# Lecture Forces The Laws of Motion by jennyyingdi

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```									  Lecture 19:
Introduction to
Solids & Fluids
Questions of Yesterday
1) A solid sphere and a hoop of equal radius and mass are both rolled
up an incline with the same initial velocity. Which object will travel
farthest up the inclined plane?
a) the sphere
b) the hoop
c) they’ll both travel the same distance up the plane
d) it depends on the angle of the incline

2) If an acrobat rotates once each second while sailing through the air,
and then contracts to reduce her moment of inertia to 1/3 of what is
was, how many rotations per second will result?
a) once each second
b) 3 times each second
c) 1/3 times each second
d) 9 times each second
Practice Problem
A 10.00-kg cylindrical reel with the radius of 0.500 m and a frictionless
axle starts from rest and speeds up uniformly as a 5.00 kg bucket falls
into a well, making a light rope unwind from the reel. The bucket starts
from rest and falls for 5.00 s.

10.0 kg               What is the linear acceleration
of the falling bucket?

0.500 m                      How far does it drop?

What is the angular acceleration of the reel?

Use energy conservation principles to determine
5.00 kg                 the speed of the spool
after the bucket has fallen 5.00 m
4 States of Matter
Solid           Fluid              Gas              Plasma

Definite                        Expands to fill    Expands to fill
Definite
Shape &                           any volume        any volume
Volume
Volume                         Takes shape of     Takes shape of
Takes shape of
Molecules                           Container         Container
container
close together                    Molecules even     Made up of ions
Molecules
& slow                          farther apart     Fastest of all
farther apart
& even faster      matter states
& faster
Density

Distance between molecules

Density

M                Density
r=       The amount of matter (mass)
V          in a given volume
Solids
Applying force can change shape & size (deform)
When force is removed -> original shape & size

Remind you of anything?
Definite                                         x=0

Shape &
SOLIDS are ELASTIC
Volume
FS = -kx

STRESS = ELASTIC MODULUS x STRAIN

Force per                              Measure of
Elasticity of
Unit Area                              Deformation
material
Length Elasticity
Elastic Modulus and Induced strain depends on type of stress

F                                F
A     L0                               L0

F                    DL       F
L0       DL                      L0

STRESS = ELASTIC MODULUS x STRAIN

F = Y DL
Relative
SI Units =                                 Length Change
N/m2 =             A     L0
Pascal (Pa)                             Young’s Modulus
LENGTH
Volume Elasticity

F                               F               DV

V0                           V0

STRESS = ELASTIC MODULUS x STRAIN

F = -B DV
Relative
SI Units =                           Volume Change
N/m2 =             A       V0
Pascal (Pa)                           Bulk Modulus
VOLUME
Pressure
Uniform force F is acting over entire
surface area A in a direction
perpendicular to surface area
F
PRESSURE (P)
Perpendicular Force per unit area
A
P= F
A
When would this
situation occur?
Pressure
Uniform force F is acting over entire
surface area A in a direction
perpendicular to surface area
F
PRESSURE (P)
Perpendicular Force per unit area
A
P= F
A
Fluids are NOT elastic ->
But…
Fluids do exerted force
Pressure
Force exerted by fluid on a
submerged object is always
PERPENDICULAR
to surface of object             F

Fluids exert PRESSURE on
submerged objects and the                                 A
walls of their container

Fluids are NOT elastic ->
But…
Fluids do exerted force
Pressure
If a fluid is at rest in a container what
y
0
It is in EQUILIBRIUM, so…
The net FORCE acting on any                 F1   F2
portion of fluid is ZERO                               y1

F1(y1) = -F2(y1)                                   y2

all points at the same DEPTH
must be at the same PRESSURE

P1(y1) = P2(y1)
Pressure
The net FORCE acting on any                  y
portion of fluid is ZERO
0
all points at the same DEPTH
P1A
must be at the same PRESSURE
y1
∑Fy = P2A - P1A - Mg = 0
y2

r= M
V   Mg    P2A

P2 = P1 + rg(y1 - y2)
Pressure
What is the pressure at the surface
of the fluid
(open to the air)?                   P0A        y

Gas making up atmosphere                                       0
exerts pressure on fluid
h

P2 = P1 + rg(y1 - y2)                                          y1

PA
P = P0 + rgh

Pressure P at depth h below the surface of a liquid
open to the atmosphere is greater than atmosphere pressure
(P0 = 1.013*105 Pa) by the amount rgh
Pressure
What if you change the pressure
exerted at the surface?
F
P = P0 + rgh

PASCAL’S PRINCIPLE
A change in pressure applied to an
enclosed fluid is transmitted
undiminished to every point of the
fluid and the walls of the container
Pressure
PASCAL’S PRINCIPLE
A change in pressure applied to an enclosed fluid is transmitted
undiminished to every point of the fluid and the walls of the
container
What happens if you apply a force F1 to one side of this
apparatus?

F1
A1

A2

F2
Pressure
PASCAL’S PRINCIPLE
A change in pressure applied to an enclosed fluid is transmitted
undiminished to every point of the fluid and the walls of the
container

F1/A1 = F2/A2

F1      Dx1
Dx2
F2
Pressure
F1/A1 = F2/A2
How does Dx1 compare to Dx2?

Fluids have a definite volume (incompressible) -> DV = 0

F1Dx1 = F2Dx2
What does
this tell you
work done       F1     Dx1
Dx2
on the fluid?                             F2
Buoyancy
What allows an object to float
in a fluid?

Is the object in equilibrium?
M
V
Buoyancy
What allows an object to float
in a fluid?                  B

Is the object in equilibrium?
m
∑Fy = B - mobjg = 0                     V
mg
Buoyancy
What allows an object to float
in a fluid?                      P1A

Is the object in equilibrium?
M
∑Fy = B - mobjg = 0                           V
Mg       P2A
∑Fy = P2A - P1A - Mfluidg = 0

B = P2A - P1A = rfluidVfluidg

BUOYANT
FORCE
Buoyancy
What allows an object to float
in a fluid?                             B

Is the object in equilibrium?
m
∑Fy = B - mobjg = 0                                   V
mg
∑Fy = P2A - P1A - Mfluidg = 0
If the object is in
B = P2A - P1A = rfluidVfluidg      equilibrium with the fluid…

BUOYANT                           robj     Vfluid
=
FORCE                             rfluid   Vobj
Buoyancy
B = rfluidVfluidg
What if the object is rising or sinking?

B                          B                          B
a                          a
m                        m                          m
V                          V                          V
mg a = 0                mg                           mg

B = mobjg              B - mobjg > 0           B - mobjg < 0
robj     Vfluid
=          (rfluid-robj)Vobjg > 0    (rfluid-robj)Vobjg < 0
rfluid   Vobj
Properties of an Ideal Fluid
An Ideal Fluid is…
NONVISCOUS
no internal friction between adjacent layers
INCOMPRESSIBLE
constant density

Ideal Fluid Motion is…
velocity, density, pressure at each point is constant in time

WITHOUT TURBULENCE
Angular velocity about center of each element is zero
All points can translate but not rotate
Ideal Fluid Motion
How does v1 compare to v2?
A2             v2
Dx2

Mass is conserved
A1                v1
DM1 = DM2

Dx1 = v1Dt          r1A1v1 = r2A2v2
Fluid is incompressible
Volume of fluid leaving 1 =
Volume of fluid entering 2         A1v1 = A2v2
in the same time interval
Equation of Continuity
Ideal Fluid Motion
P2A2
Is energy conserved in an ideal
fluid?                                v2
Dx2

P1A1                                     Dy2
v1

DM1 = DM2          Dx1   Dy1

What is the work done on the fluid?
W1 = P1A1Dx1                 W2 = -P2A2Dx2

W = P1V - P2V
Ideal Fluid Motion
P2A2
W = P1V - P2V
A2              v2
Dx2

P1A1                                       Dy2
v1

Dx1    Dy1

Is the energy of the fluid changing?
What types of energy are present?

Wfluid = DKE + DPE
P1 + (1/2)rv12 + rgy1 = P2 + (1/2)rv22 + rgy2
Ideal Fluid Motion
P2A2
W = P1V - P2V
A2            v2
Dx2

P1A1                                   Dy2
v1

Dx1   Dy1

BERNOULLI’S EQUATION
The sum of pressure, kinetic energy per unit volume,
and potential energy per unit volume
is equal at all points along the streamline
P1 + (1/2)rv12 + rgy1 = P2 + (1/2)rv22 + rgy2
Questions of the Day
1) Two women of equal mass are standing on the same hard wood
floor. One is wearing high heels and the other is wearing tennis
shoes. Which statement is NOT true?
a) both women exert the same force on the floor
b) both women exert the same pressure on the floor
c) the normal force that the floor exerts is the same for both women

2) A boulder is thrown into a deep lake. As the rock sinks deeper and
deeper into the water what happens to the buoyant force?
a) it increases
b) it decreases
c) it stays the same

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