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Room Frequency BA. Clicker Question. 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?. The pivot The end of the beam
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Room Frequency BA Clicker Question 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? The pivot The end of the beam The point where the sign is attached Any of the above None of the above In static equilibrium, the net torque is zero around any axis of rotation!
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
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 andτ=Iα=0 Not enough information to solve for T, Fwx, Fwy (2 constraint equations and 3 unknowns)
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? At the beam hinge At the beam center of gravity At the attachment of the sign’s rope 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?
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 y mXg mYg mZg x Step #2: Coordinate System 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 y mXg mYg mZg x 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. FA=mAg FB=mBg y mXg mYg mZg x CCW= +
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,….
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.
Atoms/Volume High Atoms/Volume Low Atomic KE Low Atomic KE High Counter examples exist!!!
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!
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.
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! Usuallya cube or cylinder is easiest. To find the mass of fluid minside the imaginary volume V, we use the concept of mass density.
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 saymass 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.
Room Frequency BA Clicker Question What are the units of mass density? kg/m kg/m2 kg/m3 kg None of the above For mass density, it’s mass/volume!
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? 0.1 1 100 1000 1,000,000 Both mass and volume units have to be converted!
