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PRESSURE, ARCHIMEDES’ PRINCIPLES, UPTHRUST & LAWS OF FLOATATION

PRESSURE

Pressure is defined as the perpendicular force per unit area acting on a surface. It is a scalar quantity & measured in N/m2 or Pascal (pa).It can also be defined as the force per unit area, which is calculated by taking the total force and dividing it by the area over which the force acts. Force and pressure are related but different concepts. A very small pressure, if applied to a large area, can produce a large total force.

P = F     ……………………………..1.    Where P-pressure, F- force (N) & A-area (m2­)    

   A

NB: 1 bar = 105 N/m2 = 105 pa

Example – A force of 40N acts on an area of 5m2. What is the pressure exerted on the surface?

Solution

F = 40N, A = 5m2, P = ?

P =  F/A = 40/5 = 8pa

 

Pressure in Liquid

Pressure in liquid has the following properties

  1. Pressure increases with depth
  2. Pressure depend on density
  3. Pressure at any point in the liquid acts equally in all direction
  4. Pressure at all points at the same level within a liquid is the same
  5. It is independent of cross-sectional area

P = hℓg ……………………..2.

where p-pressure, h-height & g-acceleration due to gravity

Pascal’s principle : Pressure applied to an enclosed fluid is transmitted undiminished to every part of the fluid, as well as to the walls of the container. The operation of the hydraulic press & the car brakes system is based on this principle.

The ideal press consists of two pistons of areas ( a , A ) enclosed between them incompressible liquid as in figure

 

When a small force ( f ) acts on the small piston ( a ) , it exerts a pressure ( p = f/a ).  

To keep the large piston  (A) at equilibrium with the small one (a) a load = F is placed on the large piston .

P = f/a = F/A

EVALUATION

  1. Define pressure
  2. State five characteristics of pressure in liquid

 

Archimedes’ Principle & Upthrust

Archimedes’ principle is a law that explains buoyancy or upthrust. It states that When a body is completely or partially immersed in a fluid it experiences an upthrust, or an apparent loss in weight, which is equal to the weight of fluid displaced. According to a tale, Archimedes discovered this law while taking a bath. An object experiences upthrust due to the fact that the pressure exerted by a fluid on the lower surface of a body being greater than that on the top surface, since pressure increases with depth. Pressure, p is given by p = hρg, where:
h is the height of the fluid column
ρ (rho) is the density of the fluid
g is the acceleration due to gravity
Let us confirm this principle theoretically. On the figure on the left, a solid block is immersed completely in a fluid with density ρ. The difference in the force exerted, d on the top and bottom surfaces with area a is due to the difference in pressure, given by
d = h2aρgh1aρg = (h2h1)aρg
But (h2h1) is the height of the wooden block. So, (h2h1)a is the volume of the solid block, V.
d = Vρg
Upthrust = Vρg
In any situation, the volume of fluid displaced (or the volume of the object submerged) is considered to calculate upthrust, because (h2 h1) is the height of the solid block only when it is completely immersed. Furthermore, the pressure difference of the fluid acts only on the immersed part of an object.
Now, moving back to Vρg. Since V is the volume of fluid displaced, then the product of V, ρ and g is the weight of the fluid displaced. So, we can say that
Upthrust = Weight of the fluid displaced

Compare this conclusion with the statement above summarising Archimedes’ principle. Are they the same? Well, not totally. The “apparent loss in weight” was not mentioned.

See also  ATMOSPHERIC PRESSURE

In the figure on the left, there are arrows on the top and bottom of the solid block. The downward arrow represent the weight of the block pulling it downwards and the upward arrow represent the upthrust pushing it upwards. If one were to measure the weight of the solid block when it is immersed in the fluid, he will find that the weight of the block is less than that in air. There is a so-called “apparent loss in weight”, because the buoyant force has supported some of the block’s weight.

NB: 1. When an object is wholly immersed, it displaces its volume of fluid. So up thrust = weight of fluid displaces. = Volume of fluid displaced x its density x g = volume of object x density of fluid x g

2   When the object is partially immersed e.g. if ¼ of its volume (v) is immersed then the up thrust is given by v/4 x density of liquid x g.

 

Determination of Relative Density by Archimedes’ Principle

  1. Relative density of solid

The body is weighed in air w1, and then when completely immersed in water w2

Relative density of solid

             = Weight of solid in air

                          Weight of equal volume in water

=          w1

W1-W2

  1. Relative density of liquid

A solid is weighed in air (w1), then in water (w2) and finally in the given liquid (w3)

Relative density of liquid  = apparent loss of weight of solid in liquid

apparent loss of weight of solid in water.

See also  INTRODUCTION TO PHYSICS

= W1 – W3

W1 – W2

 

Example – The mass of a stone is 15g when completely immersed in water and 10g when completely immersed in liquid of relative density 2.0 . What is the mass of the stone in air?

Solution:

Relative density = upthrust in liquid

                                       upthrust in water

let W represents the mass of the stone in air

2 = w – 10

w – 15

2(w – 15) = w –10

2w – 30 = w – 10

2 w – w = -10 + 30

w = 20g

 

Law of Floatation

A floating object displaces its own weight of the fluid in which it floats or an object floats when the upthrust exerted upon it by the fluid is equal to the weight of the body. When an object is floating freely (i.e. neither sinking nor moving vertically upwards), then the upthrust must be fully supporting the object’s weight. We can say
Upthrust on body = Weight of floating body. By Archimedes’ principle,
Upthrust on body = Weight of fluid displaced. Therefore, Weight of floating body = Weight of fluid displaced

This result, sometimes called the “principle of floatation”, is a special case of Archimedes’ principle

 

EVALUATION

  1. State the law of floatation.
  2. State Archimedes’ principle.

 

WEEKEND ASSIGNMENT

  1. A force of 40N acts on an area of 10m2. What is the pressure exerted on the surface?  (a) 8pa (b) 4pa (c) 400pa (d) 10pa
  2. What is the height of a cylindrical iron if the density is 7900kglm3 the mass is 700kg and the radius is 0.1m  [a) 2.918cm [b] 2.819m © 3.418m
  3. Density is defined as the ratio of mass to (a) pressure (b) area (c) volume
  4. Relative density is the ratio of mass of a substance to ———— (A) mass of equal volume of water (b) volume of a substance (c) density
  5. Pressure can be measured in the following except (a) bar  (b) N/m2 (c) pascal (d) Nm2

 

THEORY

  1. Differentiate between force & pressure.
  2. What is the pressure due to water at the bottom of a tank which is 20cm deep and is half of water? (Density of water = 103kg/m3 and g = 1om/s2 )

 

See also

DENSITY & RELATIVE DENSITY

VECTOR & SCALAR QUANTIT

FRICTION

MOTION IN NATURE

MEASUREMENT OF MASS

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