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Archimedes Principle A body immersed wholly or partially in fluid (Liquid or gas) experiences an up thrust force (buoyant force) equal to the weight of the fluid displaced Proof : Suppose a liquid cylinder of volume (VL) in a resting liquid of density (L). The cylinder is subjected by 2 forces ……… i. The horizontal forces in all directions cancel each other. ii. The vertical forces: a) Its weight WL = mg = LgVL ………… (1 b) Up thrust force on the cylinder which results from difference in pressure on both bases of the cylinder P (upper surface) = Pa + gh1 P (lower surface) = Pa + gh2 rP = gh Fb = rP X A = gh X A Fb = LgVimmersed ………………………………………… (2)

Archimedes Principle

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Page 1: Archimedes Principle

Archimedes Principle

A body immersed wholly or partially in fluid (Liquid or gas) experiences an up thrust force (buoyant force) equal to the weight of the fluid displaced  

Proof:

Suppose a liquid cylinder of volume (VL) in a resting liquid of density (L).

The cylinder is subjected by 2 forces ………

 i. The horizontal forces in all directions cancel each other.

 ii. The vertical forces:

a) Its weight     WL = mg = LgVL   ………… (1 

b)      Up thrust force on the cylinder which results from difference in pressure on both bases of the cylinder

P (upper surface) = Pa + gh1

 P (lower surface) = Pa + gh2

rP = gh

Fb = rP X A =  gh X A

Fb = LgVimmersed ………………………………………… (2)

From 1 & 2

                       Fb = the weight of liquid displaced.

Buoyancy

Buoyancy arises from the fact that fluid pressure increases with depth and from the fact that the increased pressure is exerted in all directions

Page 2: Archimedes Principle

so that there is an unbalanced upward force on the bottom of a submerged object.

Since the "water ball" at left is exactly supported by the difference in pressure and the solid object at right experiences exactly the same pressure environment, it follows that the buoyant force on the solid object is equal to the weight of the water displaced.

Equal Volumes Feel Equal Buoyant Forces

Suppose you had equal sized balls of cork, aluminum and lead, with respective specific gravities of 0.2, 2.7, and 11.3. If the volume of each is 10 cubic centimeters then their masses are 2, 27, and 113 gm.

Each would displace 10 grams of water, yielding apparent masses of -8 (the cork would accelerate upward), 17 and 103 grams respectively.

The behavior of the three balls would certainly be different upon release from rest in the water. The cork would bob up, the aluminum would sink, and the lead would sink more rapidly. But the buoyant force on each is the same because of identical pressure environments and equal water displacement. The difference in behavior comes from the comparison of that buoyant force with the weight of the object.

Page 3: Archimedes Principle

Floating object: The floating object is in equilibrium (acceleration a = 0), so the total force acting must be zero. Gravity still acts on the object, so there must be an equal force upwards, exerted on the floating object by the water.

Submerged object: When an object is underwater, we know that it doesn't feel as heavy. A force exerted on it by the water reduces its apparent weight. If an object is completely submerged, the displaced volume of water is equal to the volume of the object. The buoyancy force will reduce the apparent weight of the object.

Floating objectFB = fluid Vdisplaced g

Submerged object

FB = fluid Vobject g

“The magnitude of the buoyancy force is equal to the weight of the displaced fluid”

Density

RELATIVE DENSITY OF A SOLID:

Relative density of an object = mass of object...……………………

Mass of same volume of water

The concept of relative density comes by comparing the density of an object to the density of water.

Page 4: Archimedes Principle

It is calculated by weighing the object in air and then placing it in water. The amount of water that has been displaced when the object was placed in it is then weighed.

This comparison when calculated, gives the relative density of the object.

RELATIVE DENSITY OF A LIQUID:

The relative density of liquid measures:

Mass of liquid

mass of equal volume of water

Also,

Apparent loss in weight of object in liquid

Apparent loss in weight of object in water

An object immersed in meth [methanol] will experience less up thrust on it than if it was immersed in water. It is due to the fact that meths is less dense than water so the meth’s displaced by the object weighs less than the same volumes of water yet the volume displaced by it is greater than the volume displaced by water. This is again, due to the reason that meths is less dense than water therefore less heavier, so more volume of it will make for e.g., a 2N weight than the volume of 2N of water.

Hence, to displace this larger volume of meths the object floats lower in meth than in water.

By comparing the density of an object with that of the fluid it is to be immersed in, it can be found whether it will float or sink in that fluid.

Page 5: Archimedes Principle

The object will only float if its density is same or less than that of the fluid.

For e.g.:

Wood, petrol and ice will float in water Hot water will float up in cold water Hydrogen gas will float upwards in air Hot air will float upwards in cold air Steel will float in liquid mercury but sink in water

CALCULATIONS ON ARCHIMEDES' PRINCIPLE AND FLOTATION:

Archimedes' principle applies to all objects immersed in liquid regardless of whether they are floating or not

The Law of flotation, however, only applies to floating objects.

Both the Law and the Principle are concerned with weight of objects and fluids. However, when solving problems, one is often dealing with volumes. The connection between the weight of a substance and its volume is:

Mass = volume x density

Weight = mass x g

Therefore,

|Weight = volume x density x g|

THE LAW OF FLOTATION:

A floating object displaces its own weight of the fluid in which it floats.

Pressure

Consider a container full of water:

Page 6: Archimedes Principle

The volume of water is V = A×h, so the mass of water is m = V = A h, and the weight of the water is mg = g A h downwards. The weight of the water is the force exerted by the water on the bottom of the container. This force is spread out over the whole surface area of the bottom.

We call the force divided by the area the pressure:

p =

F

A

The SI unit of pressure is the Pascal (Pa), which is a Newton per square meter (N/m2).