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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?
Clicker Question Room Frequency BA
A) The pivotB) The end of the beamC) The point where the sign
is attachedD) Any of the aboveE) None of the above
In static equilibrium, the net torque is zero around any axis of rotation!
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• 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
Announcements
Step #1: Force Diagram
Step #2: Coordinate System
x
y
CCW= +
From last week…
x
y
CCW= +
Step #3: Static Equilibrium condition F=ma=0 and τ=Iα=0
0, xxnet maF cosTFwx
0, yynet maF gmgmTF sBwy sin
Not enough information to solve for T, Fwx, Fwy (2 constraint equations and 3 unknowns)
0 Inet ))(sin()4/3)(()2/)(( LTLgmLgm sB
0 Inet ))(sin()4/3)(()2/)(( LTLgmLgm sB
sin4)32( gmm
T sB
0, xxnet maF cosTFwx
0, yynet maF gmgmTF sBwy sin
sincos
4)32(
cosgmm
TF sBwx
4)32(
)(sin)(gmm
gmmTgmmF sBsBsBwy
Notice here that there is no dependence on the hinge force!
A) At the beam hingeB) At the beam
center of gravityC) At the attachment
of the sign’s ropeD) At the beam end
where T is acting
Where would I best choose the axis of rotation to be if I am not given T, and just want Fhinge,y?
Using the beam end, means that my net torque equation does not contain the unknown variable T, but does depend on Fhinge,y :
Clicker Question Room Frequency BA
net = −Fhinge,yL +mBg(L / 2)+mSg(L / 4) = 0
CCW+
Why no Fhinge,x dependence?
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 FB=mBg
mXg mYg mZg
Step #2: Coordinate System x
y
CCW= +
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 FB=mBg
mXg mYg mZg
x
y
CCW= +
FA=mAg FB=mBg
mXg mYg mZg
x
y
CCW= +
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.
10
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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Atomic KE Low Atomic KE High
Atoms/Volume LowAtoms/Volume High
Counter examples exist!!!
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Simplest Case: Static Fluids
Are static fluids really “static”?
Static fluids have no flow.
There is flow when “large” numbers of atoms all move in the same direction.
Uhh,… what’s flow?
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 μm by 1 μm by 1 μm we have about 3 x 1010 atoms of water
If we are working with volumes larger than a few nm3, we don’t need to worry about individual atoms.
In volume 1 cm by 1 cm by 1 cm we have about 3 x 1022 atoms of water
In volume 1 nm by 1 nm by 1 nm we have about 30 atoms of water
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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
There are many types of densities; another common one is number density, defined as the number of atoms N per volume V.
In words, we say mass density is defined as the mass of material per volume.
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.
ρnumber =NV
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What are the units of mass density?
Clicker Question Room Frequency BA
A) kg/mB) kg/m2
C) kg/m3
D) kgE) None of the above
For mass density, it’s mass/volume!
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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?
Clicker Question Room Frequency BA
A) 0.1B) 1C) 100D) 1000E) 1,000,000
Both mass and volume units have to be converted!
1 kgm3 1
gcm 3 g
1 kg1000 g
g100 cm1 m
⎛⎝⎜
⎞⎠⎟3
1000g
cm 3
