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Physics 201: Lecture 20

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Title: Physics 201: Lecture 20


1
Physics 201 Lecture 20
  • Statics
  • Stable and Unstable Equilibrium
  • Elastic Properties of Solids
  • Moduli
  • Tensile (Youngs)
  • Shear
  • Bulk

2
Statics
  • Make a diagram and the indicate all the forces
  • as vectors
  • In general, we can use the two equations
  • to solve any statics problem.
  • When choosing axes about which to calculate
    torque,
  • the choice can make the problem easy....

3
Example Ladder against smooth wall
  • Bill (mass M) is climbing a ladder (length L,
    mass m) that leans against a smooth wall (no
    friction between wall and ladder). A frictional
    force F between the ladder and the floor keeps it
    from slipping. The angle between the ladder and
    the wall is ?.
  • What is the magnitude of F as a function of
    Bills distance up the ladder?

?
L
m
Bill
F
4
Ladder against smooth wall...
  • Consider all of the forces acting. In addition to
    gravity and friction, there will be normal forces
    Nf and Nw by the floor and wall respectively on
    the ladder.

Nw
L/2
?
  • Again use the fact that FNET 0
    in both x and y directions
  • x Nw F
  • y Nf Mg mg

m
mg
d
Mg
F
Nf
5
Ladder against smooth wall...
  • Since we are not interested in Nw, calculate
    torques about an axis through the top end of the
    ladder, in the z direction.

axis
Nw
L/2
?
m
  • Substituting Nf Mg mg andsolving for F

mg
a
d
Mg
F
Nf
a
6
Example Ladder against smooth wall...
  • For a given coefficient of static friction
    ?s,the maximum force of friction F that can
    beprovided is ?sNf ?s g(M m).
  • The ladder will slip if F exceedsthis value.

?
m
  • Moral
  • Brace the bottom of ladders!
  • Dont make ? too big!

d
F
7
Potential Energy Diagrams
  • Consider a block sliding on a frictionless
    surface, attached to an ideal spring.
  • F -dU/dx -slope

F
x
U
F
x
x
0
8
Equilibrium
  • F -dU/dx -slope
  • So F 0 if slope 0.
  • This is the case at the minimum or maximum of
    U(x).
  • This is called an equilibrium position.
  • If we place the block at rest at x 0, it wont
    move.

m
x
U
x
0
9
Equilibrium
  • If small displacements from the equilibrium
    position result in a force that tends to move the
    system back to its equilibrium position, the
    equilibrium is said to be stable.
  • This is the case if U is a minimum at the
    equilibrium position.
  • In calculus language, the equilibrium is stable
    if the curvature (second derivative) is positive.

F
m
x
U
F
x
0
10
Equilibrium
U
  • Suppose U(x) looked like this
  • This has two equilibrium positions, one is
    stable ( curvature) and one is unstable (-
    curvature).
  • Think of a small object sliding on the U(x)
    surface
  • If it wants to keep sliding when you give it a
    little push, the equilibrium is unstable.
  • If it returns to the equilibrium position when
    you give it a little push, the equilibrium is
    stable.
  • If the curvature is zero (flat line) the
    equilibrium is neutral.

unstable
neutral
stable
x
0
11
Solids
  • Have definite volume
  • Have definite shape
  • Molecules are held in specific locations
  • by electrical forces
  • vibrate about equilibrium positions
  • Can be modeled as springs connecting molecules

12
Liquid
  • Has a definite volume
  • No definite shape
  • Exist at a higher temperature than solids
  • The molecules wander through the liquid in a
    random fashion
  • The intermolecular forces are not strong enough
    to keep the molecules in a fixed position

13
Gas
  • Has no definite volume
  • Has no definite shape
  • Molecules are in constant random motion
  • The molecules exert only weak forces on each
    other
  • Average distance between molecules is large
    compared to the size of the molecules

14
Question
  • Are atoms in a solid always arranged in an
    ordered structure?
  • Yes
  • No

15
Deformation of Solids
  • All objects are deformable, i.e It is possible to
    change the shape or size (or both) of an object
    through the application of external forces
  • Sometimes when the forces are removed, the object
    tends to its original shape, called elastic
    behavior
  • Large enough forces will break the bonds between
    molecules and also the object

16
Elastic Properties
  • Stress is related to the force causing the
    deformation
  • Strain is a measure of the degree of deformation
  • The elastic modulus is the constant of
    proportionality between stress and strain
  • For sufficiently small stresses, the stress is
    directly proportional to the strain
  • The constant of proportionality depends on the
    material being deformed and the nature of the
    deformation
  • The elastic modulus can be thought of as the
    stiffness of the material

17
Youngs Modulus Elasticity in Length
  • Tensile stress is the ratio of the external force
    to the cross-sectional area
  • For both tension and compression
  • The elastic modulus is called Youngs modulus
  • SI units of stress are Pascals, Pa
  • 1 Pa 1 N/m2
  • The tensile strain is the ratio of the change in
    length to the original length
  • Strain is dimensionless

18
Shear Modulus Elasticity of Shape
  • Forces may be parallel to one of the objects
    faces
  • The stress is called a shear stress
  • The shear strain is the ratio of the horizontal
    displacement and the height of the object
  • The shear modulus is S
  • A material having a large shear modulus is
    difficult to bend

19
Beams
20
Bulk Modulus Volume Elasticity
  • Bulk modulus characterizes the response of an
    object to uniform squeezing
  • Suppose the forces are perpendicular to, and acts
    on, all the surfaces -- as when an object is
    immersed in a fluid
  • The object undergoes a change in volume without a
    change in shape
  • Volume stress, DP, is the ratio of the force to
    the surface area
  • This is also the Pressure
  • The volume strain is equal to the ratio of the
    change in volume to the original volume

21
Notes on Moduli
  • Solids have Youngs, Bulk, and Shear moduli
  • Liquids have only bulk moduli, they will not
    undergo a shearing or tensile stress
  • The liquid would flow instead
  • The negative sign is included since an increase
    in pressure will produce a decrease in volume --
    B is always positive
  • The compressibility is the reciprocal of the bulk
    modulus

Ultimate Strength of Materials
  • The ultimate strength of a material is the
    maximum stress the material can withstand before
    it breaks or factures
  • Some materials are stronger in compression than
    in tension
  • Linear to the Elastic Limit

22
Arches
  • Which of the following two archways can you
    build bigger, assuming that the same type of
    stone is available in whatever length you desire?
  • Post-and-beam (Greek) arch
  • Semicircular (Roman) arch
  • You can build big in either type
  • Low ultimate tensile strength of sagging stone
    beams

Stability depends upon the compression of the
wedge-shaped stones
23
Gothic Arch
  • First used in Europe in the 12th century
  • Extremely high
  • The flying buttresses are needed to prevent the
    spreading of the arch supported by the tall,
    narrow columns

24
Problem
A plastic box is sliding across a horizontal
floor. The frictional force between the box and
the floor causes the box to deform. To describe
the relationship between stress and strain for
the box, you would use
  • Youngs modulus
  • Shear modulus
  • Bulk modulus
  • None of the above

25
Solution
FORCE
MOTION
FRICTION
SO THE SHEAR MODULUS IS THE CHOICE!
26
Moduli Values
27
Prestressed Concrete
  • If the stress on a solid object exceeds a certain
    value, the object fractures
  • The slab can be strengthened by the use of steel
    rods to reinforce the concrete
  • The concrete is stronger under compression than
    under tension
  • A significant increase in shear strength is
    achieved if the reinforced concrete is
    prestressed
  • As the concrete is being poured, the steel rods
    are held under tension by external forces
  • These external forces are released after the
    concrete cures
  • This results in a permanent tension in the steel
    and hence a compressive stress on the concrete
  • This permits the concrete to support a much
    heavier load
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