Equilibrium and Elasticity

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Chapter 11

• Equilibrium and Elasticity

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Goals for Chapter 11

• To study the conditions for equilibrium of a body
• To understand center of gravity and how it
  relates to a bodys stability
• To solve problems for rigid bodies in equilibrium

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Goals for Chapter 11

• To analyze situations involving tension,
  compression, pressure, and shear
• Real materials are not truly rigid. They are
  elastic and do deform to some extent.
• To investigate what happens when a body is
  stretched so much that it deforms or breaks

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Conditions for equilibrium

• First condition The sum of all external forces
  is equal to zero
• ?Fx 0 ?Fy 0 ?Fz 0
• Second condition The sum of all torques about
  any (and ALL!) given point(s) is equal to
  zero.

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Conditions for equilibrium
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Conditions for equilibrium

• First condition The sum of all external forces
  is equal to zero
• ?Fx 0 ?Fy 0 ?Fz 0
• Second condition The sum of all torques about
  any (and ALL!) given point(s) is equal to
  zero.

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Center of gravity

• Treat bodys weight as though it all acts at a
  single pointthe center of gravity

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Center of gravity

• If ignore variation of gravity with
  altitudeCenter of Gravity (COG) is same as
  center of mass.

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Center of gravity

• We can treat a bodys weight as though it all
  acts at a single pointthe center of gravity

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Center of gravity
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Center of gravity
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Center of gravity Stability
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Walking the plank

• Uniform plank (mass 90 kg, length 6.0 m)
  rests on sawhorses 1.5 m apart, equidistant from
  the center.
• IF you stand on the plank at the right edge, what
  is the maximum mass m so that the plank doesnt
  move?

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Walking the plank

• Ask WHAT WILL HAPPEN?
• It will rotate CLOCKWISE!
• If it rotates, where will it rotate around? ID
  Axis!
• It will rotate around S!

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Walking the plank

• What is the maximum mass m for stability?

Origin at c
Center of Gravity at s
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Walking the plank
Rcm M(_at_0) m (_at_ ½ L) m( ½ L) ½ D M
m (Mm) Center of Gravity is the point
where the gravitational torques are balanced.
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Walking the plank
Rcm M(_at_0) m (_at_ ½ L) m( ½ L) ½ D M
m (Mm) So m M D 90 kg (1.5/6-1.5)
30 kg (L D)
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Solving rigid-body equilibrium problems

• Make a sketch create a coordinate system and
  draw all normal, weight, and other forces.
• Specify a positive direction for rotation, to
  establish the sign for all torques in the
  problem. Be consistent.
• Choose a convenient point as your reference
  axis for all torques. Remember that forces
  acting at that point will NOT produce a torque!
• Create equations for SFx 0 SFy 0 and St 0

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Solving rigid-body equilibrium problems

• Car with 53 of weight on front wheels and 47 on
  rear. Distance between axles is 2.46 m. How far
  in front of the rear axle is the center of
  gravity?

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Solving rigid-body equilibrium problems

• Suppose you measure from the REAR wheels.
• Forces at the axis of rotation create NO torque!

St 0 gt 0.47w x 0 meters wLcg 0.53w x
2.46m 0 So.. Lcg 1.30 m
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Solving rigid-body equilibrium problems

• Suppose you measure from the FRONT wheels.
• Forces at the axis of rotation create NO torque!

St 0 gt - 0.47w x 2.46 meters w Lcg
0.53w x 0 meters 0 So.. Lcg 1.16 m L
L 1.16 1.30 2.46 m
Lcg
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Ladder Problems!

• Lots of variables
• Length of ladder L
• Mass of ladder (and its center of gravity)
• Angle of ladder against wall
• Normal force of ground
• Static Friction from ground
• Normal force of wall
• Static Friction of wall
• Weight of person on ladder
• Location of person on ladder

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Ladder Problems!

• Lots of variables
• Length of ladder L
• Mass of ladder (and its center of gravity)
• Angle of ladder against wall
• Normal force of ground
• Static Friction from ground
• Normal force of wall
• Static Friction of wall
• Weight of person on ladder
• Location of person on ladder

Wall (normal force and friction!)
cog of ladder
Ground(normal force and friction!)
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Ladder Problems!

• Equilibrium involves ?Fx 0 ?Fy 0 ?tz
  0
• 3 Equations!
• Solve for up to 3 unique unknowns!
• Use key words/assumptions tonarrow choices
• uniform ladder
• just about to slip
• frictionless

Wall (normal force and friction!)
Ground(normal force and friction!)
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Ladder Problems!
fs (wall)
Wall (normal force and friction!)
N (wall)
Length L
N (ground)
Weight (ladder)
Ground(normal force and friction!)
Angle q
fs (ground)
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Ladder Problems!
Torques about point 1
Ground forces create no torque about point 1!
Length L
N (ground)
Angle q
fs (ground)
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Ladder Problems!
Torques about point 1
Line of Action
Weight drawn from center of gravity Torque
Wladder x d Clockwise! - Wladder x (1/2 L
sinq)
Length ½ L
Weight (ladder)

Angle q
Lever Arm
d
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Ladder Problems!
Line of Action
fs (wall)
Torques about point 1
N (wall)
Angle q
Lever Arm Lsinq
Line of Action
Length L
Lever Arm Lcosq
Angle q
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Ladder Problems!

• ?Fx 0 fs (ground) N (wall)
• ?Fy 0 N (ground) fs (wall) Wladder
• ?t1 0 N (wall) Lsinq CCW fs
  (wall) Lcosq CCW
• - Wladder Lsinq CW

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Will the ladder slip?

• 800 N man climbing ladder 5.0 m long that weighs
  180 N. Find normal and frictional forces that
  must be present at the base of the ladder.
• What is the minimum coefficient of static
  friction at the base to prevent slipping?
• What is the magnitude and direction of the
  contact force on the base of the ladder?

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Will the ladder slip?

• 800 N man climbing ladder 5.0 m long that weighs
  180 N. Find normal and frictional forces that
  must be present at the base of the ladder.
• What is the minimum coefficient of static
  friction at the base to prevent slipping?

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Will the ladder slip?

• SFx 0 gt
• fs n1 0

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Will the ladder slip?

• SFx 0 gt
• fs n1 0
• SFy 0 gt
• n2 wperson wladder 0

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Will the ladder slip?

• SFx 0 gt
• fs n1 0
• SFy 0 gt
• n2 wperson wladder 0
• St (about ANY point) 0 gt
• n1(4m) wp(1.5m) wl(1.0m) 0

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Will the ladder slip?

• SFx 0 gt
• fs n1 0
• SFy 0 gt
• n2 wperson wladder 0
• St (about ANY point) 0 gt
• n1(4m) wp(1.5m) wl(1.0m) 0
• So
• n1 268 N

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Will the ladder slip?

• SFx 0 gt
• fs n1 0
• SFy 0 gt
• n2 wperson wladder 0
• St (about ANY point) 0 gt
• n1(4m) wp(1.5m) wl(1.0m) 0
• So
• n1 268 N ms(min) fs/n2 .27

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Will the ladder slip? Example 11.3

• 800 N man climbing ladder 5.0 m long that weighs
  180 N. Find normal and frictional forces that
  must be present at the base of the ladder.
• What is the minimum coefficient of static
  friction at the base to prevent slipping?
• What is the magnitude and direction of the
  contact force on the base of the ladder?

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Example 11.4 Equilibrium and pumping iron

• Given weight w and angle between tension and
  horizontal, Find T and the two components of E

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Strain, stress, and elastic moduli

• Stretching, squeezing, and twisting a real body
  causes it to deform.
• Stress is force per unit area
• Measured in N/m2 or Pascals or lbs/sq. in. or
  PSI
• 1 PSI 7000 Pa
• Typical tire pressure 32 PSI 200 kPA
• Same units as PRESSURE in fluids/gases
• Stress is an internal force on an object that
  produces (or results from) a Strain

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Strain, stress, and elastic moduli

• Stretching, squeezing, and twisting a real body
  causes it to deform.
• Stress is force per unit area
• Strain is the relative change in size or shape of
  an object because of externally applied forces.
• Strain is the fractional deformation due to the
  stress.
• Strain is dimensionless just a fraction.
• Linear strain Dl/l0 (tensile strain)

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Strain, stress, and elastic moduli

• Stretching, squeezing, and twisting a real body
  causes it to deform.
• Stress is force per unit area
• Strain is the fractional deformation due to the
  stress.
• Elastic modulus is stress divided by strain.
• The direct (linear) proportionality of stress and
  strain is called Hookes law
• STRESS / STRAIN Elastic Modulus

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Strain, stress, and elastic moduli

• The direct (linear) proportionality of stress and
  strain is called Hookes law.
• Similar to spring/rubber bands stretching
• Apply external force F, see spring/band deform a
  distance Dx from the original length x.
• Apply that force F over the entire cross
  sectional area of the rubber band (or thickness
  of the spring wire)
• F/A creates a STRESS

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Strain, stress, and elastic moduli

• The direct (linear) proportionality of stress and
  strain is called Hookes law.
• Similar to spring/rubber bands stretching
• Apply that force F over the entire area of the
  rubber band (or thickness of the spring wire)
  F/A creates a STRESS
• Deformation of band in length Dx/x is the
  resulting Strain
• F kDx
• F/A Stress kDx/A YDx/x Y X Strain

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Strain, stress, and elastic moduli

• The direct (linear) proportionality of stress and
  strain is called Hookes law.
• Similar to spring/rubber bands stretching
• Tensile Stress/Strain Y Youngs Modulus
• F/A Stress kDx/A YDx/x Y X Strain
• Y kx/A Force/Area (Pascals)

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Tensile and compressive stress and strain

• Tensile stress F? /A tensile strain ?l/l0
  (stretching under tension)
• Youngs modulus is tensile stress divided by
  tensile strain, and is given by Y (F?/A)(l0/?l)

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Tensile and compressive stress and strain

• Tensile stress F? /A tensile strain ?l/l0
  Youngs modulus is tensile stress divided by
  tensile strain, and is given by Y (F?/A)(l0/?l)
• Compressive stress compressive strain defined
  similarly.

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Some values of elastic moduli
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Tensile stress and strain

• Body can experience both tensile compressive
  stress at the same time.

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Bulk stress and strain

• Pressure in a fluid is force per unit area p
  F?/A.
• Bulk stress is pressure change ?p bulk strain
  is fractional volume change ?V/V0.
• Bulk modulus is bulk stress divided by bulk
  strain and is given by B ?p/(?V/V0).

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Sheer stress and strain

• Sheer stress is F/A and sheer strain is x/h.
• Sheer modulus is sheer stress divided by sheer
  strain, and is given by S (F/A)(h/x).

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Elasticity and plasticity

• Hookes law applies up to point a.
• Table 11.3 shows some approximate breaking
  stresses.