Transcript Chapter 1

Chapter 1
Forces Maintaining Equilibrium or
Changing Motion
Objectives
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Define force
Classify forces
Define friction force
Define weight
Determine the resultant of two or more
forces
• Resolve a force into component forces
acting at right angles to each other
Objectives
• Determine whether an object is in static
equilibrium, if the forces acting on the
object are known
• Determine an unknown force acting on an
object, if all the other forces acting on the
object are known and the object is in static
equilibrium
What Are Forces?
• Forces enable us to start moving, stop
moving, and change directions
• We manipulate the forces acting on us to
maintain our balance in stationary
positions (e.g. cyclist, diver)
• Simply defined, a force is a push or a pull
• Forces are exerted by objects on other
objects
What Are Forces?
• Forces come in pairs; the force exerted by one
object on another is matched by an equal but
oppositely directed force exerted by the second
object on the first—Action and reaction
• Forces accelerate objects—Start, stop, speed
up, slow down, or change direction
• SI unit is N (Newton)
• 1N force is the force that would accelerate a 1kg
mass 1m/s2
• 1N = .225lb
• 1lb = 4.448N
What Are Forces?
• Describe the force a shot-putter exerts on a shot
at the instant shown in Figure 1.1
• Factors we want to know:
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Size
Point of application
Direction (line of action)
Sense (pushes or pulls along the line of action)
• Force is known as a vector
– Mathematical representation of anything described by
size or direction
Internal Forces
• Forces that act within the object or system
whose motion is being investigated
• Muscles pull on tendons, which pull on
bones
• At joints, bones push on cartilage
• Tensile forces—Pulling forces acting on
the ends of an internal structure
• Compressive forces—Pushing forces
acting on the ends of an internal structure
Internal Forces
• Sometimes the tensile or compressive forces
acting on a structure are greater than the
internal forces the structure can withstand—
Structure fails or breaks—Muscle pulls, tendon
ruptures, ligament tears, bone breaks
• Muscles can only produce internal forces—They
are incapable of producing changes in the
motion of the body without external forces
• Body is only able to change its motion if it can
push or pull against some external object (e.g.
ground reaction force)
External Forces
• Forces that act on an object as a result of its
interaction with the environment
• Non-Contact forces—Objects not touching (e.g.
gravity, magnetic, electrical)—Gravity is primarily
studied in sports biomechanics
• Gravity (g) accelerates objects downward at
9.81m/s2 no matter how large or small the
object, and any place on earth (ignoring air
resistance)
External Forces
• Weight—In biomechanics, force of gravity
acting on an object—Don’t confuse with
weight as commonly thought of being in
lbs from a bathroom scale
• W = mg
• W = weight (measured in N)
• m = mass (measured in kg)
• g = acceleration due to gravity
• How much do you weigh?
External Forces
• Contact forces—Occur between objects in contact with
each other
• Objects in contact can be solid or fluid (e.g. air
resistance and water resistance are fluid)
• Forces between solid objects studied primarily in sports
biomechanics—An athlete and some other object
(opponent, ground, implement)
• To put a shot, an athlete must be holding the shot
• To jump up, an athlete must be in contact with the
ground and push down
• To run forward, an athlete must push backward against
ground
External Forces
• Contact forces can be resolved into parts or
components
• Normal contact force (Normal reaction force)—
Line of action of the force is perpendicular to the
surfaces in contact
• Acts upward on runner and downward on
ground
• Friction—Line of action of friction is parallel to
the two surfaces in contact and opposes motion
or sliding between the surfaces
• Acts forward on runner and backward on ground
Friction
• Dry friction—Acts between the non-lubricated surfaces of
solid objects or rigid bodies and acts parallel to the
contact surfaces
• Arises due to interactions between the molecules of the
surfaces in contact
• Static friction—When dry friction acts between two
surfaces that are not moving relative to each other
• Limiting friction—Maximum static friction that develops
just before the two surfaces begin to slide
• Dynamic friction (Sliding friction, Kinetic friction)—When
dry friction acts between two surfaces that are moving
relative to each other
Friction and Normal Contact Force
• Self Experiment 1.1
• How does adding
books to the pile
cause the static
friction force to
increase?
• Friction force is
proportional to the
normal contact force
Friction and Normal Contact Force
• Self Experiment 1.2
• Is friction force only a
horizontal force?
• Is the normal contact
force always vertical
and related to the
weight of the object
that friction acts on?
Friction and Surface Area
• Self Experiment 1.3
• Does increasing or
decreasing area in
contact affect friction
force?
Friction and Surface Area
• Force pushing the surfaces together remains the same,
the force is spread over a greater area
• Pressure between the surfaces decreases (pressure
equals force/area)
• Individual forces pushing each of the molecules together
at the contact surfaces decreases
• Interactions between the molecules decreases
• Friction decreases
• An increase in surface area increases the number of
molecular interactions, but the decrease in pressure
decreases the magnitude of these interactions—The net
effect is zero and friction is unchanged
Friction and Contacting Materials
• Friction and Contacting Materials
• What about the nature of the materials in
contact?
– Rubber soled shoes vs. leather soled shoes
Friction and Contacting Materials
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Fs = µsR
F d = µ dR
Fs = static friction force
µs = coefficient of static friction
µd = coefficient of dynamic friction
– Number that accounts for the different effects
that materials have on friction
• R = normal contact force
Friction in Sport and Human
Movement
• Static friction is larger than dynamic friction
• More difficult to start an object moving than to
keep it moving
• Locomotion requires frictional forces
• Materials used for the soles of athletic shoes
have large coefficients of friction
– Wax skis to decrease coefficient of friction
– Tape, chalk, and sprays improve grip which increases
coefficient of friction between hands and implement
Addition of Forces: Force
Composition
• Several external forces can act on us in
sports activities
– Gravity, friction, normal contact force
– In most sport and exercise situations more
than one external force will act on an
individual
– How do we add these forces to determine
their effect on a person?
– What is the net force or resultant force?
Addition of Forces: Force
Composition
• Full description of force includes
magnitude and direction (force is a vector)
– Resultant force—Vector addition of two or
more forces
– Net force—Vector addition of all the external
forces acting on an object
Colinear Forces
• Same line of action—May act in the same direction or in opposite
directions
• Visually we can think of forces as arrows:
– Length of the shaft = magnitude of force
– Orientation of the arrow = line of action
– Arrow head = direction of action
• When forces act along the same line (colinear), they can
be added using regular algebraic addition
– Add forces to the right (positive) and forces to the left (negative)
• Sample Problem 1.1 (text p. 28)
– Add forces upward (positive) and forces downward (negative)
Concurrent Forces
• Do not act along the same line, but do act
through the same point
• Gymnast example (text p. 29)
– How large is the force of gravity that acts on the
gymnast?
– Force of gravity acting on an object is the object’s
weight (W = mg)
– What is the net external force acting on the gymnast?
– We can represent the forces graphically
Concurrent Forces
• Resultant directed “upward and slightly to
the left” (not very precise description)
• Calculate resultant horizontal and vertical
forces separately
• Left and downward (negative)
• Right and upward (positive)
Trigonometric Technique
• Right triangle—90° angle formed by
horizontal resultant force and vertical
resultant force
• Pythagorean theorem
– A2 + B2 = C2
– A and B represent the two sides that make up
the right angle and C represents the
hypotenuse (the side opposite the right angle)
– Solve for C to determine the resultant force
Trigonometric Technique
• Other relationships between sides of right
triangle
– If we know the lengths of any two sides, we
can determine the length of the other side,
and the size of the angle between the sides
– Conversely, if we know the length of one side
of a right triangle and the measurement of
one of the angles other than the right angle,
we can determine the lengths of the other
sides and the measurement of the other angle
Trigonometric Technique
• Relationships exist
between the lengths
of the sides of a right
triangle and the
angles in a right
triangle
Trigonometric Technique
• These relationships can be expressed as
ratios of one side to another for each size
of angle that may exist between two sides
of a right triangle
• Sinθ = opposite side/hypotenuse
• Cosθ = adjacent side/hypotenuse
• Tanθ = opposite side/adjacent side
• “Some Of His Children Are Having Trouble
Over Algebra”
Trigonometric Technique
• Theta (θ) represents the angle
• Opposite refers to the length of the side of the
triangle opposite the angle theta
• Adjacent refers to the length of the side of the
triangle adjacent to the angle theta
• Hypotenuse refers to the length of the side of
the triangle opposite the right angle
• Sin refers to sine
• Cos refers to cosine
• Tan refers to tangent
Trigonometric Technique
• If the sides of the right triangle are known,
then arcsine, arccosine, and arctangent
functions are used to compute the angle
– θ = arcsin (opposite/hypotenuse)
– θ = arccos (adjacent side/hypotenuse)
– θ = arctan (opposite side/adjacent side)
Trigonometric Technique
• Gymnast example
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θ = arctan (opposite side/adjacent side)
θ = arctan (50N/10N)
θ = arctan(5)—arctan usually abbreviated tan-1
θ = 78.7°
• Angles in a right triangle add up to 180°; one
angle is 90°, so the sum of the other two angles
is 90°
• In the gymnast example, what would the other
angle be?
Trigonometric Technique
• Sample Problem 1.2 (text p. 34)
– Use the Pythagorean theorem to compute the
size of the resultant force
– Use the arctangent function to determine the
angle of the resultant force with the horizontal
Resolution of Forces
• What if the forces acting on the object are
not colinear and do not act in a vertical or
horizontal direction?
– Shot-put (Figure 1.1)
– Athlete exerts a 100N force on a 4kg shot at
an angle of 60° above the horizontal
• What is the net force acting on the shot?
– First determine the weight of the shot
Resolution of Forces
• Weight of the shot (40N) acts vertically
(downward)
• Force from the athlete (100N) acts
horizontally and vertically (pushing upward
and forward)
– Calculate horizontal and vertical components
of 100N resultant force separately
Resolution of Forces
• Trigonometric technique
– The 100N force represents the hypotenuse
acting upward and to the right 60° above
horizontal
– The side of the triangle adjacent to the 60°
angle represents horizontal component
– Cosθ = adjacent side/hypotenuse
– (100N)Cos60 = adjacent side
– Horizontal force component = 50N
Resolution of Forces
• Side of the triangle opposite the 60° angle
represents the vertical component
– Sinθ = opposite side/hypotenuse
– (100N)Sin60 = opposite side
– Vertical force component = 86.6N
• Determine net force acting on the shot by adding
all the horizontal forces to get the resultant
horizontal force and then adding all the vertical
forces to get he resultant vertical force
Resolution of Forces
• Only horizontal force acting on the shot is 50N
• Two vertical forces acting on the shot
– Weight of the shot acting downward (-40N)
– Vertical component of the 100N force acting upward
(+86.6N)
– Add these forces to get 46.6N acting upward
• Use Pythagorean theorem to get the net force
acting on the shot (Figure 1.13)
– (50N)2 + (46.6N)2 = C2
– C = √4672N2
– C = 68.4N
Resolution of Forces
• Determine angle between hypotenuse
(68.4N) and adjacent (50N) sides
– θ = arctan (opposite side/adjacent side)
– θ = arctan (46.6N/50N)
– θ = 43°
• Sample Problem 1.3 (text p. 38)
– Calculate horizontal and vertical components
of biceps force separately
Static Equilibrium
• If an object is at rest then the forces acting
on the object are in equilibrium and the
object is described as being in a state of
static equilibrium
– Gymnastics coach may want to know how
strong a gymnast must be to hold a certain
position such as an iron cross
– We need to know what external forces are
acting on the gymnast
Free-Body Diagrams
• Example 1
– A 50kg woman in skates is standing still on
ice
– What external forces act on her?
– Force of gravity acts vertically (downward)
through the woman’s center of gravity—an
imaginary point in space through which the
force of gravity acts on an object
– Reaction force from ice acts vertically
(upward)
Free-Body Diagrams
• Free body diagram is a mechanical
representation of an athlete—Only the
object in question is drawn along with all of
the external forces that act on it
• Any horizontal forces?
– On ice, friction is very small, and because the
woman is standing still, the frictional forces
under her skates must be zero
Static Analysis
• If an object is not moving, it is in static
equilibrium; acceleration is zero, and the
sum of all external forces acting on the
object is zero (ΣF=0)
• Reaction force of ice exerted on skater
• ΣF = R + W = R + (-500N) = 0
• R = reaction force from the ice
• W = weight of skater
• R = 500N
Static Analysis
• Reaction force from the ice is equal to the
weight of the skater but acting in the
opposite direction
• Example 2
– 80kg weightlifter has lifted 100kg barbell over
his head and is holding it still; he and the
barbell are in static equilibrium
– What is the reaction force from the floor that
must act on the weightlifter’s feet?
Static Analysis
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Three external forces
1. Reaction force from the floor acting upward
2. Weight of the athlete acting downward
3. Weight of the barbell acting downward on
the athlete’s hands
Static Analysis
• Reaction force exerted by the hands on
the barbell
– ΣF = R + W = R + (-1000N) = 0
– R = 1000N
– Hands exert 1000N upward force on barbell
• Reaction force from floor exerted on
athlete
– ΣF = R + W = R + (-1800N) = 0
– R = 1800
Static Analysis
• Example 3
– A 20kg child is sitting on a swing, and the child’s
parent is exerting a 40N horizontal force on the child
to the right and 10N upward force; the child is not
moving and is in a state of static equilibrium
– What is the resultant force exerted by the swing on
the child?
– Use one equation for the horizontal forces (ΣFx = 0)
and one for the vertical forces (ΣFy = 0)
Static Analysis
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ΣFx = Rx + 40N = 0
Rx = -40N (left)
ΣFy = RY + 10N + (– 200N) (weight of child) = 0
RY = +190N (upward)
Determine the magnitude of the resultant using
the Pythagorean theorem
– (40N)2 + (190N)2 = C2
– C = √37,700N2
– C = 194N
Static Analysis
• Must describe the angle the force makes
with horizontal
• θ = arctan (opposite side/adjacent side)
• θ = arctan (190N/40N) = 78°
• Resultant force exerted by the swing on
the child is a force of 194N acting
backward and upward on the child at an
angle of 78° from the horizontal
Summary
• Forces are pushes or pulls
– Vector quantities described by size and direction of
action
– Use arrows to represent forces graphically
– Internal forces hold things together and cannot cause
changes in motion without external forces
– Most common external forces are gravity and contact
forces
– Friction and normal reaction force are two
components of a contact force
Summary
• Forces are added using vector addition
– Accomplished using graphical techniques or
algebra if the forces are resolved into
horizontal and vertical components
• If the net external force acting on an object
is zero, the object is standing still or
moving in a straight line at a constant
velocity and is in static equilibrium