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.
Why is this impossible?
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
Friction
 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 is FUN
F =  N
Friction depends on
 Nature of materials in contact ()
 Force squeezing bodies together
(N)

Known as the normal contact
(reaction) force
Coefficient of Friction ()
Nature of materials in contact
 Coefficient
of friction ()
 value
serves as an index of the
interaction between two surfaces in
contact.
F

N
Factors Affecting Friction
 The greater the coefficient of friction, the greater the
friction
 The greater the normal contact force, the greater the
friction
Reaction Force
 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
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
•your examples???
Curling
Bode Miller, Silver Medalist 2002
Shark Skin Suits in Swimming
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
Vector represented with an arrow
 sense (push or pull along the line of action)
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?
Note: link contains
practice problems

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=ma
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)
 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
a) Weightlifter
b) bar
Fig 1.19,
p 42
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
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
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 + b 2 = c 2
(-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
Trigonometry: SOH, CAH, TOA
Key Equations:
 sin  = opp/hyp
 cos  = adj/hyp
 tan  = opp/adj
Pythagoras theorem
 a2 + b2 = c2
hyp (c)
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
 tan  = opp/adj
 cos 30 = 100/hyp
 tan 30 = opp/100
 hyp = 100/.866
 opp = .5774 x 100
 hyp = 115.47 N
 opp = 57.74 N
hyp (c)
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
 cos  = adj/hyp
 sin 35 = opp/120
 cos 35 = adj/120
 opp = 120 X .5736
 adj = 120 X .8192
 opp = 68.83 N
 adj = 98.30 N
hyp (c)
opp

adj
Michelle Kwan
Note orientation of
stance leg.
Michelle Kwan
Note orientation of
stance leg.
Michelle Kwan
Note orientation of
stance leg.
Ground Reaction Force
Foot Contact
Ground Reaction Force
Toe Off
Resolution of Forces
 Forces are not colinear and not Hor & Vert?
 ie figure 1.1: forces on shot
100N
 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
Resolution of Forces
 Forces are not colinear and not concurrent
 ie figure 1.1: forces on shot
100N
 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
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
 Cos angle = adj/hyp
 Cos 60 = adj/hyp
 Cos 60*hyp = adj
 (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