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Equilibrium
Chapter 4
Chapter Topics
• Discuss equilibrium, including:
– The concept of static equilibrium (SE)
– Force and moment conditions for SE
– Center of mass
– Segmental method for human center of mass
– Hydrostatics and flotation
Bodies at Rest
• A body at rest or in uniform motion is in
equilibrium
– External forces must give zero resultant.
Behold, a new definition for Force!
• A vector quantity that tends to cause movement in
material objects
Types of Equilibrium
• Stable
– A slight disturbance generates a restoring
force to return the equilibrium
• Unstable
– A slight disturbance leads to an increasing
departure from equilibrium
• Neutral
– A disturbance simply moves the object to a
new position
Stable/Unstable/Neutral?
Friction
• Many common objects are in equilibrium because of the
effects of friction forces
• Example?
Friction and Equilibrium
• Objects at rest on a surface are held by
static friction.
• Surfaces moving with respect to each other
are met with sliding friction or kinetic
friction.
• Which is greater?
Friction and Equilibrium cont’d
• Examples of friction:
– Slowing of a golf ball across a green
– Stationary object on an inclined plane
– Walking and not falling over
Friction and Equilibrium cont’d
• Examples of low friction:
– Ice skating
– Skiing
Coefficient of Friction
• Depends on the surfaces of both of the bodies in contact
with one another
• Simple experiment: at what angle does an object start to
slide down a slope
– Mechanical principle:
Question
• Is friction essential for a body to be in equilibrium?
Moments
• Moment of a force: a measure of the turning
effect of a force
Moments of Force
• Moment of force about a point is the tendency of
the force to turn the body to which it is applied
about that point
– Also known as torque
– F d
– d = length of moment arm
• Perpendicular distance between point and line of
action of the force
– Given in Nm
Example (page 81)
• A man supports a weight of 250 N with his
arms at right angles to his body. He holds
the weight with both hands level with his
shoulders. If is arms are 75 cm long, what is
the moment of the force?
Example cont’d
• Moment = F x d
= 250 N x 0.75 m
= 187.5 Nm
BE CAREFUL!
• The book’s answer is incomplete. A moment is a vector,
so the answer should have a direction!
Practice
The following problems would be useful to review as
you prepare for the first exam.
Page 49: Problems 1, 2, 4, 6
Page 50: Problems 7, 8, 11b, 13
Page 73: Problem 1
Page 74: Problem 5
Page 75: Problems 11, 13, 17a, 17b
Page 104: Problems 1, 2
Page 106: Problem 8
Answers are in the back of the book
Point of Application of Resultant Force
Where does the resultant force act?
• Solution comes from sum of moments for equilibrium
Where does the resultant force act?
• What one force at what one location would produce
equilibrium by balancing all the distributed forces?
Sum of moments
• Choose a point, and sum the clockwise and ccw moments
about that point. For example, sum moments about point
A, taking the opposite sense of P+Q to assume
equilibrium
(P+Q) x AC = Q x AB
Example and Case Study, pg. 82-
83
Center of Gravity (CoG)
• A point within or near the object through
which the resultant weight of the object
passes
Center of Mass (CoM)
• A point at which the object's mass can be
assumed, for many purposes, to be
concentrated
Whole Body CoM
• How would you calculate it?
• Giovani Alfonso Borelli (1608-1679)
• Case study 4.3
Consistency of CoM as a % of height
• Page 85:
• Height range: 158-188 cm
• CoM range 54.4-55.89 %ht
Force Couples
• What’s the sum of the horizontal forces?
• Do they therefore have no mechanical effect?
Couples
• A pair of equal parallel forces acting in
opposite senses, and not in the same line, is
called a couple
• Object will tend to rotate and is therefore not
in equilibrium
Falling Over
CoM of a Stationary Body
• Objects will balance on pivots if CoG is
directly over the pivot.
• Works mainly for straight objects.
• Example: a cricket bat
CoM of a Stationary Body cont’d
• Position of CoG for a bat:
%100.1
L
L
Similar expression for body segments
• total arm COM = 0.53 from proximal
– GH to ulnar styloid
• total leg COM = ____ from proximal
– Greater trochanter to medial malleolus
Equilibrium Under Three Forces
• Parallel forces:
– The sum of the forces must be zero
– Equilibrium requires that BOTH ΣF and ΣM each
equal zero
Equilibrium Under Three Forces
cont’d
• A uniform beam AB, 6 m long, weighing 400
N, is supported at the end A at point C. Point
C is 2 m from B. Find the reaction forces at
supports A and C.
Equilibrium Under Three Forces
cont’d
• R1 + R2 = 400 N
• ΣM about A
Moment due to 400N force at G:
400 x AG = 400 x 3 = 1200 Nm CW
Moment due to R2 at C:
R2 x AC = R2 x 4m CCW
Equilibrium Under Three Forces
cont’d
• Equate the moments (for equilibrium)
1200 = 4R2
R2 = 1200/4 = 300 N
R1 = 400 - R2 = 400 - 300 = 100 N
Now, solve the problem by summing
moments to determine R1
Equilibrium Under Three Forces
cont’d
• Nonparallel forces:
– Consider three forces D, E, and F in equilibrium.
Equilibrium Under Three Forces
cont’d
• Triangle forces rule:
– Three nonparallel forces in equilibrium
– Represented in size and direction by three sides
of the triangle taken in order
Alternative
• Measure the horizontal and vertical components of
each vector
Hydrostatics and Flotation
Pressure as a Function of Depth
• Increasing depth gives increasing pressure.
• Liquid density is:
• Pressure at depth h in N/m2
given by:
ρ kg/m3
P=ρgh
Upthrust on Immersed Body
• Pressure increases with depth.
• The underside of the object experiences
greater force than the top side.
Upthrust on Immersed Body cont’d
• Principle of Archimedes:
– When a solid body is wholly or partially
immersed in fluid, it experiences an upthrust
equal to the weight of the mass of displaced
fluid.
Upthrust on Immersed Body cont’d
• For floating body in water:
– Upthrust force = Weight of body
– U = W
• For sinking body in water:
– Upthrust force < Weight of body
– U < W
Factors: Density and Shape
• Less dense than water: always float
• More dense than water: only float if shaped such that the
object can displace at least its own weight in water
Specific Gravity
• SG = Mass of certain volume of substance
Mass of equal volume of water
• SG = Weight of certain volume of substance
Weight of equal volume of water
Specific Gravity cont’d
• Water at temperature 4°C
Density is 999.97 kg/m3
• Human body: close, but not homogenous!
Center of Mass in Humans
Estimating Center of Mass
• Use the segmental method if you know:
– Position of the end points of all of the body’s
segments
– Mass of each segment
– Location of CoM within each segment
Estimating Center of Mass cont’d
• 14 body segments:
– Trunk
– Head and neck
– Right and left thighs
– Right and left lower legs
Estimating Center of Mass cont’d
• 14 body segments cont’d
– Right and left feet
– Right and left upper arms
– Right and left lower arms
– Right and left hands
• Each segment has its own CoM
• Calcuation is based on the moment attributable to the
weight of each segment about the x and y axes.
Free-body Diagrams
• Include forces acting on the body only
• Exclude forces that the body exerts on its
surroundings and internal forces
• What is the system?
• How can you determine
R1 & R2 ?
• Draw the FBD of:
– The bucket
– The left arm
– The head
Calculation of Joint Moments
• Just like with summing forces with a free-
body diagram, calculate and sum moments
• Consider Figure 4-20.
FBD: foot segment ΣM about a
Summary
• Certain physical conditions are necessary to maintain
static equilibrium
– Force balance: ΣF=0
– Moment balance: ΣM=0
• Additional principles associated with equilibrium
– Center of mass
– Hydrostatics

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equilibrium diagrams

  • 2. Chapter Topics • Discuss equilibrium, including: – The concept of static equilibrium (SE) – Force and moment conditions for SE – Center of mass – Segmental method for human center of mass – Hydrostatics and flotation
  • 3. Bodies at Rest • A body at rest or in uniform motion is in equilibrium – External forces must give zero resultant.
  • 4. Behold, a new definition for Force! • A vector quantity that tends to cause movement in material objects
  • 5. Types of Equilibrium • Stable – A slight disturbance generates a restoring force to return the equilibrium • Unstable – A slight disturbance leads to an increasing departure from equilibrium • Neutral – A disturbance simply moves the object to a new position
  • 7. Friction • Many common objects are in equilibrium because of the effects of friction forces • Example?
  • 8. Friction and Equilibrium • Objects at rest on a surface are held by static friction. • Surfaces moving with respect to each other are met with sliding friction or kinetic friction. • Which is greater?
  • 9. Friction and Equilibrium cont’d • Examples of friction: – Slowing of a golf ball across a green – Stationary object on an inclined plane – Walking and not falling over
  • 10. Friction and Equilibrium cont’d • Examples of low friction: – Ice skating – Skiing
  • 11. Coefficient of Friction • Depends on the surfaces of both of the bodies in contact with one another • Simple experiment: at what angle does an object start to slide down a slope – Mechanical principle:
  • 12. Question • Is friction essential for a body to be in equilibrium?
  • 13. Moments • Moment of a force: a measure of the turning effect of a force
  • 14. Moments of Force • Moment of force about a point is the tendency of the force to turn the body to which it is applied about that point – Also known as torque – F d – d = length of moment arm • Perpendicular distance between point and line of action of the force – Given in Nm
  • 15.
  • 16. Example (page 81) • A man supports a weight of 250 N with his arms at right angles to his body. He holds the weight with both hands level with his shoulders. If is arms are 75 cm long, what is the moment of the force?
  • 17. Example cont’d • Moment = F x d = 250 N x 0.75 m = 187.5 Nm
  • 18. BE CAREFUL! • The book’s answer is incomplete. A moment is a vector, so the answer should have a direction!
  • 19. Practice The following problems would be useful to review as you prepare for the first exam. Page 49: Problems 1, 2, 4, 6 Page 50: Problems 7, 8, 11b, 13 Page 73: Problem 1 Page 74: Problem 5 Page 75: Problems 11, 13, 17a, 17b Page 104: Problems 1, 2 Page 106: Problem 8 Answers are in the back of the book
  • 20. Point of Application of Resultant Force
  • 21. Where does the resultant force act? • Solution comes from sum of moments for equilibrium
  • 22. Where does the resultant force act? • What one force at what one location would produce equilibrium by balancing all the distributed forces?
  • 23. Sum of moments • Choose a point, and sum the clockwise and ccw moments about that point. For example, sum moments about point A, taking the opposite sense of P+Q to assume equilibrium (P+Q) x AC = Q x AB
  • 24. Example and Case Study, pg. 82- 83
  • 25. Center of Gravity (CoG) • A point within or near the object through which the resultant weight of the object passes
  • 26. Center of Mass (CoM) • A point at which the object's mass can be assumed, for many purposes, to be concentrated
  • 27. Whole Body CoM • How would you calculate it? • Giovani Alfonso Borelli (1608-1679) • Case study 4.3
  • 28.
  • 29. Consistency of CoM as a % of height • Page 85: • Height range: 158-188 cm • CoM range 54.4-55.89 %ht
  • 30. Force Couples • What’s the sum of the horizontal forces? • Do they therefore have no mechanical effect?
  • 31. Couples • A pair of equal parallel forces acting in opposite senses, and not in the same line, is called a couple • Object will tend to rotate and is therefore not in equilibrium
  • 33. CoM of a Stationary Body • Objects will balance on pivots if CoG is directly over the pivot. • Works mainly for straight objects. • Example: a cricket bat
  • 34.
  • 35. CoM of a Stationary Body cont’d • Position of CoG for a bat: %100.1 L L
  • 36. Similar expression for body segments • total arm COM = 0.53 from proximal – GH to ulnar styloid • total leg COM = ____ from proximal – Greater trochanter to medial malleolus
  • 37. Equilibrium Under Three Forces • Parallel forces: – The sum of the forces must be zero – Equilibrium requires that BOTH ΣF and ΣM each equal zero
  • 38. Equilibrium Under Three Forces cont’d • A uniform beam AB, 6 m long, weighing 400 N, is supported at the end A at point C. Point C is 2 m from B. Find the reaction forces at supports A and C.
  • 39.
  • 40. Equilibrium Under Three Forces cont’d • R1 + R2 = 400 N • ΣM about A Moment due to 400N force at G: 400 x AG = 400 x 3 = 1200 Nm CW Moment due to R2 at C: R2 x AC = R2 x 4m CCW
  • 41. Equilibrium Under Three Forces cont’d • Equate the moments (for equilibrium) 1200 = 4R2 R2 = 1200/4 = 300 N R1 = 400 - R2 = 400 - 300 = 100 N
  • 42. Now, solve the problem by summing moments to determine R1
  • 43. Equilibrium Under Three Forces cont’d • Nonparallel forces: – Consider three forces D, E, and F in equilibrium.
  • 44.
  • 45. Equilibrium Under Three Forces cont’d • Triangle forces rule: – Three nonparallel forces in equilibrium – Represented in size and direction by three sides of the triangle taken in order
  • 46. Alternative • Measure the horizontal and vertical components of each vector
  • 48. Pressure as a Function of Depth • Increasing depth gives increasing pressure. • Liquid density is: • Pressure at depth h in N/m2 given by: ρ kg/m3 P=ρgh
  • 49. Upthrust on Immersed Body • Pressure increases with depth. • The underside of the object experiences greater force than the top side.
  • 50.
  • 51. Upthrust on Immersed Body cont’d • Principle of Archimedes: – When a solid body is wholly or partially immersed in fluid, it experiences an upthrust equal to the weight of the mass of displaced fluid.
  • 52. Upthrust on Immersed Body cont’d • For floating body in water: – Upthrust force = Weight of body – U = W • For sinking body in water: – Upthrust force < Weight of body – U < W
  • 53. Factors: Density and Shape • Less dense than water: always float • More dense than water: only float if shaped such that the object can displace at least its own weight in water
  • 54. Specific Gravity • SG = Mass of certain volume of substance Mass of equal volume of water • SG = Weight of certain volume of substance Weight of equal volume of water
  • 55. Specific Gravity cont’d • Water at temperature 4°C Density is 999.97 kg/m3 • Human body: close, but not homogenous!
  • 56. Center of Mass in Humans
  • 57. Estimating Center of Mass • Use the segmental method if you know: – Position of the end points of all of the body’s segments – Mass of each segment – Location of CoM within each segment
  • 58. Estimating Center of Mass cont’d • 14 body segments: – Trunk – Head and neck – Right and left thighs – Right and left lower legs
  • 59. Estimating Center of Mass cont’d • 14 body segments cont’d – Right and left feet – Right and left upper arms – Right and left lower arms – Right and left hands • Each segment has its own CoM
  • 60. • Calcuation is based on the moment attributable to the weight of each segment about the x and y axes.
  • 61.
  • 62.
  • 63.
  • 64.
  • 65. Free-body Diagrams • Include forces acting on the body only • Exclude forces that the body exerts on its surroundings and internal forces
  • 66. • What is the system? • How can you determine R1 & R2 ?
  • 67. • Draw the FBD of: – The bucket – The left arm – The head
  • 68. Calculation of Joint Moments • Just like with summing forces with a free- body diagram, calculate and sum moments • Consider Figure 4-20.
  • 69. FBD: foot segment ΣM about a
  • 70. Summary • Certain physical conditions are necessary to maintain static equilibrium – Force balance: ΣF=0 – Moment balance: ΣM=0 • Additional principles associated with equilibrium – Center of mass – Hydrostatics