# Buoyancy Worksheet

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"Buoyancy Worksheet"

```					  Archimedes’ Principle
Boyle’s Law
Bernoulli’s Principle
Newton’s Laws of Motion
 The ball and the boat weigh the same; WHY DOES
THE BALL SINK, BUT THE BOAT FLOAT?
Oi… math…
 What determines whether it will sink or float?
 MASS
 The amount of matter contained in an object
 VOLUME
 The amount of 3-dimensional space an object occupies
 ****DENSITY****
 The ratio of mass-to-volume of an object
 Mass divided by Volume
 M/V = D
EUREKA!!!!!!!
 To a king, a crown
 Is the crown real or fake, though?
 Density of Gold: appx. 1206 lbs/ft3
 We can weigh the crown
 (place it on a scale!)
 How to find the volume of the crown?
Standard volumes
 Cube: l3
 Rectangular Prism: l*w*h
 Cylinder (Circular Prism): pi*r2*l
 Triangular Prism: ½ b*h*l
 Sphere: 4/3*pi*r3
Archimedes’ Principle
 Any floating object displaces its own weight of fluid.
 Any object, wholly or partially immersed in a fluid, is
buoyed up by a force equal to the weight of the fluid
displaced by the object.
   Buoyancy = weight of displaced fluid.
   Buoyancy is an upward force by a fluid acting on an
object immersed in that fluid.
   Archimedes' principle does not consider the surface
tension (capillarity) acting on the body.
   The weight of the displaced fluid is directly proportional to
the volume of the displaced fluid (if the surrounding fluid
is of uniform density).
What the heck?
   Hydro = water
   Phobos = fear
   Philius = love
   Hydrophilic = water-loving
   Hydrophobic = water-fearing

 The fluid doesn’t necessarily “push” up on the object, but rather
is pulled down by gravity, more than is the object, and, since
“two objects cannot occupy the same space at the same time,” the
object is “lifted” by the surrounding fluid. Ah… gravity.

 <<ponos = work>>
Standard Densities
 Approximate average density of a human body:
64.7283 lbs/ft3
 Density of Pure Water at Sea-Level at 4 degrees
Celcius: 62.4264 lbs/ft3
 Approximate density of air at sea level: 0.09485 lbs/ft3
Or… we could use a table!
 http://www.coolmagnetman.com/magconda.htm
 http://www.coolmagnetman.com/magconda.htm
 Suppose a 7-lb weight is lowered into water, displacing
3 lbs worth of that water.
 Then, the weight would “SEEM” to weigh 4 lbs.
Sink or Swim?
 To calculate net-buoyancy, you need to know a number of things:
 Mass and volume of the Object
 Mass and volume of the DISPLACED FLUID
 The difference between the two
 Mass of Fluid – Mass of the Object = Net-Buoyancy
 Net-Buoyancy means “resulting buoyancy”
 M F – MO = B
 B > 0 means the object will rise
 B = 0 means the object will float
 B < 0 means the object will sink
What else?
 Archimedes’ Principle also explains why balloons float
in the air.
 Here, if the AIR displaced equals (or exceeds) the
weight of the balloon, the balloon will float.
 THIS IS NOT THE REASON AIRPLANES FLY!! We’ll
talk about that later, but it has a tiny bit to do with
Bernoulli’s Principle and a LOT to do with
“conservation of momentum.”
Conversions
(yes, I’ll give these again!)
 1 ft3 = 12*12*12 in3 = 1 728 in3
 1 in3 = 1/1 728 ft3 = 0.000 578 7 ft3

 1 lb = 0.4545 kg
 1 kg = 2.2 lbs

 1 in = 2.54 cm
 1 cm = 0.3937 in

 1 in = 0.0254 m
 1 m = 39.37 in
Now for some examples:
 What is the weight of a cube of aluminum, with edge-
length of 1ft? (The density of aluminum is about 169
lb/ft3.)
 What is the weight of water for the same volume?
 What is the NET-BUOYANCY?
 WILL THE CUBE SINK, FLOAT, OR RISE IN WATER?
Now for some examples:
 What is the weight of a ball of air, with radius 1ft? (The
density of air is about 0.095 lb/ft3.)
 What is the weight of water for the same volume?
 What is the NET-BUOYANCY?
 WILL THE BALL SINK, FLOAT, OR RISE IN WATER?
Now for some examples:
 What is the weight of a rectangular bar of GOLD, with
a length of 1ft, width of 0.5 ft, and height of 0.25 ft?
(The density of gold is about 1206 lb/ft3.)
 What is the weight of water for the same volume?
 What is the NET-BUOYANCY?
 WILL THE CUBE SINK, FLOAT, OR RISE IN WATER?
I ain’t sayin’ she a gold digga’…
 Btw- a “standard” gold bar is approximately 6”*3”*2”, so has
25.17525 lbs. This is approximately 402.804 oz., the unit by
which the value of Gold is measured. (Of course, there are
different kinds of ounces, Troy and Avoirdupois, with vary
with respect to the number of grams to which each is
equivalent, but never mind all that.)
 The morning of 8 April 2010, the price of gold was \$1146.85
per Troy oz.
 That means the above gold bar would be worth about
\$461 955.77!!
 Suppose you took that gold (from the last example)
and pounded it into a hemisphere (that’s half of a
hollow ball, by the way) with a radius of… say… 3 ft.
 Let pi equal 22/7
 What is the weight of the gold?
 What is the weight of the water DISPLACED BY THE
GOLD HEMISPHERE??
 What is the net buoyancy?
 Will the gold semi-sphere sink, float, or rise?
Well… what really happens?
 If an object is “lighter” than air; i.e., the density is less
than that of the surrounding air, it will rise in the air.
If it is “heavier” than water; i.e., the density is greater
than that of water, it will sink in the water.
 BUT, when the density of the object is between that of
air and that of water, it will be partially submerged and
partially exposed.
 We can calculate HOW MUCH OF THE OBJECT is
submerged!!
Here’s where it gets tricky…
 The hemisphere in the previous example weighed
150.75 lbs (the same as the gold brick), but had a
volume of 56.5487 ft3. That same volume of water
would weigh 3530.1318 lbs.
 That means BN = 3379.3818 lbs!
 MOST of the hemisphere would be out of the water.
The only part of the hemisphere IN the water will be
the 150.75 lbs-worth of water.
 That volume (150.75 / 62.4264) = 2.4148 ft3
 In other words, a little bit of the hemisphere would be
below the water, the rest would be above.
Let’s try a simpler example
 Imagine a cube, with volume of 1 ft3, of a certain density,
say 62.4264 lb/ft3, for example. If placed in water, what
would happen?
   Float JUST below the surface
   Suppose we have a cube of the same volume, but with a
density of 64.7283 lb/ft3. Now what happens?
   Sinks.
   What if the density is 60?
   Mf = 62.4264, Mo = 60, BN = 2.4264
   What is the volume of 60lbs of water?
   0.9611 ft3, so that’s how much of the box would be under
water.
Gold Balloon!
 What would need to be the volume of a balloon made
from that 0.125 ft3 of gold, if it were to float in air?
 It weighs 150.75 lbs.
 The density of air is 0.09485 lb/ft3
 1589.3416 ft3!!
 It would need, for example, a sphere with radius of
7.2395 feet AND it would need to be “filled” with a
vacuum
Next Steps
 Definitions: Mass, Volume, Density, Buoyancy,
Archimedes’ Principle
 Be able to state the definitions and principles
 Be able to calculate, given two of M, V, or D, the missing
value
 Be able to determine whether a given object will sink,
float, or rise in a given fluid
 Please complete the Archimedes’s Principle/Buoyancy
Worksheet for homework.
Boyle’s Law
 Imagine Bubble-Wrap.
 As you squeeze the bubbles, what happens?
 The bubbles are sealed, so have a constant MASS of air
 As you squeeze the bubble, you decrease the VOLUME of
the bubble
   This, in turn, increases the PRESSURE of the air inside the
bubble
   The pressure, pushing in all directions from each point of
gas inside, acts on the walls of the bubble
   When the pressure is great enough, the bubble pops!
   Boyle’s Law concerns the relationship of a constant mass of
a fluid under changing VOLUME and PRESSURE
conditions
Under Pressure
 “Force per unit area, where the force is Normal
(perpendicular) to the surface.”
 A ratio of the force to the unit area, where the force is
Normal to the surface
 Pounds-per-Square-Inch (psi) is the CMS standard
unit for pressure.
 lbs / in2
Pressure Drop
 Consider a rectangular-prism filled with a fluid; the prism
is 12” long x 8” wide x 4” tall.
   Suppose the fluid exerts 10 psi.
   How many pounds of FORCE is exerted in each direction?
   Calculate the AREA of each face
   12*8 = 96, 8*4 = 32, 4*12 = 48
   Multiply by the PRESSURE, psi
   96*10 = 960, 32*10 = 320, 48*10 = 480
   The top could lift 960 lbs, the sides can resist 320 lbs, and
the front can resist 480 lbs!
Boyle’s Law
 Assuming the temperature is held constant, the
volume of a constant mass of a fluid varies inversely
with its pressure.
 P1 * V1 = P2 * V2
 As the volume decreases, pressure increases
 As the volume increases, pressure decreases
Lift Kit
 Boyle’s Law explains hydraulics and pneumatics
 Hydraulics- machines and tools which use a liquid to
do work, especially actuators and turbines
 Pneumatics- machines and tools which use gasses to
do work, especially actuators and turbines
 LIQUIDS ARE HARDER TO COMPRESS THAN
GASSES!
 That is why hydraulics are used in many applications
as actuators, where pneumatics are not.
Two examples
Example 1
 Suppose a sample of Oxygen gas occupies a volume of 1
ft3 at a pressure of 14.5 psi. What volume will it occupy
at a pressure of 20 psi?
 14.5 * 1 = 20 * V2
 14.5 / 20 = V2 = 0.725 ft3

 http://www.unit-conversion.info/pressure.html
 http://www.unit-conversion.info/pressure.html
Example 2
 A sample of Carbon Dioxide occupies a volume of 3.5
ft3 at 18.125 psi. What pressure would the gas exert if
the volume was decreased to 2 ft3?
 18.125 * 3.5 = P2 * 2
 63.4375 / 2 = P2 = 31.71875 psi
Next Step
 Homework!
 Please complete the Boyle’s Law worksheet for
homework, due at the beginning of class tomorrow
BERNOULLI’S PRINCIPLE
 Take a piece of paper and hold with two hands, just
 Now exhale strongly. (You might need to angle-down a
little and/or blow very hard to get the desired result.)
In a nutshell
 Bernoulli states that a faster-moving body of air will
exert less pressure.
 In other words, if a fluid is accelerated, the pressure
will be lower for that accelerated section of fluid.
 Similarly, if a fluid is decelerated, the pressure will be
higher for that decelerated section of fluid.

 http://hyperphysics.phy-astr.gsu.edu/hbase/pber.html
 http://hyperphysics.phy-astr.gsu.edu/hbase/pber.html

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