ARCHIMEDES’ PRINCIPLE
Ch9 (Ch11 Hons.)
Pressure
•
•
•
•
Pressure = Force (N) / Area (m2)
Unit is pascal (Pa) which is VERY small.
Also use atm, or psi (pounds/inch2)
Pressure is useful as it expresses the
strength of materials (Ex. a container can
withstand 150 psi.
• As usual, knowing two variables allows
you to calculate the third. ( P = F/a )
Pressure at depth
• FLUID – a liquid or a gas.
• Pressure is higher:lower at the bottom of a fluid.
P = P0 + ρ g h
Where
P0 is atmospheric pressure
P is pressure at depth,
ρ is density of fluid,
g is acc. Due to gravity
h is the height below the fluid
top surface
Who is Archimedes?
• Archimedes (287-212 BC), pre-eminent
Greek mathematician and inventor, who
wrote important works on plane and solid
geometry, arithmetic, and mechanics.
Reason for buoyant force….
The law
• Archimedes' Principle, states that
when an object is totally or partially
immersed in a fluid, it experiences
an upward buoyant force equal to
the weight of the fluid displaced.
The principle is most frequently
applied to the behaviour of objects
in water, and helps to explain
floating and sinking, and why
objects seem lighter in water. It
also applies to balloons in the air.
UPTHRUST AND
BUOYANT FORCE
The key word in the principle is “upthrust” (or
buoyant force), which refers to the force acting
upward to reduce the apparent weight of the object
when it is in the water or under water.
for example, a metal block with a volume of 100 cm3
is dipped in water, it displaces an equal volume of
water, which has a weight of approximately 1 N. The
block therefore seems to weigh about 1 N less.
SINKING AND FLOATING OBJECTS
The reading of
spring balance is 2.7 N
The reading of
spring balance is 1.7 N
What is the reading of spring balance
if the wood is attached to it ?
ZERO
Density and Buoyancy
From Archimedes’s Principle :
Buoyant Force = Weight of fluid displaced
= mg
(note : F = ma)
= Vg
(note :  = m )
V
Thus FB =  V g
Where ……
FB
= Buoyant Force or Upthrust

= Density of fluid
V
= Volume of fluid displaced or
the volume of the object if it is totally submerged in
the fluid.
Buoyant Force and Flotation
Buoyant force = weight  the object floats and stationary
Buoyant force > weight  the object moves up
Buoyant force < weight  the object moves down
The Law of Flotation
A floating object displaces its own weight of fluid
in which it floats.
THINK!!!!!
warm fresh
water
cold fresh
water
warm sea
water
cold sea
water
1. Why the depth of ship immersed in the water different?
Fresh water less dense than sea water and warm
water less dense than coldwater so warm fresh
water need to be displaced more to keep the
uptrust force equal with weight of the boat so it
still can float.
2. If the modeling clay is formed into a ball, it will sink.
But when it is formed into a hull it will float. Why?
-
BECAUSE…..
APPLICATIONS
Hot air balloon
1. rises upwards
(Upthrust > Weight of hot air (helium gas) +
weight of airship fabric + weight of gondola +
weight of passengers.)( balloon expand)
2..descends
(Upthrust < Weight of hot air (helium gas) +
weight of airship fabric + weight of gondola +
weight of passengers.)(balloon shrinks)
3. stationary
(Upthrust = Weight of hot air (helium gas) +
weight of airship fabric + weight of gondola +
weight of passengers.)( balloon size uncanged)
PLIMSOLL LINE OF THE SHIP
The density of sea water varies with location
and season. To ensure that a ship is loaded
within safe limits , the Plimsoll line marked
on the body of the ship acts as a guide.
If ballast tanks empty Upthrust > weight  submarine rises to surface
If ballast tanks full  Upthrust < weight  submarine sinks to bottom
SUBMARINE
Hydrometer
An hydrometer is an instrument
used to measure the density of a
liquid.
lead shot to make it float upright
In a liquid of lesser density , the hydrometer is more submerged.
The hydrometer floats higher in a liquid of higher density.
Archimedes Principle problems
• Use fact that weight of displaced fluid = Fb
• Remember ρ = m / V
• Fg (object) / Fb = ρo / ρ f
• Where Fg (object) is weight of object (or weight in
air)
• Fb is buoyant force
• ρ is density (of object or fluid)
Example
1. The weight of the rock in air is 0.85N. When it is
completely submerged in water, its weight is 0.45N.
What is the buoyant force acting on the rock when it is
completely submerged in the water ?
Solution :
Buoyant force = Actual weight – Apparent weight
= 0.85 – 0.45
= 0.4N or 0.4 kg m/s2
2. A concrete slab weight 180N. When it is fully
submerged under the sea its apparent weight
is 105N.
Calculate the density of the sea water if the
volume of the sea water displaced by the
concrete slab is 4800 cm3. [ g = 9.8 m/s2 ]
Solution :
Buoyant force = actual weight – apparent weight
= 180 – 105
= 75N
 According to Archimedes’s principle
Buoyant force = weight of sea water
displaced
Therefore,
F = ρVg
so…. ρ = F / Vg
= 75kg m/s2 / (4800 x 10-6 m2 x 9.8m/s2 )
= 1530.61 kg / m-3
Example
 A rectangular ferry boat is 15m long by 6m
wide. A 2500kg truck drives onto the ferry. If
the density of water is 1000kg/m3 . Find out
how far down the hull of the ferry sinks.
 Δ Fb = Δ weight of water displaced.
mt g = Vw ρw g
mt = Vw ρw = (l w h) ρw , we need to solve h
h = mt / l w ρw = 2500kg/ 15m 6m 1000kg/m3
=
Homework
Regular: P343 Concepts Q1-7 Write out in full
sentences. You must read the section on
Archimedes principle and use the ideas within.
Honors: Question 11 – 17 P352 (Concepts)
2nd day homework
 HONORS: P 353 Q2 (8300Lb), 5 (317m2)
(density), 10 (1.3 million lb),12 (32N),13
(4.76 X104),14 (24) (pressure), 19 (1.26x
105Pa), 21 (3.5x 106 N), 23 (1.2X10 5 Pa)
(pressure with depth)
 Regular : P327 Q 2 (2.7X102), 3 (1.2X10 3
Pa, 6.0 X10 -2 N . P330 Q1 (1.11X108 Pa),
2 (8.25X10 5 Pa), 3 (2.7X10 4 Pa) , 4 (1.03
X105 Pa, 1.05X105 Pa). P331 Q1,2,3,4
Pascal’s principle
 Honors P329 , reg. P326
 Any change in pressure in an enclosed,
static fluid is transmitted equally to all parts
of the fluid and the container walls.
 What does static mean?
 Examples of Pascal’s principle include a
hydraulic car jack, a hydraulic ram etc.
–
P1
P2
Ex: car jack has small piston area = 1.0 X10-6 m2
and a large piston of 0.50 X10-4 m2.
– How much can it lift if the small piston is pushed
down with a force of 180N?
– Given: Pascal says P1 = P2 ,So F1 / A1 = F2 / A2
– Rearrange to get F2 on its own F2 = A2 F1 / A1
– So substitute to solve….
– Tonight for homework do P
Important general rule…..
 Fluids flow from areas of ______ pressure
to areas of ____ pressure.
 Pressure in a fluid at any point pushes
equally in all directions
Bernoulli’s Principle
 Daniel Bernoulli (Swiss) in the early 1700’s
observed that air and other fluids do not
behave as one would expect. Specifically:
 “A fluid moving at a high velocity exhibits
____ pressure . A fluid moving at a low
velocity exhibits ____ pressure.”
 This principle explains many phenomena:
 Being pushed into path of speeding vehicle,
a carburetor and many more…..
Shower curtain pushed in
House roof coming off in tornado
Curve balls in any ball sport
Lift from an airfoil (wing)
Where to find Bernoulli’s
Principle in your book..
 Honors: P 343- P344
 Regular : P335