Review of Lecture 12

1
Review of Lecture 12

• Rolling objects
• The K.E. of rolling objects
• Forces on rolling objects
• The yo-yo
• Torque revisited
• Angular Momentum
• Newtons 2nd Law in Angular form

2
Review of Lecture 12

• Angular Momentum of a system of particles
• Angular Momentum of a rigid, rotating object
• Conservation of Angular Momentum

3
Equilibrium

• An object is in equilibrium if
• The linear momentum P of its center of mass is
  constant
• The angular momentum L about its center of mass
  (or about any other point) is constant

4
Equilibrium

• The following objects are in equilibrium
• A book resting on a table
• A hockey puck sliding across the ice at a
  constant speed
• The rotating blades of a ceiling fan (turning at
  a constant angular speed)
• The wheel of a bicycle traveling in a straight
  line at a constant speed

5
Equilibrium

• In Chapter 12, we are concerned with objects
  where the linear momentum P and angular momentum
  L are not only constant, but zero
• This special case of equilibrium is called static
  equilibrium

6
Equilibrium

• Lets examine a domino in some different
  positions
• In (a) we can see that the domino is precariously
  balanced just on a supporting edge
• The force of gravity is directly through the
  supporting edge therefore the moment arm is
  zero
• As a result there is no torque and the domino is
  stable

7
Equilibrium

• But it is an unstable equilibrium as the
  slightest rotation particularly to the right
  will cause the domino to fall over

8
Equilibrium

• In this case you can see that the domino is also
  stable
• It is more stable than in case (a) as any
  rotation must be sufficient to cause the vector
  Fg to move past the supporting edge of the domino

9
Equilibrium

• More stable yet is the block
• In this case there must be a considerable
  rotation in order to cause the vector Fg to move
  past an edge

10
Equilibrium

• The study of equilibrium is very important in
  engineering practice
• The design engineer must ensure that the item in
  question a building, a bridge, landing gear for
  an aircraft will remain stable under load (the
  external forces and torques that may be applied
  to it)

11
Equilibrium

• We know that Newtons 2nd Law (in translational
  form) is
• So if an object is in translational equilibrium,
  then P is constant and therefore Fnet 0

12
Equilibrium

• Similarly, we know that Newtons 2nd Law (in
  rotational form) is
• So if an object is in rotational equilibrium,
  then L is constant and therefore tnet 0

13
Equilibrium

• So we can sum this up by saying that for an
  object to be in static equilibrium it must be
  that
• The vector sum of all of the external forces that
  act on the body must be zero
• The vector sum of all of the external torques
  that act on the body, measured about any possible
  point, must also be zero

14
Equilibrium

• If an object is just in equilibrium it must be
  that P and L are both constant but they are
  not necessarily zero
• For static equilibrium we also have the
  requirement that the linear momentum P of the
  body must be zero

15
Equilibrium

• Note that these are vector equations, so to be
  true they must hold along each axis of the
  coordinate system

Balance of Forces
Balance of Torques
16
Equilibrium

• We will simplify matters some by considering only
  situations where the following three conditions
  are true

Balance of Forces
Balance of Torques
17
Equilibrium

• Imagine a hockey puck moving across the ice (a
  frictionless surface) at a constant speed
• The puck is clearly in equilibrium, but it is not
  in static equilibrium as it is still moving
• Thus to be in static equilibrium an objects
  linear momentum P must also be zero

18
Checkpoint 1

• Given properly adjusted (nonzero) forces, in
  which of these 6 situations can the rod possibly
  be in static equilibrium?

19
The Center of Gravity

• The gravitational force on an extended body is
  the vector sum of the gravitational forces acting
  on all of the individual elements (the atoms) of
  the body
• Rather than considering all of those individual
  elements, we can say
• The gravitational force Fg on a body effectively
  acts at a single point, called the center of
  gravity (cog) of the body

20
The Center of Gravity

• By effectively we mean that, if gravity could
  be turned off on the individual elements of the
  body, it could be replaced by the sum of those
  forces (Fg ) acting at the center of gravity
• The result would be the same net force and torque
  (about any point) on the body

21
The Center of Gravity

• There has been an implicit assumption that we
  have made as we have gone along so far that the
  bodys center of gravity is at the same position
  as its center of mass
• This is true only if the acceleration of gravity
  g is the same everywhere for the body
• For everyday objects, this is true as g varies
  only slightly over the surface of the earth and
  only slightly with altitude

22
The Center of Gravity

• Imagine an extended object as shown where we have
  a mass element mi with the force of gravity Fgi
  migi acting on it over a moment arm xi
  relative to the origin O
• The result is the torque

23
The Center of Gravity

• The net torque for all of the mass elements of
  the body would then be
• Now lets look at the body where the total force
  of gravity is acting on the center of gravity

24
The Center of Gravity

• Here we have the force of gravity Fg acting at
  the center of gravity
• The moment arm is xcog and as a result the torque
  (relative to the origin O) is

25
The Center of Gravity

• We can see that the total gravitational force is
  nothing more than the sum of the gravitational
  forces on the individual mass elements mi
  thus
• So we can rewrite the previous equation to be

26
The Center of Gravity

• Now recall our definition for the center of
  gravity that the torque due to the force Fg
  acting at the center of gravity is equal to the
  net torque due to all of the forces Fgi acting
  on the individual mass elements
• This gives us

27
The Center of Gravity

• Substituting migi for Fgi gives us
• Now if we can assume that all of the
  accelerations gi are the same, we can cancel the
  term gi from each side of the equation

28
The Center of Gravity

• The summation on the left turns into nothing more
  than the mass of the object we therefore
  get
• But by definition, the right side is nothing more
  than the center of mass therefore

29
The Center of Gravity

• Checkpoint 2

30
Examples of Static EquilibriumSample Problem 12-1

• Suppose we have a uniform beam of length L and
  massm 1.8 kg which is at rest on two scales as
  shown
• A uniform block of mass M 2.7 kg is resting on
  the beam with its center a distance of L/4 from
  the beams left end
• What do the scales read?

31

• We begin with a free-body diagram
• We also concentrate the masses so they are
  acting at the indicated points

32
Sample Problem 12-1

• The balance of forces equation gives us

33
Sample Problem 12-1

• If we place the origin at the left end of the
  beam, the balance of torques equation gives us

34
Sample Problem 12-1

• We now have two equations with two unknowns
• Solving the 2nd (balance of torques) equation for
  Fr we have

35
Sample Problem 12-1

• Substituting that result into the balance of
  forces equation we get

36
Checkpoint 3

• The rod is in static equilibrium
• Can you find F1 and F2 by balancing the forces?
• To find F2 using a single equation, where must
  you place the axis of rotation?
• F2 turns out to be 65 N what is F1?

37
Sample Problem 12-3

• Here we have a safe that is hanging by a rope
  from a boom
• The boom consists of a hinged beam and a
  horizontal cable which connects the beam to the
  wall

38

• The safe has a massM 430 kg the beam has a
  mass m 85 kg
• The specified dimensions area 1.9 m and b
  2.5 m
• The mass of the cable and rope are negligible
• What is the tension Tc in the cable?

39

• In order to simplify our equations, we have
  placed the origin at the hinge point
• Starting with the balance of torques equation
  this time we get

40

• Entering the known values and solving for Tc we
  get

41

• Find the magnitude F of the net force on the beam
  from the hinge
• We start by using the balance of force equations
  first for the horizontal forces

42

• We know that since the boom is in static
  equilibrium, the horizontal forces must be
  balanced thusor

43

• We also know that the vertical forces must also
  be in balance thusor

44
Sample Problem 12-3

• Now that we have both Fh and Fv we can use the
  Pythagorean theorem to calculate the magnitude of
  the force on the beam from the hinge

45
Checkpoint 4

• The system is in static equilibrium
• The mass of the rod is 5 kg and it is 1 meter in
  length the angle between the rod and the rope is
  30º

46
Indeterminate Structures

• So far we have dealt with problems that allow for
  solutions with three independent equations (2
  balance of force and 1 balance of torque) in
  three unknowns
• But suppose a problem has more than 3 unknowns
  what happens then?
• Given the constraints we have at the moment the
  problem is not solvable it is indeterminate and
  the structure is called an indeterminate structure

47
Indeterminate Structures

• Any time we have more unknowns than equations we
  have an indeterminate problem
• These kinds of problems crop up all the time (not
  necessarily with elephants though)

48
Indeterminate Structures

• Another example is that of the forces (all
  different) on the four tires of a car that has
  been un-symmetrically loaded
• The reason we have a problem in these cases is
  because we have assumed (very quietly) that the
  object is perfectly rigid
• In reality however, no object is perfectly rigid
• All objects will deform however slightly when
  a force is applied to it

49
Indeterminate Structures

• Imagine a table where one leg is a little shorter
  than the other three
• With little or no weight on it the table will
  wobble until some enterprising person places some
  folded up paper under the short leg
• However if you were to place a very heavy weight
  on the table it would deform until all four legs
  were touching the ground and supporting some (not
  necessarily equal) portion of the weight

50
Indeterminate Structures

• This deformation of an object when force is
  applied is called elasticity and is what we shall
  study next

51
Checkpoint 5

• A horizontal uniform bar of weight 10 N hangs
  from two wires that exert forces F1 and F2
• Which arrangements, if any, are indeterminate?

52
Elasticity

• The atoms in a metallic solid (such as an iron
  nail) are arranged in a three-dimensional lattice
• Each atom has a well-defined equilibrium distance
  from each of its neighbors

53
Elasticity

• The inter-atomic forces between the atoms can be
  modeled as tiny springs
• Because the lattice is very rigid (the metal is
  not easily deformed), the spring constants are
  very large (e.g., the springs are very stiff)

54
Elasticity

• Other kinds of materials dont have this lattice
  arrangement
• Plastics for example are long, flexible chains of
  molecules where each chain is only loosely bound
  to its neighbors
• But back to our metals even though we might
  call it rigid we know that it is possible to
  get it change its dimensions a little by pulling
  or pushing on it or perhaps twisting or bending
  it

55
Elasticity

• These changes may be very slight it will depend
  on the material
• For example, a vertical steel rod 1 m long and 1
  cm in diameter can easily support the weight of a
  subcompact car and stretch only by 0.5 mm (0.05)
  in doing so
• And when the weight is removed the rod will
  return to its original length

56
Elasticity

• Now suppose we were to hang two subcompact cars
  from that same rod
• In this case the rod would lengthen even more,
  but this time it would be permanently stretched
  and would not return to its original length when
  the weight is removed

57
Elasticity

• Finally, suppose we were to hang three subcompact
  cars from that same rod
• In this case the rod will break but just before
  it does, the elongation of the rod will be less
  than 0.2
• Now lets look at the details

58
Elasticity

• There are three ways in which a solid may change
  its dimensions in reaction to an applied force
• The first is that it is stretched (or compressed)

59
Elasticity

• The second case is when the material has a force
  applied perpendicular to its axis such as we
  might deform a deck of cards or a book

60
Elasticity

• And the third case is when the force is applied
  uniformly from all directions such as by a
  fluid under high pressure

61
Elasticity

• Common to all three situations is that a stress
  a deforming force per unit area produces a
  strain a unit deformation
• More formally, the three cases are called
• Tensile stress,
• Shearing stress, and
• Hydraulic stress

62
Elasticity

• While the details of the three cases vary, it is
  true that over the range of engineering
  usefulness stress and strain are proportional
  to each other
• The constant of proportionality is called the
  modulus of elasticity, so that stress
  modulus x strain

63
Elasticity

• The figure at the right shows how the
  relationship between stress and strain is linear
  up to a point
• After that point (called the yield strength)
  things get more complicated
• The ultimate strength of a material is when it
  fails

64
Elasticity

• A test specimen, such as illustrated below, is
  used in a special machine to determine the
  characteristics of various materials

65
Tension and Compression

• For the first case where an object is stretched
  or compressed along one axis, the stress is
  defined as F/A where F is the magnitude of the
  force applied perpendicularly to the area A
• The strain (unit deformation) is defined as?L/L
  which you will note is a dimensionless quantity
• Strain is often expressed as a fraction
  (sometimes as a percentage) of the objects
  original length

66
Tension and Compression

• Expressed as a formula we havewhere E is the
  symbol used to represent the modulus of
  elasticity for tensile and compressive stresses
  it is called Youngs modulus

67
Tension and Compression

• Note that because strain is a dimensionless
  quantity the units of the modulus are the same as
  that of stress, e.g., force per unit area

68
Tension and Compression

• For our purposes, we will assume that Youngs
  modulus is the same for tension and compression
• Note however that does not mean that the yield
  strength and ultimate strength are the same in
  both modes
• Concrete for example is very strong in
  compression but considerably weaker in tension
  so much so that it is almost never used that way
• Properties of some common materials are shown in
  Table 12-1 on page 317

69
Shearing

• The situation for shearing forces is much the
  same as for tensile forces the difference is
  that the force vector lies in the plane of the
  area rather than being perpendicular to it

70
Shearing

• The strain in this case is the dimensionless
  quantity ?x/L where the dimensions are as shown
  at the right

71
Shearing

• Expressed as a formula we havewhere G is the
  symbol used to represent the shear modulus

72
Hydraulic Stress

• The third and final case is when the stress is
  from a fluid pressure p which is equivalent to
  force per unit area
• The strain in this case is ?V/V where V is the
  original volume of the object and ?V is the
  absolute value of the change in volume

73
Hydraulic Stress

• The modulus in this case case is represented by
  the symbol B and is called the bulk modulus
• The formula that expresses the relationship
  between the hydraulic pressure (the stress) and
  the change in volume (the strain) is

74
Hydraulic Stress

• The bulk modulus for water is 2.2 x 109 N/m2 for
  steel it is 16 x 1010 N/m2
• The pressure at the bottom of the Pacific ocean
  (4000 m) is 4 x 107 N/m2
• So the fractional compression of a volume of
  water at that depth is about 1.8, but for steel
  it is about 0.025
• Solids are generally less compressible than
  liquids due to their more rigid lattice structure

75
Elasticity Problems

• Sample Problem 12-5
• Sample Problem 12-6
• Checkpoint 6

76
Sample Problem 12-5

• A structural steel rod has a radiusR 9.5 mm
  and a length L 81 cm
• A 62 kN force stretches the rod along its length
• What are the stress on the rod, and the
  elongation and strain of the rod?

77
Sample Problem 12-5

• We will begin by assuming that the rod is clamped
  in some fashion at one end and that the force is
  applied uniformly across the face of the rod and
  parallel to the length of the rod
• The area of the face of the rod will beA pR2

78
Sample Problem 12-5

• We can now use (part of) the formula for tensile
  stress

79
Sample Problem 12-5

• Note that the yield strength for structural steel
  is 2.5x108 N/m2 so the force on this rod is
  dangerously close to breaking it
• To calculate the elongation (?L) we can now use
  the whole formula for tensile stress rearranged a
  bit

80
Sample Problem 12-5

• Plugging in the values we get

81
Sample Problem 12-5

• Finally, the strain is the ratio of the change in
  length to the original length, which results in

82
Sample Problem 12-6

• Back to our wobbly table
• Three of the tables legs are exactly 1.00 m in
  length the fourth leg is longer by d 0.50 mm
• A heavy steel cylinder with a mass M 290 kg is
  placed on the table (which has negligible mass by
  comparison) so that all four legs now touch the
  floor and the table no longer wobbles
• Assume that the tables surface doesnt flex and
  remains level

83
Sample Problem 12-6

• For this to happen the legs must be compressed by
  different amounts in order to remove the wobble
• The table legs are made of wood with a Youngs
  modulus E 1.3x1010 N/m2
• What are the magnitudes of the forces on the four
  table legs from the floor?

84
Sample Problem 12-6

• This is just like the picture we saw earlier
  except that we now have a steel cylinder instead
  of a pachyderm
• We assume that each of the short legs (lets call
  them 1 3) is compressed by the same amount, but
  that the long leg (4) must be compressed more

85
Sample Problem 12-6

• Lets call the compression of the three identical
  legs ?L3 and the compression of the fourth leg
  ?L4
• For the tabletop to be level, it must be that

86
Sample Problem 12-6

• We can use the formula which relates stress to
  strain and Youngs modulus to help us here it
  is rearranged a bit

87
Sample Problem 12-6

• Using the relationship we established earlier for
  the changes in length, we can then say that
• This is fine, but we have two unknowns (F4 and
  F3) so we need another equation

88
Sample Problem 12-6

• For the other equation we will first note that
  the system is in static equilibrium
• We can therefore use one of our equilibrium
  equations balance of forces
• In other words, the forces from the 3 shorter
  legs plus the force from the one longer leg must
  equal the gravitational force of the
  tablecylinder

89
Sample Problem 12-6

• Now we have two equations and two unknowns so
  we can go ahead and solve them
• We will start by solving for F4 in terms of F3

90
Sample Problem 12-6

• We can then substitute that into our other
  equation and solve for F3
• Some turning of the algebra crank is required,
  but you will eventually get to

91
Sample Problem 12-6

• Substituting in the supplied values yields
• Now substituting that result into the other
  equation gives us

92
Sample Problem 12-6

• Now that we know the forces, we can calculate the
  compression of the legs

93
Checkpoint 6

• We have a horizontal block suspended by two wires
  A and B which are identical except for their
  original lengths
• The COM is closer to wire B than to wire A

94
Checkpoint 6

• Measuring torques about the blocks COM, is the
  magnitude of the torque due to wire A greater
  than, equal to, or less than the magnitude of the
  torque due to wire B?

95
Checkpoint 6

• Which wire exerts more force on the block?
• If the wires are now equal in length, which one
  was originally shorter?

96
Next Class

• Homework Problems Chapter 125, 13, 21, 36, 41,
  81
• Read sections 15-1 through 15-9