Physics 106P: Lecture 1 Notes

```Exam III
Physics 101: Lecture 17
Fluids
Physics 101: Lecture 17, Pg 1
States of Matter
 Solid
Hold Volume
Hold Shape
 Liquid
Fluids
Hold Volume
 Gas
Physics 101: Lecture 17, Pg 2
Qualitative Demonstration of
Pressure

Force due to molecules of fluid colliding with container.
 Impulse Fav t = p

Average Pressure = F / A
y
p y
average vertical force  f y 

t
 mv y 


t
Physics 101: Lecture 17, Pg 3
Atmospheric Pressure
 Basically
weight of atmosphere!
 Air molecules are colliding with you right
now!
 Pressure = 1x105 N/m2 = 14.7 lbs/in2!
 Example: Sphere with r = 0.1 m
Magdeburg Spheres demo
A = 4 p r2 = .125 m2
F = 12,000 Newtons (over 2,500 lbs)!
Can demo
Physics 101: Lecture 17, Pg 4
Pascal’s Principle
A change in pressure at any point in a confined
fluid is transmitted everywhere in the fluid.
 Hydraulic Lift

P1 = P2
F1/A1 = F2 / A2
F1 = F2 (A1/A2)
lift demo
Compare the work done by F1 with the work done
by F2
A) W1 > W2
B) W1 = W2
C) W1 < W2

Physics 101: Lecture 17, Pg 5
Gravity and Pressure

Two identical “light” containers are filled with water.
The first is completely full of water, the second
container is filled only ½ way. Compare the pressure
each container exerts on the table.
1
A) P1 > P2
B) P1 = P2
2
C) P1 < P2
Physics 101: Lecture 17, Pg 6
Pascal’s Principle (Restated)
1. Without gravity: Pressure of a confined fluid is
everywhere the same.
2. With gravity: P = Patm + r g h
Density r = M/V
Pressure of a fluid is everywhere the same
at the same depth. [vases demo]
In general: in a confined fluid, change in pressure is
everywhere the same.
Physics 101: Lecture 17, Pg 7
Dam ACT
B
A
A
Two dams of equal height prevent water from
entering the basin. Compare the net force due to
the water on the two dams.
A) FA > FB
B) FA=FB
C) FA< FB
Physics 101: Lecture 17, Pg 8
Pressure and Depth
Barometer: a way to measure
atmospheric pressure
For non-moving fluids, pressure depends only on
depth.
p1=0
p2 = p1 + rgh
Patm - 0 = rgh
p2=patm
h
Measure h, determine patm
example--Mercury
r = 13,600 kg/m3
patm = 1.05 x 105 Pa
 h = 0.757 m = 757 mm = 29.80” (for 1 atm)
Physics 101: Lecture 17, Pg 9
Checkpoint
Is it possible to stand on the roof of a five story (50
foot) tall house and drink, using a straw, from a
glass on the ground?
1.No
2.Yes
Physics 101: Lecture 17, Pg 10
Archimedes’ Principle
 Determine
force of fluid on immersed cube
Draw FBD
» FB = F2 – F1
» = P 2 A – P1 A
» = (P2 – P1)A
» =rgdA
» =rgV
» = (Mfluid/V) g V
» = Mfluid g
 Buoyant
force is weight of displaced fluid!
Physics 101: Lecture 17, Pg 11
Fb
Archimedes Example
A cube of plastic 4.0 cm on a side with density = 0.8
g/cm3 is floating in the water. When a 9 gram coin is
placed on the block, how much does it sink below
the water surface?
FNet = m a = 0
Mg mg
h
Fb – Mg – mg = 0
r g Vdisp = (M+m) g
M = rplastic Vcube
Vdisp = (M+m) / r
= 4x4x4x0.8
h A = (M+m) / r
= 51.2 g
h = (M + m)/ (r A)
= (51.2+9)/(1 x 4 x 4) = 3.76 cm
[coke demo]
Physics 101: Lecture 17, Pg 12
Summary
 Pressure
is force exerted by molecules
“bouncing” off container P = F/A
 Gravity/weight
affects pressure
P = P0 + rgd
 Buoyant
force is “weight” of displaced
fluid. F = r g V
Physics 101: Lecture 17, Pg 13
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