EquilibriumChapter 4

EquilibriumChapter 4

Chapter TopicsChapter 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 RestBodies 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!Behold, a new definition for Force!

• A vector quantity that tends to cause movement in material objects

Types of EquilibriumTypes 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?Stable/Unstable/Neutral?

FrictionFriction

• Many common objects are in equilibrium because of the effects of friction forces

• Example?

Friction and EquilibriumFriction 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’dFriction 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’dFriction and Equilibrium cont’d

• Examples of low friction:

– Ice skating

– Skiing

Coefficient of FrictionCoefficient 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:

QuestionQuestion

• Is friction essential for a body to be in equilibrium?

MomentsMoments

• Moment of a force: a measure of the turning effect of a force

Moments of ForceMoments 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)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’dExample cont’d

• Moment = F x d

= 250 N x 0.75 m

= 187.5 Nm

BE CAREFUL!BE CAREFUL!

• The book’s answer is incomplete. A moment is a vector, so the answer should have a direction!

PracticePracticeThe 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 ForcePoint of Application of Resultant Force

Where does the resultant force act?Where does the resultant force act?

• Solution comes from sum of moments for equilibrium

Where does the resultant force act?Where does the resultant force act?

• What one force at what one location would produce equilibrium by balancing all the distributed forces?

Sum of momentsSum 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

Example and Case Study, pg. 82-83

Center of Gravity (CoG)Center of Gravity (CoG)

• A point within or near the object through which the resultant weight of the object passes

Center of Mass (CoM)Center of Mass (CoM)

• A point at which the object's mass can be assumed, for many purposes, to be concentrated

Whole Body CoMWhole Body CoM

• How would you calculate it?

• Giovani Alfonso Borelli (1608-1679)

• Case study 4.3

Consistency of CoM as a % of heightConsistency of CoM as a % of height

• Page 85:

• Height range: 158-188 cm

• CoM range 54.4-55.89 %ht

Force CouplesForce Couples

• What’s the sum of the horizontal forces?

• Do they therefore have no mechanical effect?

CouplesCouples

• 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 OverFalling Over

CoM of a Stationary BodyCoM 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’dCoM of a Stationary Body cont’d

• Position of CoG for a bat:

%100.1L

L

Similar expression for body segmentsSimilar 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 ForcesEquilibrium Under Three Forces

• Parallel forces:

– The sum of the forces must be zero

– Equilibrium requires that BOTH F and M each equal zero

Equilibrium Under Three Forces cont’dEquilibrium 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’dEquilibrium 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’dEquilibrium 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

Now, solve the problem by summing moments to determine R1

Equilibrium Under Three Forces cont’dEquilibrium Under Three Forces cont’d

• Nonparallel forces:

– Consider three forces D, E, and F in equilibrium.

Equilibrium Under Three Forces cont’dEquilibrium 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

AlternativeAlternative

• Measure the horizontal and vertical components of each vector

Hydrostatics and FlotationHydrostatics and Flotation

Pressure as a Function of DepthPressure 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 BodyUpthrust on Immersed Body

• Pressure increases with depth.

• The underside of the object experiences greater force than the top side.

Upthrust on Immersed Body cont’dUpthrust 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’dUpthrust 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 ShapeFactors: 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 GravitySpecific 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’dSpecific Gravity cont’d

• Water at temperature 4°C

Density is 999.97 kg/m3

• Human body: close, but not homogenous!

Center of Mass in HumansCenter of Mass in Humans

Estimating Center of MassEstimating 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’dEstimating 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’dEstimating 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 DiagramsFree-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 MomentsCalculation 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 aFBD: foot segment ΣM about a

SummarySummary

• 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