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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

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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

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

• Example: push from side and front = at angle

### Force

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

### 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

• Examples

• Contraction of muscles

• Do not accelerate the body

### 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)

### 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

• 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 - 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

• 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.

### 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

### 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)

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

### 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

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

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

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

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

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

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

Examples of manipulating  and Nc

shoe design

### Friction Manipulation

• Examples of manipulating  and

• Nc

• shoe design

• grips (gloves, tape, sprays, chalk)

• skiing: decrease for speed, increase for safety

• curling

### 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 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

• 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

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

• Given = Normal force of 1000N

• C of F = 0.02

### 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

### 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

• Concept of Net External Force

Must add all forces acting on an object together

### 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 diagram

• 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

• 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)

• 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] 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

• 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 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 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 analysis

• Weightlifter (mass 80 kg)

• 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 analysis

• Weightlifter (80 kg)

• 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 analysis

• Weightlifter (80 kg)

• 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

• 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 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 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)

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

• 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 Kinetics

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

• Graphic method.

• Trigonometric method.

Graphic Method

hyp (c)

opp

### Trigonometry: SOH, CAH, TOA

Key Equations:

• sin  = opp/hyp

• Pythagoras theorem

• a2 + b2 = c2

Check this site.

hyp (c)

opp

### Vector Composition

Sample:

 = 30 degrees

• Find opp and hyp

hyp (c)

opp

• use

hyp (c)

opp

### Vector Composition

• cos 30 = 100/hyp

• hyp = 100/.866

• hyp = 115.47 N

• tan 30 = opp/100

• opp = .5774 x 100

• opp = 57.74 N

hyp (c)

opp

### Vector Resolution

Sample

•  = 35 degrees

• hyp = 120 N

hyp (c)

opp

### Trigonometric Calculations

Use

• sin  = opp/hyp

hyp (c)

opp

### Vector Resolution

• sin  = opp/hyp

• sin 35 = opp/120

• opp = 120 X .5736

• opp = 68.83 N

• adj = 120 X .8192

### Michelle Kwan

Note orientation of

stance leg.

### Michelle Kwan

Note orientation of

stance leg.

### Michelle Kwan

Note orientation of

stance leg.

### 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 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

### 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

• (0.500)*(100N) = 50N

### 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 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

### Solution

• Fx = Rx + 40N = 0

• Rx = -40N

• Fy = Ry + 10N + (-200N) = 0

• Ry = 190N