The Archimedes’ principle states that; When a body is partially or totally immersed in a fluid, it experiences an upthrust force equal to the weight of the fluid displaced.
The law of flotation
It is a special case of the Archimedes’ Principle which states that: A floating object displaces it’s own weight of the fluid in which it is floating.
Explaining upthrust force from the Archimedes’ principle
Upthrust force, also known as buoyant force, is the upward force exerted by a fluid (liquid or gas) on an object immersed in it. This force acts vertically upward and opposes the weight of the object. If the upthrust is greater than the object’s weight, the object floats; if it is less, the object sinks. Thus, Archimedes’ principle explains why objects behave differently in fluids depending on the amount and weight of fluid they displace.
Objects will weigh less in water than in air. Take a spring balance and hang some mass on it. Determine the weight of the mass and then push the mass up slightly with your hand. What have u observed?
When you place some upward force on a mass hanging on the spring, it’s weight is seemed to reduce as observed by lesser leading of the spring balance.
When you apply a force upward on the object hanging on a spring balance, you are providing some force that is acting opposite to the weight of the object. Weight is always acting downward on a straight line that is directed towards the center of the earth.
When you push the object upward, you are reduce the overall resultant downward force by providing some force acting opposite to the weight.
From the law of addition of forces, when two forces are acting in opposite direction on the same object, then one force is considered positive force and the other one taken as negative force . The total resultant force acting on the object is the algebraic sum of the forces acting on that object Consider the setup below that shows some weight acting on an object hanging freely on air.
We consider the force acting on the object which is it’s weight as W and any force applied upward as U as shown.
The resultant force will be given as W’=W-U. Where W’ represents the reduced weight.
If U is greater than W’, then the object will accelerates upward, otherwise it will accelerates downward with reduced force.
The downward acceleration force is balancing with tensional forces on the spring causing some extension, hence the object remains on the spring balance but causing it to extend in length.
The Archimedes’ principle, Upthrust Force
When an object is immersed in a fluid, the upward forces on the object are provided by pressure in the fluid. That is why objects weighs less in water because some weight of the object is being cancelled out by the upward forces in water. This upward forces produced by fluid on an object is known as the upthrust force. It is the same force that causes object to float in water.
However, it is important to note that, for heavier objects falling in air, the upthrust by air is soo small such that it cannot be notices. We say that upthrust of air on an object is negligible.
showing upthrust with paper and stone
If you release a piece of paper and a stone from some distance above the ground, you will notice that the stone reaches the ground faster than the paper. This is because upthrust force on paper is comparable to that of paper, because a piece of paper has very small weight. However, the stone weight is much more than the upthrust that can be provided by paper hence the total resultant downward forces is larger than that of paper hence causing more acceleration downward.
Later on, we will see that upthrust fall is a characteristic of both volume of the object and density of the fluid.
cause of upthrust
Consider a cylindrical solid of cross-section area A which is totally immersed in a fluid of density ρ as shown.
The pressure due to liquid column is usually given by P=ρgh.
Pressure at the top of the solid will be given by, PT = h1ρg.
Where h1 is the height of the liquid column above the top of the object.
Pressure at the lower end of the object will be given by
Pb=h2ρg where h2 is the height of the liquid above the lover surface of the cylinder .
The pressure at the top of the cylinder will provide downward force exerted by the liquid up on the object.
From the pressure laws, F=pressure P x Area A.
i.e F=PA.
Taking the area of the cylinder at the top, the force from the liquid acting on that surface is Given by F=PT x A=h1ρgA.
Similarly, pressure at the bottom is given as F=PB x A=h2ρgA.
The total resultant upwardward force F is this given as
F=F2-F1
Hence F=h2ρgA-h1ρgA
Factoring out the common factors: F=ρgA (h2-h1)
Let h be the difference between liquid column on top and the one at bottom h2 such that h=h2-h1
Hence F=ρgAh
But Volume is always given by V=Ah
The resultant force F is the upthrust force U and will thus be expressed as.
F=U=Aρpg=pgV
where V is the volume of the liquid displaced.
Mass of the liquid is usually given by density x volume. Hence mass m of liquid displaced will be given by m=Ahρ
Weight is usually given as Weight W=mg
Hence weight of liquid displaced will be W=U=Ahρg which represents the upthrust force we calculated earlier. This confirms the Archimedes’ principle that upthrust force is equal to the weight of the fluid it displaces.
From our mathematical arguments, it should be easy to see that Magnitude of the upthrust force is equal a function of volume of the object and density of the liquid considering. From the Archimedes’ principle, we can solve many problems that involves floating and sinking.
Example problem 1
1. A wooden block of mass 375g and density 750kgm-3 is held under water by tying it to the bottom of the container with a light thread as in the diagram below.
Determine the tension in the thread.
(Density of water e = 1000kgm-3 )
solution:
upthrust = weight of the water displaced
weight displaced = 0.5Kg x 10 = 5.0N
Upthrust exerted by water = 5.0N
Weight of the block = 3.75N
Tension = upthrust – weight
Tension = (5.0N – 3.75N) = 1.25 N
(c) A sphere suspended from a spring balance in air has its weight recorded as 6N when submerged half-way in water, the spring balance reads 4.2 N. Calculate the volume of the sphere.
upthrust = weight in air – weight in water
u = 6.0 – 4.2 = 1.80N
mass of water displaced = 0.18kg
Example problem 2
A metallic object weights 30.0N in air and 26.0 N when immersed in water. Calculate:
(a) volume of the object
(b) density of the object
(Take g=10Nkg-1)
solution
(a)
but upthrust = weight of water displaced by the object
hence, weight of water displaced =4.0 N
We need to find the volume of water that is equal to this weight:
we know that density of water=1000kgm-3
hence volume of the stone will be given as 0.0004m3 or 4.0 x 10-4m3
(b) we finds the density of the stone given it’s weight and having calculated it’s volume
mass of the stone = 30.0N/10Nkg-1 = 3.0Kg
Exam practice Question
(a) i) State the law of flotation. (1 mark)
(ii) Fig. 6 shows a piece of cork held with a light thread attached to the bottom of
a beaker. The beaker is filled with water.
(I) Indicate and label on the diagram the forces acting on the cork. (3 marks)
II) Write an expression showing the relationship between the forces. (1 mark)
……………………………………………………………………………………………………
……………………………………………………………………………………………………
III) If the thread breaks name another force which will act on the cork. (1 mark)
……………………………………………………………………………………………………
……………………………………………………………………………………………………
b) A solid displaces 8.5 cm3 of liquid when floating on a certain liquid and 11.5 cm3 when
fully submerged in the liquid. The density of the solid is 0.8 gcm3
Determine:
i) The upthrust on the solid when floating. (3 marks)
………………………………………………………………………………………………….…
……………………………………………………………………………………………………
……………………………………………………………………………………………………
ii) The density of liquid. (3mrks)
………………………………………………………………………………………………….…
……………………………………………………………………………………………………
……………………………………………………………………………………………………
iii) The upthrust on the solid when fully submerged (3 marks)
………………………………………………………………………………………………….…
……………………………………………………………………………………………………
……………………………………………………………………………………………………
- Select Integrating products of trigonometric functions
Integrating products of trigonometric functions - Grade3 mathematics Exam
Leave a Reply