Chapter 1

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# Forces: Maintaining Equilibrium or Changing Motion - PowerPoint PPT Presentation

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

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

Vector Composition
• 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