#### Transcript Review of buoyant force

```Physics Warm Up: Agenda
Copy these assignemts into your binder
December 2
WarmUp: Agenda/ Review of buoyant force
InClass: Archimedes Worksheet
Homework: Read and take notes p358-362 DUE NEXT CLASS
December 3-4
WarmUp: Specific Heat of Copper
Lab: Specific Heat
Homework: Answer q 5p374 DUE NEXT CLASS
December 5-6
WarmUp: Heat of Fusion
Lab: Heat of Fusion
Homework: When the air temperature is 22.2°C many people find it a comfortable
temperature, yet the same people often find a swimming in 22.2°C water too cold to
be comfortable. Use specific heat to explain the reason for the difference in sensation
of temperature of the air and the water. Explain it to somebody and bring in a written
WHEN YOU FINISH WRITING DOWN THE HOMEWORK, PLEASE GET OUT YOUR BUOYANCY
HOMEWORK FROM LAST WEEK, QUESTIONS 1 AND 2 P324 SO THAT WE CAN GO OVER THE
QUESTIONS
Buoyant force p324
There was some difficulty with problems 1 and 2 of the
homework last week. We didn’t go over it in class.
Question 1: A piece of metal weighs 50.0N in Air. 36.0N in water, and
41.0N in oil. Find the density of
a. The metal
Two pages back are the equations used in the sample problem
Fnet = (ρfVf – ρoVo) g
Fnet = (ρf – ρo) V•g
Fg(object)
ρo V•g
=
FB
ρf V•g
Fg(object) ρf
= ρo
FB
50.0N • 1.00g/cm3
14.0N
where Vnet is the apparent weight, and the subscripts f and o stand for
The fluid and the object respectively
Because the object is submerged, the volumes are equal and we
can simplify
Use the ratio (you can find this on p322 if you forgot where it
comes from) and solve for the density of the object (ρo).
Substitute values from the question
= 3.57g/cm3
= 3.57x103kg/m3
Buoyant force p324
There was some difficulty with problems 1 and 2 of the
homework last week. We didn’t go over it in class.
Question 1: A piece of metal weighs 50.0N in Air. 36.0N in water, and
41.0N in oil. Find the density of
Then B asks you to find the density of the oil.
Fg(object)
FB
ρf =
=
ρo V•g
ρf V•g
We can use the same simple ratio, but now solve for the density
of the fluid, (ρf)
ρo FB
Fg(object)
3kg/m • 9.0N
3.57x10
3
ρf =
50.0N
Substitute values from the question
= 640x103kg/m3
Archimedes Principle
Question 2 from that homework is easy if you just
remember that the buoyant force acting on a submerged
object is equal to the weight (force) of the fluid it
displaces.
The mass of this water
Verify it:
On the demo table is a can with a spout.
Fill the with just enough water to run out of the spout.
Weigh the small beaker.
When the spout stops dripping, put the beaker under it.
Add the 100g weight to the water and collect the displaced water.
Weigh the beaker with the water and calculate the weight of the displaced water. Record it
Weigh the 100g weight in air by hanging it from the hook on
the balance.
Fill the 250mL beaker and place it on the arm of the balance.
Weigh the 100g weight in water and calculate the difference.
The difference is the buoyant force.
It should be very close to the weight of the water displaced.
Should equal the
difference in weight
Buoyant force p324
2. An empty rubber balloon has a mass of 0.0120kg. The balloon is filled with helium at 0°C,
1atm pressure and a density of 0.181kg/m3. The filled balloon has a radius of 0.500m.
a. What is the buoyant force acting on the balloon
V= 4/3 πr3
The buoyant force acting on the balloon is equal to
V= 4/3 π(0.5m)
the weight of the air it displaces (p320).
The mass of the displaced air is the volume of the
V= .524m3
balloon times the density of the. Multiplying that by
Fg(air)= v• ρf • g
acceleration due to gravity gives its weight.
Fg(air)=0.524m3 • 1.29kg/m3 • 9.8 m/s2 B. The net force will be the difference
between the balloon’s weight and the
buoyant force. The weight of the balloon is
Fg(air)= 6.62kgm/s2 = 6.62N
the density of the helium times the volume
of the balloon added to the mass of the
B.
empty balloon.
Fg(balloon)=9.8 m/s2 (0.524m3 )(0.181kg/m3)+9.8 m/s2(0.0120kg)
Fg(balloon)=1.05N
Fnet= Fg(air) - Fg(balloon)
Fnet= 6.62N- 1.05N
Fnet= 5.57N
Pressure & Pascal’s Law
Pressure is force per area. p = f/a
Liquid Pressure = (depth)(density)
Units of pressure:
pascal = n/m2
Atmosphere,
mm of Hg (in manometer)
kg/cm2 (mass units).
Common pressures:
1 atm =
100 kPa
760 mm of Hg
10 m of H2O
1 kg/cm2
p = h
Blaise Pascal
Tap water pressure =
4 atm
400 kPa
4 kg/cm2
Car tire 2 atm
Pressure cont’d
Total Force = (pressure)(area)
TF = pA
.
Two
Sample
Problems
Solutions are on the next page
Find the pressure needed to push water to the top
of Tower 2, a height of 8.0 meters. (Use mass units).
The density of water is 1.0 g/cm3.
Mainmachinery
Find the Total Force on
the filled school dam whose
dimensions are 5.0m by 2.0m.
The average depth is 1.0m.
Castle attack
Solutions
Find the pressure needed to push water to the top of Tower 2, a height of 8.0 meters.
(Use mass units). The density of water is 1.0 g/cm3.
1
2
3
4
p = h
= (8.0m)(100cm/m)(1.0g/cm3 2 )
= 800 g/cm2
Find the Total Force on the filled school dam whose dimensions are 5.0m by 2.0m.
The average depth is 1.0m.
1
2
3
4
f = pA
= (100g/cm2)(10.0m2)(104cm2/m2) *
= 1 X 107 g = 1 X 104 kg or 10 tons!
Note: 1kg = 1000g
1 metric ton = 1000kg
1
2
3
4
A = LW
= (5.0m)(2.0m)
= 10.0 m2
* There are 104 cm2 in a m2.
1
2
3
4
p = h
= (1.0m)(100cm/m)(1.0g/cm3 2 )
= 100 g/cm2
Pascal’s
Law
The pressure on a confined fluid is transmitted in all directions.
Pascal’s Vases show pressure
depends only on depth & density.
Pascal Pressure in Rocket (1)
pascalrocket.mov
Hiero’s Fountain Demo
h2 is higher
than h1, so the
pressure is greater
in the right system
which pushes the
water up into the
fountain.
50mL of water to the
funnel of the
fountain on the
demo table!
Hydraulic Lift: Demo: Syringes
Try this with
different sized
syringes.
The pressure is transmitted undiminished in all directions.
Hydraulic Brakes
The applied pressure to the master cylinder is transmitted
equally to all four brake pistons.
Hydraulics lifts a House! (2)
Oh, Pascal! Thanks to Mark Shisler
hydraulics.mov
An Uplifting Experience: Demo
A strong rubber balloon inflated beneath a car or truck
can lift 15 metric tonnes of load. The pressure is low,
but the surface area is large. Total Force = (press)(area).
Demo: A garbage bag
blown up by a vacuum
cleaner can lift a massive
person.
```