Transcript No Slide Title

Chapter 4
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 (an acceleration of the body)
– Characteristics of a force





magnitude
Vector represented with an arrow
direction
point of application.
line of action
sense (push or pull along the line of action)
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, maintain integrity
– Garfield slide
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)
Friction

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- when surfaces are not moving
relative to each other
 Dynamic friction- when surfaces move
relative to each other

Friction: create graph on board

Maximum static friction - maximum amount of
friction that can be generated between two
static surfaces
» fixed value based on
– nature of materials in contact
– force holding the bodies together (normal contact force)

Dynamic friction - constant magnitude friction
during motion
» always less than maximum static friction
Reaction Force

Normal reaction force (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
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 is FUN

Coefficient of friction (mu) - value that serves
as an index of the interaction between two
surfaces in contact.
Examples of manipulating mu and Nc
shoe design
grips (gloves, tape, sprays, chalk)
skiing: decrease for speed, increase for safety
your examples???
Friction is FUN

Coefficient of friction (mu) - number that
serves as an index of the interaction between
two surfaces in contact.
Homework for next day: perform the 4 (four) self-experiments
on pages 53 to 56. Submit (next day) typed answers to the questions
posed in the experiments.
Ask for clarification of any experimental result that is confusing.
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
» ie. 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
» ie. Add force of 100N and 100N
– 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
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
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 4.8 & 4.9 in text)
» separate algebraic summation of horizontal and
vertical
– Pythagorean thereom to solve resultantl
Resolution of Forces

Forces are not colinear and not Hor & Vert?
» ie figure 4.1: 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
100N
wt
Click here for a
Quiz on the
concepts of
net force & friction
Trigonometry: SOH, CAH, TOA
Click on Trigonometry above to go to a Penn State website on trig
Key Equations:
sin  = opp/hyp
 cos  = adj/hyp
 tan  = opp/adj

 Pythagoras
 a2
+
b2
Pythagorean
Humour
=
hyp (c)
thereom
c2
opp

adj
Check this site.
Vector Composition
Sample:
  = 30 degrees
 adj = 100 N
 Find opp and hyp
hyp (c)
opp

adj
Vector Composition
use
 cos  = adj/hyp
 tan  = opp/adj

hyp (c)
opp

adj
Vector Composition




cos  = adj/hyp
cos 30 = 100/hyp
hyp = 100/.866
hyp = 115.47 N




hyp (c)
tan  = opp/adj
tan 30 = opp/100
opp = .5774 x 100
opp = 57.74 N
opp

adj
Vector Resolution
Sample
  = 35 degrees
 hyp = 120 N
 Find opp and adj
hyp (c)
opp

adj
Trigonometric Calculations
Use
 sin  = opp/hyp
 cos  = adj/hyp
hyp (c)
opp

adj
Vector Resolution




sin  = opp/hyp
sin 35 = opp/120
opp = 120 X .5736
opp = 68.83 N




hyp (c)
cos  = adj/hyp
cos 35 = adj/120
adj = 120 X .8192
adj = 98.30 N
opp

adj
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
Resolution of Forces

Forces are not colinear and not concurrent
» ie figure 4.1: forces on shot
– 100 N from shot-putters hand
– mass of shot = 4 kg = -40 N
» What is the resultant force on shot?
– Draw Components of 100N force


solve graphically: tedious & imprecise
trigonometric technique
100N
60o
wt
-40N
Recall

Concept of Net External Force
Recall
Concept of Net External Force
 Newton’s First Law of Motion

Recall
Concept of Net External Force
 Newton’s First Law of Motion

» acceleration = 0 if net external force = 0

Newton’s Second Law of Motion
Recall
Concept of Net External Force
 Newton’s First Law of Motion

» acceleration = 0 if net external force = 0

Newton’s Second Law of Motion
» a=F/m
Static equilibrium:
object is at rest (no motion)
forces are in equilibrium (sum to 0 in all directions)
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 (75 kg): at rest (a = 0) in front of class
 sagittal plane view
F=ma
F=0
 Wt & vGRF
Wt + vGRF = 0
 CofM, at feet
vGRF = - Wt
 750N, ?????
vGRF = - (-750)
-&+
vGRF = 750 N

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?

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=ma
F=0
Wt + bar + vGRF = 0
vGRF = - Wt - bar
vGRF = - (-800) - (-1000)
vGRF = + 1800 N
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?

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?

Fx = m ax
Fy = m ay