Clicker Question
Room Frequency BA
In the problem about finding the tension in a support wire
for a beam and sign in static equilibrium, where is the
correct point to use as the axis of rotation when calculating
the net torque?
A) The pivot
B) The end of the beam
C) The point where the sign
is attached
D) Any of the above
E) None of the above
In static equilibrium, the net torque is zero around any axis of rotation!
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Announcements
• Print out and bring Lab Manual #5 to your lab
meeting this week. There is no pre-lab!
• CAPA assignment #12 is due on Friday at 10 pm.
• Start reading Chapter 10 on Fluids
• Dr. Paul will not have a review meeting this
Wednesday evening (November 9)
• Midterm scores should be uploaded on CU Learn by
end of today/early tomorrow. Solutions are posted
on CULearn.
• I will be in the Physics Help Room 1:45pm to 3:45pm
today
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From last week…
Step #1: Force Diagram
Step #2: Coordinate System
y
x
CCW= +
y
x
CCW= +
Step #3: Static Equilibrium condition F=ma=0 and τ=Iα=0
Fnet , x  ma x  0  Fwx  T cos
Fnet, y  may  0  Fwy  T sin   mB g  ms g
Not enough information to solve for T, Fwx, Fwy
(2 constraint equations and 3 unknowns)
 net  I  0  (mB g )( L / 2)  (ms g )(3L / 4)  (T sin  )( L)
 net  I  0  (mB g )( L / 2)  (ms g )(3L / 4)  (T sin  )( L)
Notice here that there is no
dependence on the hinge force!
(2mB  3ms ) g
T
4 sin 
Fnet , x  ma x  0  Fwx  T cos
Fwx
(2mB  3ms ) g  cos 
 T cos 


4
 sin  
Fnet, y  may  0  Fwy  T sin   mB g  ms g
Fwy
(2mB  3ms ) g
 (mB  ms ) g  T sin   (mB  ms ) g 
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Clicker Question
Room Frequency BA
Where would I best choose the axis of rotation to be if
I am not given T, and just want Fhinge,y?
A) At the beam hinge
B) At the beam
center of gravity
C) At the attachment
of the sign’s rope
D) At the beam end
where T is acting
CCW+
Using the beam end, means that my net torque
equation does not contain the unknown
variable T, but does depend on Fhinge,y :
Why no Fhinge,x dependence?
 net  Fhinge,y L  mB g(L / 2)  mS g(L / 4)  0
Another static equilibrium example! See the physical setup in class.
You are given the scale reading of B as mB, the distance D1,
D2, and D3, and the masses of blocks X and Y; you are asked
to find the scale reading of A (the bar is massless).
Step #1: Force Diagram of beam
FA=mAg
mXg
FB=mBg
mYg
mZg
y
Step #2: Coordinate System
CCW= +
x
Before choosing the axis of rotation, think about knowns and
unknowns and what you are asked for!
You don’t know mZ ; you are asked for the reading of scale A.
FA=mAg
mXg
FB=mBg
mYg
mZg
y
CCW= +
x
Clicker Question
Room Frequency BA
Which choice of axis of rotation on the beam depends on the
desired unknown, but does not depend on the other unknowns?
A) at Scale A
B) at block X
C) at Block Y
D) at Block Z
At block Z, the torques from scale B and the weight of Mz are zero.
FA=mAg
mXg
FB=mBg
mYg
mZg
y
CCW= +
x
The Phases of Matter
Solids: Shape independent of “container”
Liquids: Shape depends on “container” and
surface
Gases: Shape only depends on “container”
Plasmas: Shape depends on “container”,
surface, electrodes, plasma itself, magnetic
fields,….
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Atomic View of the Phases of Matter
Solids: Atoms locked together, close and rigidly.
Atoms per volume high
Liquids: Atoms close together, but free to
move individually. Attraction strong enough
to keep them from flying apart. Atoms per
volume high, similar to solids
Gases: Atoms only interact weakly, mostly
just fly around hitting the container walls or
each other. Atoms per volume low.
Plasmas: Atoms broken apart into electrons
and ions which have strong electric forces
acting on them. Atoms per volume low or
medium.
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Atoms/Volume High
Atomic KE Low
Atoms/Volume Low
Atomic KE High
Counter examples exist!!!
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Simplest Case: Static Fluids
Static fluids have no flow.
Uhh,… what’s flow?
There is flow when “large” numbers of atoms all move in the
same direction.
Are static fluids really “static”?
On the atomic level no! But on the “large” number level yes!
Critical Point: In a static fluid, any small volume of
liquid is at rest, hence the net force on this small
volume is zero, by Newton’s 2nd Law!
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How large is a “large” number?
In volume 1 cm by 1 cm by 1 cm we have about 3 x 1022
atoms of water
In volume 1 μm by 1 μm by 1 μm we have about 3 x 1010
atoms of water
In volume 1 nm by 1 nm by 1 nm we have about 30 atoms
of water
If we are working with volumes larger than a few nm3, we don’t
need to worry about individual atoms.
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Back to square 1: Fluid “Mass”
When we try to apply Newton’s Laws to Fluids what do we use
for the mass?
We always consider a “small” imaginary volume inside
the body of fluid! Usually a cube or cylinder is easiest.
To find the mass of fluid m inside
the imaginary volume V, we use the
concept of mass density.
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Fluid Density
We imagine that the volume V is completely filled with some
“continuous” substance. We measure the mass m and the
volume V then calculate the mass density ρmass as m/V.
In words, we say mass density is defined as the mass of material
per volume.
There are many types of densities; another
common one is number density, defined as
the number of atoms N per volume V.
number
N

V
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Clicker Question
Room Frequency BA
What are the units of mass density?
A)
B)
C)
D)
E)
kg/m
kg/m2
kg/m3
kg
None of the above
For mass density, it’s mass/volume!
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Room Frequency BA
Clicker Question
Many mass densities are given in units of g/cm3. What is the
factor to multiply a density of 1 g/cm3 to get SI units of kg/m3?
A)
B)
C)
D)
E)
0.1
1
100
1000
1,000,000
Both mass and volume units have to be
converted!
3
kg
g
1 kg  100 cm 
g
1 3 1 3 g
g

1000

m
cm 1000 g  1 m 
cm 3
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