wire sinks and the box floats.
Experiment conclusion:
The larger the volume of a body, the greater the volume of fluid it displaces and the greater the archimedical force.
Particular cases of the law of Archimedes:
a) When the density of the body (Οc) is less than the density of the liquid (Οl), the body floats.
A resultant force appears, which acts on the body vertically, upwards, called the ascending force (Fa) which causes a partial come out of the body from the liquid.
The submerged portion displaces a volume of liquid equal to the weight of the body.
Applications of the particular cases of the law of Archimedes
Driftwood of the ships, logs, icebergs.
Particular cases of the law of Archimedes:
b) When the density of the body (Οc) is equal to the density of the liquid (Οl), the body is in equilibrium inside the liquid.
π Applications of the particular cases of the law of Archimedes
Driftwood of balloons or submarine in water.
Particular cases of the law of Archimedes:
c) When the density of the body (Οc) is higher than the density of the liquid (Οl), the body sinks (goes to the bottom of the vessel).
A resultant force appears, acting on the body vertically, downwards, called the apparent weight:
Remember
The Archimede's force, FA:
- It has a vertical direction, upwards and the point of application in the weight Center of the volume of the displaced liquid.
- Equal in module to the weight of the displaced liquid.
- It does not depend on the weight and shape of the body, the depth of immersion, the height of the liquid in the vessel.
Applications of the law of Archimedes to Fluids
1) Densimeter is a measuring instrument for liquid density. The scale is inversely graded vertically upwards so that the gradations decrease as the densimeter sinks deeper into liquids with lower densities. It enters the liquid until the weight of the displaced liquid becomes equal to its own weight. In order to maintain an upright position when immersed in liquid, its weight is concentrated towards the bottom of the densimeter, where some lead or steel balls are found. The density of the liquid is read at the level of the free surface of the liquid, on a graduated scale.
Uses of the densimeter:
At car service stations, when checking:
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2) Icebergs: blocks of ice from polar ice that have a lower density than seawater, therefore are floating on water.
3) Boats, ships, rafts are built in such a way that they can displace a volume of water as much as possible, so that the archimedean force of the water to be greater than their weight. The waterline indicates how far the vessels can sink when loaded.
4) The submarine is a vessel that floats on the surface of the water, but can also move underwater. The walls of the submarine are double. They comprise rooms (compartments) that can be filled with water. As water enters these compartments, the weight of the submarine increases and the vessel sinks. It rises to the surface by draining water from these rooms.
5) Bathyscaphe is a smaller submarine consisting of a spherical cabin made of titanium alloy steel (for crew) and a central body containing the engine and fuel tanks. It can sink to much greater depths than the submarine, even to the Pacific Ocean Marianas Pit with a depth of 11.022 m.
James Cameron's diving in 2012 with the Deepsea Challenger submarine (which can go down 150 meters / minute and has a torpedo shape, 7 meters long) lasted a total of seven hours, of which three hours were at 10.900 meters depth.
Cameron worked on the project for seven years and became the third man to reach the deepest point on Earth's oceans. The first two arrived 52 years before him: the Swiss Jacques Piccard and the American Don Walsh.
6) The balloon is built based on the law of Archimedes for gases. It consists of a rubber balloon filled with a light gas (H2 or He or hot air) and a cabin (nacelle) which is caught by the balloon with ropes and where the measuring instruments are located. In the 17th century, the Montgolfier brothers built the first hot air balloon filled with hot air.
π Solved problems
1. A cube made of cork, with a side of 0,3 dm and a density of 200 kg/m3 is immersed in water, which has a density of 1000 kg/m3.
Is required:
a) The value of the Archimedean force.
b) The value of the resulting force acting on the body in water. What is this force called?
c) What is the height of the portion of the cube that is under water?
Solution:
We write down the problem data and turn it into SI:
l = 0,3 dm = 0,03 m
Οcork = 200 kg/m3
Οwater = 1000 kg/m3
FA = ?
R = ?
h = ?
a) We calculate the volume of the cube: Vcube = l3 = (0,03 m)3 = 0,000027 m3
We apply the formula of the Archimedean force in the Law of Archimedes:
FA = Οwater β Vcube β g = 1000 kg/m3 β 0,000027 m3 β 10 N/kg = 0,27 N
b) We calculate the body weight:
G = m β g = Οcork β Vcube β g = 200 kg/m3 β 0,000027 m3 β 10 N/kg = 0,054 N
Because | FA | > | G | a resultant force appears, which acts on the body vertically, upwards, called the ascending force (Fa) which causes the partial body to come out of the liquid.
| Fa | = | FA - G | = 0,27 β 0,054 = 0,216 N
c)The submerged portion displaces a volume of liquid equal to the weight of the body:
| FA | = | G |
2. A body weighs 800 g in the air, and when immersed in glycerin it weighs 600 g. Knowing the density of glycerin of 1260 kg/m3, find out:
a) Body volume.
b) The resulting force. What does the body do immersed in this liquid?
Solution:
We write down the problem data and turn it into SI:
m = 800 g = 0,8 kg
mapparent = 600 g = 0,6 kg
Οglycerin = 1260 kg/m3
a) V body = ?
b) R = ?
We calculate body weight and its apparent weight:
G = m β g = 0,8 kg β 10 N/kg = 8 N
Gap = map β g = 0,6 kg β 10 N/kg = 6 N
We calculate the Archimedean force with its two formulas:
FA = G β Gap = 8 N β 6 N = 2 N
FA = Οglycerin β Vbody β g = 1260 β Vbody β 10
2 = 1260 β Vbody β 10
Vbody = 2/12600 = 0,0001 m3
b)Because | FA | < | G | a resultant force appears, which acts on the body vertically, downwards, called the apparent weight (Gap) which causes the body to sink into the liquid.
| Gap | = | G β FA | = 8 N β 2 N = 6 N
3. A steel sphere of 500 cm3 and a density of 7800 kg/m3 is suspended by a dynamometer in the air and then immersed in water (with a density of 1000 kg/m3). What does the dynamometer indicate when the sphere is in the air or in the water?
Solution:
We write down the problem data and turn it into SI:
Vc = 500 cm3 = 0,0005 m3
Οc = 7800 kg/m3
Οwater = 1000 kg/m3
G, Gap = ?
We calculate the mass of the sphere:
m = Οc β Vc = 7800 kg/m3 β 0,0005 m3 = 3,9 kg
We calculate the weight of the sphere:
G = m β g = 3,9 kg β 10 N/kg = 39 N
We calculate the Archimedean force:
FA = Οwater β Vc β g = 1000 kg/m3 β 0,0005 m3 β 10 N/kg = 5 N
We calculate the apparent weight (when the sphere is submerged in water):
| FA | = | G β Gap |
| Gap | = | G β FA | = 39 N β 5 N = 34 N
4. An iceberg floating in ocean water (with a density of Ο = 1010 kg/m3), has the upper part, the one above water level, with a volume V0 = 600 m3. Knowing the density of ice Ο0 = 920 kg/m3, find the total volume, V, of the iceberg.
Solution:
We write down the problem data:
V0 = 600 m3
Ο = 1010 kg/m3
Ο0 = 920 kg/m3
V = ?
a) We calculate the volume of the iceberg submerged in water:
Vsubmerged = V - V0
The submerged portion displaces a volume of liquid equal to the weight of the body:
| FA | = | G |
Ο β Vsubmerged β g = Ο0 β V β g
Ο β Vsubmerged = Ο0 β V
Ο β (V-V0) = Ο0 β V
Ο β V - Ο β V0 = Ο0 β V
V β (Ο - Ο0) = Ο β V0
