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Kinesiology 430- Exam #2- Chapter 10
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Gravity
Key Concepts:
Terms in this set (75)
Analysis Using Newton's Law
• Three approached for analysis
1. The effect of forces at an instant in time
i. Static and dynamic analysis
2. Th effects of forces applied over a period of time
3. The effects of forces applied over a distance
• Selection of analysis depend on the question
Momentum & Impulse
• Linear Momentum - "quantity of linear motion"
○ Any object in motion has mass and a non-zero velocity
○ Linear momentum (M) is calculated as the product of mass and linear velocity
§ M = mv
○ Units of measurement: (kg x m)/s
• Linear Impulse - "effect of a force over time"
○ Linear impulse is calculated as a product of average force and time over which it is applied:
§ Impulse = F x t
○ Units of measurement: N x s
• Application of an impulse causes a change in the momentum of a body
• If mass is constant, this means a change in velocity
Impulse-Momentum Relationship
Impulse-Momentum relationship:
○ F x t = m (change in velocity)
• Derived from Newton's 2nd Law
Impulse Examples
• We can produce the same change in velocity (assuming constant mass), in different ways:
○ Large force over small time
○ Small force over long time
• Consider:
○ Catching an egg (as we catch the egg apply smaller force over long time --> egg won't break)
○ Landing from a jump (land a jump with straight knees hurt since there's a large force over small time, bending knees increases the amount of time that force is applied --> hurts less)
Computing Impulse
• During "constant-speed" walking the area A_1 must be equal to area A_2
• In other words, the propulsive impulse must be equal and opposite to the braking impulse
• The integral must therefore be zero over the gait cycle
○ If A2 greater than A1 --> speeding up
○ If A1 greater than A2 --> slowing down
Must be equal to have constant speed walking
Computing Impulse 2
We do not always know the average force, but we can find the impulse using numerical integration (area under the curve)
The area A will be equal to F x t (Impulse)
Walking Impulse
• COM during walking, at the peak height (COM) it correlates with the lowest GRF, while the lowest COM height correlates the highest total GFR
Running Impulse
• Two graphs showing landing with heel strike (impact peak) versus landing on the forefoot
○ With heel strike, there's a large force applies over a small period time (rapid stop) --> impact peak
○ With forefoot, force is applied over a longer period of time so there's no impact peak
-Impulse is similar in both, but with heel strike there's an impact peak
Impulse Q and A
Momentum & Impulse 2
• Application: Determine Vertical Jump Performance
• Projectile Motion Review- Velocity at take-off dictates jump height
Squat Jump
• Start in squatted position
• Push up against ground and jump into the air
When F= 0, up in the air
Squat Jump- Free Body Diagrams
Squat Jump- GRF & BW
There's a portion where you are on your toes, about to take off where BW force is greater than GRF (At the end of push-off you have a net positive impulse)
Countermovement Jump
• Start standing than squat into squat jump
• Let's us jump higher versus a normal squat jump
Impulse in Countermovement Jump
• Same initial BW force
-Than GRF decreases less than BW as the person squats
Impulse Countermovement Jump- At which point are you at the bottom of the countermovement jump?
Impulse in Counter-movment Jump
• Negative change in momentum as person is moving down into their squat, v = negative, the next section where's there's a peak in GRF (green) change in the momentum in the opposite direction as negative velocity decreases to a stop (positive change in velocity/momentum) --> when the positive area = negative area --> bottom of countermovement jump
• The last part (before jump) there's movement in the positive direction before takeoff(positive impulse) , there's a slight slow down right before takeoff when you're on your tip-toes (small negative GRF)
-At the end of push-off you have a net positive impulse
Vertical GRF Graph-Using Impulse to Find Velocity
(F x t)GRF - (F x t)BW = m (delta Velocity)
1. Plot GRF
2. Subtract FBW from FGRF
3. Calculate impulse at each time point
4. Calculate net impulses (between vf and vi)
5. Divide by mass to get velocity
6.Find final velocity value
Vertical GRF Graph-Using Acceleration to find Velocity
(F x t)GRF - (F x t)BW = m (delta Velocity)
1. Plot GRF
2. Subtract BW from GRF
3. Divide by mass to get acceleration
4. Integrate to get velocity
5.Find final velocity value
Net Impulse - Squat Jump vs Countermovement Jump
• The scribbled portion is the net impulse
-The squat jump has a slightly higher net impulse (bigger are under the curve) which would indicate higher jump, but people choose the countermovement jump instead
• People can jump 5-10% higher with a countermovement than in a squat jump
• The net impulse after the countermovement is zero, so why does a countermovement help
• Some possible mechanisms:
○ Greater activation of muscles during propulsion
○ Recovery of stored elastic energy (in tendons)
○ CMJ is more familiar task then SJ
-People do not really just stay squatted for a few seconds, it feels less natural than CMJ
Momentum & Impulse Example #1
• A pitched ball with a mass of 0.2 kg reaches a catcher's glove travelling at a velocity of 28 m/s
○ How much momentum did the ball have?
§ Momentum = m x v = 0.2 kg x 28 m/s = 5.6 kg x m / s
○ How much impulse is required to stop the ball?
§ Impulse = f x t = m(vf- vi) =(0.2)x (0- 28) = -5.6 n x t
○ If it takes 0.03 sec to stop the ball, what was the average force applied by the catcher's glove?
-5.6/0.3 = -18.6667 N
Conservation of Momentum
• Principle of conservation of momentum:
○ In the absence of external forces, the total momentum of a given system remains constant
○ Thus, total momentum before and after an interaction remains constant
-May apply to the horizontal and/or vertical motion
Conservation of Momentum Example #1
Mi (V) = Mf (V)
40kg (5m/s) = (40 +3 kg) (V)
200 = (43)v
200/43=4.6512m/s
Auto Safety-Impulse Application
• Occupant restraint systems
○ Goal to decrease injury by reducing forces on body
○ Original systems built for 50% male or female only
○ Center for Applied Biomechanics University of Virginal Leaders in accident biomechanics for development of improved safety
-Air bags slows you down so it is less force applied on the bod
Impulse Question #2
M= 60 kg x 25 m/s = 1500 kg x m/s
Ft = F x (0.5s) = 1500 = -3000 N
Impulse Question #3
1500/0.004=-375000 N
Impulse Question #4
Impulse Question #5
Impulse Question #6
Friction
• When two surface are in contact, a friction force will be present if the two objects attempt to slide relative to each other
• The friction force always opposes the relative motion between the two objects
• Positive effects: foot on the ground, designing winter boots to maximize friction
-Undesirable effect: ski on snow (want to reduce friction to go fast)
Friction Example #1: Box sitting on the floor
-If there's no sliding force, there's no friction
-If you start to push the box, you can slide the box if you overcome static friction
Friction Example #2: Pushing Box with Small Force
If I push the box with a small force (less than max friction force), the box will not move so net F = 0, FSLIDE = FFRICTION
Friction Example #3: Pushing Box with Large Force
If I push hard enough, the box will being to slide (when I surpass the limiting value of friction, FLIM)
Factors Limiting Friction
• Factors influencing the limiting value of friction
○ Normal force presenting two objects together (R)
○ Properties of the two surfaces in contact (smooth, rough, etc)
§ Characterized by coefficient of friction (u )
○ Not affected by masses of objects (not directly)
○ Not affected by area of contact
• Mathematically, limiting value of friction is the product of coefficient of friction and normal force
FLIM = u x R
R= normal force
Static Friction
• Friction exists in two states:
○ Static - FSLIDE exactly counterbalanced by FFriction, up to limiting (maximal) value FLIM
-Siding occurs when FSLIDE exceeds FLIM
Dynamic Friction
• Dynamic Friction Force: is less than maximum static friction force
-In the dynamic case, FSLIDE is greater than or equal to FFRiction to keep something moving
Manipulating Friction
• For human movement, body size and movement technique (resulting in acceleration of the body) usually determine R (Normal force)
• Therefore, we manipulate coefficient of friction to modify FLIM
What are some examples where we try to manipulate (maximize or minimize) friction to our advantage?
• Maximize
○ Surface- shoe sole friction basketball or soccer
○ Winter boots with snowy sidewalk (prevent falls in elderly, etc)
○ ....
• Minimize
○ Curling rock- ice surface
○ Cross-country or downhill ski and snow - using wax
○ .....
Why do you wear different shoes for different activities or occasions?
• Fashion
• Comfort
• Friction
○ u(coefficient of friction) = 0.3 (dress shows on wood floor)
○ R = 725 N
○ FLIM = u x R = (0.30)(725) = 217.5 N
• Horizontal GRF can easily exceed 217.5 N in sport activities
• How will a "sticker" shoe affect your ability to "push off" and "cut" in basketball, volleyball, etc?
Stickier shoe will allow ore push off without slipping
Friction exmple: Case #1
W = 100 N
u = 0.5
R =?
Fslide= 35 N
What is Ffriction?
FLIM = 100 x 0.5 = 50N
Case I: Fslide = 35 N, which is less than FLIM, so FFriction = 35 N
Friction Example: Case 3
W = 100 N
u = 0.5
R =?
Udynamic= 0.4
what is Friction force?
Case III: uDynamic = 0.4
FFriction = 0.4 x 100 = 40 N
Friction Example: Case 2
W = 100 N
u = 0.5
R =?
Fslide= 75 N
What is Ffriction?
Case II: FSlide = 75 N, so box moves with FS > FLIM, Ffriction has to be less than FLIM (= < 50 N)
Dynamic vs Static Friction
Friction
• The dynamic friction force is lower than the static friction force; thus, there are really two different friction coefficients: us and uK
• Static case: FLIM = us x R
○ There is no sliding FSLIDE less than or equal to FLIM
○ And FFRICTION = FLIM
• Dynamic case: FFRICTION = uK x R (K stands for kinetic)
○ There is sliding FSLIDE > FLIM
-And FFRICTION less than or equal to FSLIDE and FFRICTION < FLIM
Friction Example#2
•A football player pushes a 670 N tackling sled where uS = 0.73 and mK = 0.58 between the sled and the grass.
•How hard must the player push to get the sled moving?
670 x 0.73 = 489.1 N
How hard must he push to keep it moving?
670 x 0.58=388.6 N
Friction Example #3: Friction on an incline
Friction
• Calculating friction force for an object on an incline
• Given:
○ W = 100 N
○ u = 0.5
○ Theta = 40 degrees
○ R = W cos (theta)
§ Parallel Force
§ = (100 N) (0.766) = 76.6 N
○ FLIM = u x R = 0.5 x 76.6 = 38.3 N
FSLIDE = W sin (Theta) = (100) x sin (40) = 64.28 N --> it will slide
Friction Example #4: Pushing a desk
It is easier to pulls a desk then push it
• When you push, you usually have a downward component to the applied force. This increases the normal force, and hence the fiction force
Friction Example #5: Pulling a desk
It is easier to pulls a desk then push it
When you pull, you usually have an upward component to the applied force. This decreases the normal force and hence the frictional force.
Work
• Another way of expressing the effect of a force
○ Mechanically, work is done on an object when a force causes a change in position
○ Work done on a body by a force is the product of the force magnitude and the displacement of the body in the direction of the force
W = F x d
○ Units of measurement: N x m or joule (J)
1 N x m = 1 J
Work : Force and Displacement Along Same Line
Work : Force and Displacement NOT Along Same Line
• Only the component of force in direction of motion (F x cos (theta) does work
Vertical component does not matter in this case
Work Example #1
• A simple example - weight lifting
○ Lifting a 750 N barbell 0.4 m from the floor
W = 750 x 0.4 = 300 J
Work Example #2: Negative Work
• What about lowering the 750 N barbell down to the floor (0.4 m below)?
W = (750 N) x (-0.4 m) =-300 J
Positive Work
• Force and displacement in same direction
Associated with concentric muscle activity
Negative Work
• Force and displacement in opposite direction
Associated with eccentric muscle activity
Work and Displacement
• Work is a scalar quantity, but it can still be positive or negative
• The sign relates to increase (+) or decrease (-) in the mechanical energy of the system
○ (+) increasing the energy of a system, (-) decreases the energy of the system
• What if you can't get the barbell to move
○ W = (750N) x (0 m) = 0 J
-If there is no displacement then mechanical work is zero, but there still may be a physiological cost
Work Example #3: Stairs
• Effectively raising body weight d, some vertical by height of the stairs
• A person weighing 700 N climb a flight of stairs containing twenty 25 cm steps
• Average force needed to raise body weight?
○ 700 N
• Total vertical displacement?
○ d = 20 x 25 = 500 cm = 5 m
W = (700 N) x (5 m) = 3,500 J
Work Example #4: Work on an incline
W = 100 x cos(35) x 4 = 327.66 J
Work Example #5
W= (9.81 m/s^2) x (10 kg) x (4 m) = 392.4 J
Work Example #6
1. NO, no displacment
2. yes
3. YES
4. NO, no force being applied, constant speed, frictionless surface
Mechanical Work
• "Direct" measures - ex.) experimental methods
○ To quantify work done system must measure
1. Force and 2. Displacement
○ Bike/rowing ergometer
§ Determine
□ Moment of force produced by flywheel braking system
□ Number of rotations
• W = fd = M (theta) Units Nm or Joules
○ Examples: Monark Bicycle Ergometer
○ Measures
§ Force from flywheel braking system
§ Radius of flywheel (25.46 cm)
§ Distance travelled - rotations (revs)
W = F (N) x Radius (m) x Rotations (revolutions) x 2pi (radians/rev)
Work Example #7: Bicycle Ergometer
• Work done on ergometer if person pedal against a 24.5 N workload for 20 s at a rate of 60 revolutions per second. Each revolution is equivalent to 6 m of linear motion.
Power
• Power is the rate at which work is performed
○ Power = work / change in time = w / (delta t)
○ Power = force x disp / change in time = F x d / (change in time)
• Since v = d/ change in time power = F x v
• Unit of measurement: J/s or watts (W)
Power Example #1
• Previous stair climbing example
○ W = 3500 J
○ Case 1 ; t = 20 s
○ Case 2 : t = 5 s
• Power is an indicator of the intensity of the task
Components of Human Power Production
• Strength (force producing ability) combined with speed of movement (muscle shortening/lengthening)
• What does it mean to be powerful?
○ Cars- main advantage of high hp engine is not in producing a higher top speed but rather in generation high acceleration (ex. 0-60 mph in short time)
○ Human sprinter- high power output capability is especially important to success
§ Tied to ability to accelerate to top speed
§ Capability of producing high muscle forces with relatively high speeds of muscle shortening
Power: Muscle Force-Velocity
• Power = Force x Velocity
Maximum muscle power is reached at about 33% of peak shortening velocity, at which point muscle force production is considerable lower than in an isometric contraction
Power- Muscle Force
At a given speed of shortening, a muscle with more fast-twitch fibers can produce more power than a muscle with more slow-twitch fibers
Power- Muscle Power and Age
Age seems to limit muscle power over time despite fitness level
Power- Efficiency Formula
• The human body has an efficiency of approximately 25% (0.25), though it differs from person-to-person and from task-to-task
○ Efficiency = mech work/met energy = mech power/ met power
• Thus we can estimate metabolic energy from mechanical work and efficiency
Energy
Energy is the capacity to do work
• Kinetic Energy (KE): energy of motion, if you're moving
○ KE = 1/2 mv2
• Potential energy (PE): energy stored by position
○ PE = wt x h = m x g x h
• Strain energy (SE): energy stored by deformation
○ SE = 1/2 k (delta X)2
○ Delta X = amount of deformation, elastic bands,
Units are the same as for work: J (joules) or N x m
Law of Conservation of Mechanical Energy: gravity the only external force
Law of Conservation of Mechanical Energy: when gravity is the only external force a body's mechanical energy remains constant
• KE + PE = Constant
• Mechanical energy is conserved in a pendulum
• KE and PE both fluctuate, but their sum remains constant
Law of Conservation: Mechanical Energy Example#1
PE = mgh = 9.81 x 10 x 2=196.2 J = KE
Law of Conservation: Mechanical Energy Example#2
196.2 J = KE = 1/2m(v^2) = 1/2(10)v^2
196.2/5=39.24 = v^2
V= 6.3 m/s
Law of Conservation: Mechanical Energy Example#3
PE = mgh = 10 x 9.81 x 1=98.1
98.1 = PE= KE = 1/2mv^2
98.1/5=19.62 = v^2
V= 4.4 m/s
Walking- Inverted Pendulum
• Walking has been molded as the motion of an inverted pendulum
• If the body functioned as a true inverted pendulum, no work would be needed to walk
○ The exchange between KE and PE in walking is only about 65% (mech energy is not fully conserved)
○ Need muscles to do some work to keep propagating forward
-This has been used as an indicator of how much mechanical work muscles need to do in walking
Work-Energy
• When a force other than gravity does work on the object, there is a change in mechanical energy
The work done by a force is equal to the change in total energy that it produces on the object
Work-Energy Example 1
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