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Chapter 1 . Forces: Maintaining Equilibrium or Changing Motion. Force. Force : a push or pull acting on a body that causes or tends to cause a change in the linear motion of the body Characteristics of a force magnitude direction point of application. line of action

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Chapter 1

Chapter 1

Forces:

  • Maintaining Equilibrium or Changing Motion


Force

Force

  • Force: a push or pull acting on a body that causes or tends to cause a change in the linear motion of the body

    • Characteristics of a force

      • magnitude

      • direction

      • point of application.

      • line of action

  • Net Force:resultant force (overall effect of multiple forces acting on a body)

    • Example: push from side and front = at angle


Force1

Force

  • Free body diagram - sketch that shows a defined system in isolation with all the force vectors acting on the system.


Why is this impossible

Why is this impossible?


Classifying forces

Classifying Forces

  • Internal Force: acts within the object or system whose motion is being investigated

    • action / reaction forces both act on different parts of the system

      • tensile-internal pulling forces when the structure is under tension

      • compressive- internal pushing (squeezing) forces act on the ends of an internal structure

    • do not accelerate the body

      • Orientate segments, maintain structural integrity


Internal forces

Internal Forces

  • Examples

    • Contraction of muscles

    • Do not accelerate the body


Classifying forces1

Classifying Forces

  • External Force: acts on object as a result of interaction with the environment surrounding it

    • non-contact - occur even if objects are not touching each other

      • gravity, magnetic

    • contact - occur between objects in contact

      • fluid (air & water resistance)

      • reaction forces with another body (ground, implement)

        • vertical (normal) reaction force

          • acts perpendicular to bodies in contact

        • shear reaction force

          • acts parallel to surfaces in contact (friction)


Forces maintaining equilibrium or changing motion

F = ma

  • Force may also be defined as the product of a body's mass and the acceleration of that body resulting from the application of the force.

  • Units of force are units of mass multiplied by units of acceleration.


Units of force

Units of Force

  • Metric system (systeme internationale -SI)

    • Newton (N)

      • the amount of force necessary to accelerate a mass of 1 kg at 1 m/s2

  • English system

    • pound (lb)

      • the amount of force necessary to accelerate a mass of 1 slug at 1 ft/s2

      • equal to 4.45 N


Weight external force

Weight (external force)

  • Weight - the amount of gravitational force exerted on a body. wt=mag.

  • Since weight is a force, units of weight are units of force - either N or lb.

  • As the mass of a body increases, its weight increases proportionally.


Weight

Weight

  • The factor of proportionality is the acceleration of gravity, which is

    -9.81m/s2 or-32 ft/s2.

  • The negative sign indicates that the acceleration of gravity is directed downward or toward the center of the earth.


Weight1

Weight

  • On the moon or another planet with a different gravitational acceleration, a body's weight would be different, although its mass would remain the same.

    • Space station example


Weight2

Weight

  • Because weight is a force, it is also characterized by magnitude, direction, and point of application.

  • The direction in which weight acts is always toward the center of the earth.


Center of weight gravity

Center of Weight (Gravity)

  • The point at which weight is assumed to act on a body is the body's center of gravity.


Friction external force

Friction (external force)

  • Component of a contact force that acts parallel to the surface in contact

    • acts opposite to motion or motion tendency

    • reflects interaction between molecules in contact

    • reflects force “squeezing” surfaces together

    • acts at the area of contact between two surfaces


Forces maintaining equilibrium or changing motion

Static friction: surfaces not moving relative to each other

Maximum static friction: maximum amount of friction that can be generated between two static surfaces

Dynamic friction: surfaces move relative to each other

constant magnitude friction during motion

always less than maximum static friction

Friction


Forces maintaining equilibrium or changing motion

F =  N

Friction depends on

Nature of materials in contact ()

Force squeezing bodies together (N)

Known as the normal contact (reaction) force

Friction is FUN


Forces maintaining equilibrium or changing motion

Nature of materials in contact

Coefficient of friction ()

value serves as an index of the interaction between two surfaces in contact.

Coefficient of Friction ()


Factors affecting friction

Factors Affecting Friction

  • The greater the coefficient of friction, the greater the friction

  • The greater the normal contact force, the greater the friction


Forces maintaining equilibrium or changing motion

Normal reaction force (NRF)

AKA - normal contact force

force acting perpendicular to two surfaces in contact.

magnitude intentionally altered to increase or decrease the amount of friction present in a particular situation

football coach on sled

push (pull) upward to slide object

pivot turn on ball of foot

Reaction Force


Friction manipulation

Friction Manipulation

Examples of manipulating  and Nc

shoe design


Friction manipulation1

Friction Manipulation

  • Examples of manipulating  and

  • Nc

  • shoe design

  • grips (gloves, tape, sprays, chalk)

  • skiing: decrease for speed, increase for safety

  • curling

  • your examples???


Curling

Curling


Bode miller silver medalist 2002

Bode Miller, Silver Medalist 2002


Shark skin suits in swimming

Shark Skin Suits in Swimming


Friction and surface area

Friction and Surface Area

  • Friction force is proportional to the normal contact force

  • Friction is not affected by the size of the surface area in contact

    • normal contact force distributed over the area in contact

  • Friction is affected by the nature of the materials in contact


Friction and surface area1

Friction and Surface Area

  • With dry friction, the amount of surface area in contact does not impact the amount of friction.

Same force acting over more area

Same force acting over smaller area


Coefficient of friction

Coefficient of Friction

  • Hot, soft, rough surfaces have higher coefficients of friction

    • Tires, concrete, etc

  • Cold, hard, smooth surfaces have lower coefficients of friction

    • Ice, marble, etc


Friction

Friction

  • Calculate the force of friction when you slide on ice.

  • Given = Normal force of 1000N

  • C of F = 0.02


Solution

Solution

Calculate the force of friction when you slide on ice.

  • Given = Normal force of 1000N

  • C of F = 0.02

  • F = µ * N

  • F = (0.02) * (1000N)

  • F = 20 N


Force2

Force

  • Force: a push or pull acting on a body that causes or tends to cause a change in the linear motion of the body (an acceleration of the body)

    • Characteristics of a force

      • magnitude

      • direction

      • point of application

      • line of action

      • sense (push or pull along the line of action)

Vector represented with an arrow


Recall

Recall

  • Concept of Net External Force

Must add all forces acting on an object together


Free body diagram

Free body diagram

  • Free body diagram - sketch that shows a defined system in isolation with all the force vectors acting on the system

    • defined system: the body of interest

    • vector: arrow to represent a force

      • length: size of the force

      • tip: indicates direction

      • location: point of application


Free body diagram1

Free body diagram

Note: link contains practice problems

  • Me, at rest in front of class

    • sagittal plane view

  • What are the names of the forces?

  • How big are the forces?

  • What direction are the forces?

  • Where are the forces applied?


Free body diagram static analysis

Free body diagram ==>static analysis

Note: link contains practice problems

  • Me (mass= 75 kg): at rest (a = 0) in front of class (assume gravity at -10m/s/s)

    • sagittal plane view

  • Wt & vGRF

  • C of M, at feet

  • 750N, ?????

  • - & +

F = m a

F = 0

Wt + vGRF = 0

vGRF = - Wt

vGRF = - (-750)

vGRF = 750 N


Addition of forces calculating the net resultant force

Addition of Forces(calculating the net [resultant] force)

  • Net force = vector sum of all external forces acting on the object (body)

    • account for magnitude and direction

    • e.g., Add force of 100N and 200N


Addition of forces calculating the net resultant force1

Addition of Forces(calculating the net [resultant] force)

  • Net force = vector sum of all external forces acting on the object (body)

    • account for magnitude and direction

    • e.g., Add force of 100N and 200N

      • act in same direction???

      • Act in opposite direction???

      • Act orthogonal to each other???

      • Act at angles to each other???


Colinear forces

Colinear forces

  • Forces have the same line of action

  • May act in same or different directions

    • ie tug of war teammates: 100N, 200N, 400N

      • show force on rope graphically


Colinear forces1

Colinear forces

  • Forces have the same line of action

  • May act in same or different directions

    • ie tug of war teammates: 100N, 200N, 400N

    • tug of war opponents: 200N, 200N, 200N

      • show force on rope graphically


Colinear forces2

Colinear forces

  • Forces have the same line of action

  • May act in same or different directions

    • ie tug of war teammates: 100N, 200N, 400N

    • tug of war opponents: 200N, 200N, 200N

    • calculate resultant of the two teams

      • show force on rope graphically

      • calculate algebraically


Free body diagram static analysis1

Free body diagram ==>static analysis

  • Weightlifter (mass 80 kg)

  • 100 kg bar overhead

  • at rest (a = 0)

  • sagittal plane view

  • What are the forces?

  • Where are the forces applied?

  • How big are the forces?

  • What direction are the forces?

  • Free body diagrams

  • Weightlifter

  • bar

Fig 1.19,

p 42


Free body diagram static analysis2

Free body diagram ==>static analysis

  • Weightlifter (80 kg)

  • 100 kg bar overhead

  • at rest (a = 0)

  • sagittal plane view

  • What are the forces?

  • Where are the forces applied?

  • How big are the forces?

  • What direction are the forces?

  • Wt & vGRF

  • CofM, at feet

  • 800N, ?????

  • - & +


Free body diagram static analysis3

Free body diagram ==>static analysis

  • Weightlifter (80 kg)

  • 100 kg bar overhead

  • at rest (a = 0)

  • sagittal plane view

  • Wt, bar & vGRF

  • CofM, on hands, at feet

  • 800N, 1000N, ?????

  • - , -, +

F = m a

F = 0

Wt + bar + vGRF = 0

vGRF = - Wt - bar

vGRF = - (-800) - (-1000)

vGRF = + 1800 N


Concurrent forces

Concurrent Forces

  • Forces do not act along same line, but do act through the same point

  • ie gymnast jumps up to grab bar. Coach stops swinging by applying force to front and back of torso.

    • 20 N posterior directed push on front of torso

    • 30 N anterior directed push on back of torso

    • 550N force from bar on gymnast’s hands

    • gymnast mass 50 kg


Concurrent forces1

Concurrent Forces

  • gymnast hanging from grab bar. Coach applies force to front and back of torso to stop swing.

    • 20 N posterior directed push on front of torso

    • 30 N anterior directed push on back of torso

    • 550N force from bar on gymnast’s hands

    • gymnast mass 50 kg

Page 29 in book


Concurrent forces2

Concurrent Forces

  • gymnast hanging from grab bar. Coach applies force to front and back of torso to stop swing.

    • 20 N posterior directed push on front of torso

    • 30 N anterior directed push on back of torso

    • 550N force from bar on gymnast’s hands

    • gymnast mass 50 kg

  • What is the resultant force?

    • Tip to tail method (fig 1.8 & 1.9 in text)

    • separate algebraic summation of horizontal and negative forces

      • Pythagorean theorem to solve resultant magnitude

      • Inverse tangent to solve direction (angle)


Forces maintaining equilibrium or changing motion

Horizontal forces:

20N – 30N = -10N

Vertical forces:

-500N + 550N = 50N

Resultant force:

a2 + b2 = c2

(-10N)2 + (50N)2 = c2

(100N2) + (2500N2) = c2

2600N2 = c2

C = 51N


Quantifying kinetics

Quantifying Kinetics

  • Vector composition - process of determining a single vector from 2 or more vectors through vector addition.

  • Resultant - single vector that results from vector composition.


Quantifying kinetics1

Quantifying Kinetics

  • Vector resolution - breaking down a resultant vector into its horizontal and vertical components.

    • Graphic method.

    • Trigonometric method.

Graphic Method


Trigonometry soh cah toa

hyp (c)

opp

adj

Trigonometry: SOH, CAH, TOA

Key Equations:

  • sin  = opp/hyp

  • cos  = adj/hyp

  • tan  = opp/adj

  • Pythagoras theorem

  • a2 + b2 = c2

Check this site.


Vector composition

hyp (c)

opp

adj

Vector Composition

Sample:

 = 30 degrees

  • adj = 100 N

  • Find opp and hyp


Vector composition1

hyp (c)

opp

adj

Vector Composition

  • use

  • cos  = adj/hyp

  • tan  = opp/adj


Vector composition2

hyp (c)

opp

adj

Vector Composition

  • cos  = adj/hyp

  • cos 30 = 100/hyp

  • hyp = 100/.866

  • hyp = 115.47 N

  • tan  = opp/adj

  • tan 30 = opp/100

  • opp = .5774 x 100

  • opp = 57.74 N


Vector resolution

hyp (c)

opp

adj

Vector Resolution

Sample

  •  = 35 degrees

  • hyp = 120 N

  • Find opp and adj


Trigonometric calculations

hyp (c)

opp

adj

Trigonometric Calculations

Use

  • sin  = opp/hyp

  • cos  = adj/hyp


Vector resolution1

hyp (c)

opp

adj

Vector Resolution

  • sin  = opp/hyp

  • sin 35 = opp/120

  • opp = 120 X .5736

  • opp = 68.83 N

  • cos  = adj/hyp

  • cos 35 = adj/120

  • adj = 120 X .8192

  • adj = 98.30 N


Michelle kwan

Michelle Kwan

Note orientation of

stance leg.


Michelle kwan1

Michelle Kwan

Note orientation of

stance leg.


Michelle kwan2

Michelle Kwan

Note orientation of

stance leg.


Ground reaction force foot contact

Ground Reaction Force Foot Contact


Ground reaction force toe off

Ground Reaction ForceToe Off


Resolution of forces

Resolution of Forces

  • Forces are not colinear and not Hor & Vert?

    • ie figure 1.1: forces on shot

      • 100 N from shot-putters hand

      • mass of shot = 4 kg

    • What is the resultant force on shot?

      • Draw Components of 100N force

        • solve graphically: tedious & imprecise

        • trigonometric technique

wt

100N


Resolution of forces1

Resolution of Forces

  • Forces are not colinear and not concurrent

    • ie figure 1.1: forces on shot

      • 100 N from shot-putters hand

      • mass of shot = 4 kg

      • W = mg = (4kg)(-10m/s/s)= -40 N

    • What is the resultant force on shot?

      • Draw Components of 100N force

        • solve graphically: tedious & imprecise

        • trigonometric technique

wt

-40N

100N

60o


Solution1

Solution

  • Given

  • Hyp = 100N

  • Angle = 60 degrees

  • Sin angle = opp/hyp

  • Sin 60 = opp/hyp

  • Sin 60*hyp = opp

  • (0.866)*(100N)= 86.6N

  • Given

  • Hyp = 100N

  • Angle = 60 deg

  • Cos angle = adj/hyp

  • Cos 60 = adj/hyp

  • Cos 60*hyp = adj

  • (0.500)*(100N) = 50N


Free body diagram static analysis4

Free body diagram ==>static analysis

  • Child on swing (20 kg)

  • Mother’s force

    • 40 N horizontal

    • 10 N upward

  • at rest (a = 0)

  • force of swing on child?

Fig 1.2 p 43


Free body diagram static analysis5

Free body diagram ==>static analysis

  • Child on swing

  • (mass = 20 kg)

  • Mother’s force

    • 40 N horizontal

    • 10 N upward

  • at rest (a = 0)

  • force of swing on child?

Fx = m ax

Fy = m ay


Solution2

Solution

  • Fx = Rx + 40N = 0

  • Rx = -40N

  • Fy = Ry + 10N + (-200N) = 0

  • Ry = 190N


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