Transcript Forces and COM

Linear Kinetics Objectives
• Identify Newton’s laws of motion and gravitation and
describe practical illustrations of the laws
• Explain what factors affect friction and discuss the role of
friction in daily activities and sports
• Define impulse and momentum and explain the relationship
between them
• Explain what factors govern the outcome of a collision
between two bodies
• Discuss the interrelationship among mechanical work,
power, and energy
• Solve quantitative problems related to kinetic concepts
Linear Kinetics Outline - The Relationship
between force and motion
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Read Chapter 12 in text
Classification of forces
Types of forces encountered by humans
Force and motion relationships – three ways to look at it:
– Instantaneous effect – Newton’s law of acceleration (F=ma)
– Force applied through time (Impulse-momentum)(Ft = mv)
• Conservation of Momentum
– Force applied through distance (work-energy) (Fd = 1/2mv2)
• Conservation of Energy
• Self-study problems
– Sample problems: #2 p 392; #3 p 396, #4 p 397, #5 p 402, #6 p 405, #7 p 408
– Introductory problems, p 411: 1,3,5,7,8,10
• Homework problems (Due Thursday, April 20)
– Additional problems, p 412: 6,8,9
Effect of forces on the system (can be total
human body, or a part of the body)
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Action vs reaction
Internal vs external
Motive vs resistive
Force resolution – horizontal and vertical
components
• Simultaneous application of forces – determining
the net force through vector summation
External forces commonly encountered by
humans
• Gravitational force (weight = mg)
• Ground Reaction Force (GRF)(Figure 12-4, p 386)
– Vertical
– Horizontal (frictional)
• Frictional force (coefficient of friction) (pp 389-395)
• Elastic force (coefficient of restitution) (pp 399-402)
• Free body diagram - force graph (p 63)
Force Plates –
Measurement of ground
reaction forces
Coefficient
of friction,
resistance to
sliding:
Cfr = Frf
/Nof
Sample Prob
# 2, p 392
Coefficient of Restitution (COR)
• COR is a measure of the liveliness of an object
• When 2 objects collide:
• When one object is stationary,
this reduces to:
• An alternative way to measure COR
is to drop a ball and measure the ht
bounced compared to ht dropped:
Coefficient of Restitution (COR)
• COR of balls dropped or thrown at a rigid wooden
surface is shown here.
• COR increases
directly with
temperature and
inversely with
impact velocity.
Coefficient of Restitution (liveliness or bounciness)
Free body diagrams:
Instantaneous Effect of Force on
an Object
• Remember the concept of net force?
• Need to combine, or add forces, to
determine net force
• Newton’s third law of motion (F = ma)
• Inverse dynamics – estimating net forces
from the acceleration of an object
• Illustrations from Kreighbaum: Figures F.4,
F.5, and F.6 (pp 283-284)
Force Applied Through a Time: ImpulseMomentum Relationship (pp 295-399)
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Force applied through a time
Impulse - the area under the force-time curve
Momentum - total amount of movement (mass x velocity)
An impulse applied to an object will cause a change in its
momentum (Ft = mv)
• Conservation of momentum (collisions, or impacts)
– in a closed system, momentum will not change
– what is a closed system?
• It is a system where net forces are zero
• Example – horizontal movement of airborne objects, or where frictional
forces are negligible
– Example: Sample problem #3, p . 396
– Second example – slide # 19 (football player jumping and catching a ball)
Impulse: area
under forcetime curve
Net impulse (Ft)
produces
a change in
momentum
(mV)
Sample problem
#4, p 397
Vertical
impulse
While
Running:
Area under
Force-time
curve
Anterioposterior
(frictional)
component
of GRF: impulse
Is area under
Force-time curve
Positive and
Negative impulse
Are equal if
Horizontal comp
Of velocity is
constant
Conservation of momentum: when net impulse is zero
(i.e. the system is closed), momentum does not change
Also, sample
prob #3, p 396
Force Applied Through a Distance: Work,
Power, Energy (pp 403-409)
• Work - force X distance (Newton-meters, or Joules)
– On a bicycle: Work = F (2r X N)
– Running up stairs: Work = Weightd (slide 21)
– On a treadmill: Work = Weightd X per cent grade (slide 22)
• Power - work rate, or combination of strength and speed
(Newton-meters/second, or watts)
– On a treadmill: P = Weightd X per cent grade/ time
– On a bicycle: P = F (2r X N) / time
– Running up stairs: P = Weightd /time (See next slide)
• Energy - capacity to do work
– kinetic, the energy by virtue of movement (KE = 1/2 mv2 )
– gravitational potential, energy of position (PE = weight x height)
– elastic potential, or strain, energy of condition (PE = Fd)
Power running up stairs:
Work rate = (weight X vertical dist) ÷ time
Sample prob
#6, p 405
Work while running on treadmill:
Note that vertical distance equals the product of running
speed, time, and %grade.
Calculating Power on a Treadmill
• Problem: What is workload (power) of a 100 kg
man running on a treadmill at 10% grade at 4 m/s?
• Solution:
– Power = force x velocity
– Force is simply body weight, or 100 x 9.8 = 980 N
– Velocity is vertical velocity, or rate of climbing
• Rate of climbing = treadmill speed x percent grade = 4 m/s x .1 = .4 m/s
– Workload, workrate, or power = 980N X .4 m/s = 392 Watts
• Note: 4 m/s = 9 mph, or a 6 min, 40 sec mile
• Calculate your workload if you are running on a treadmill
set at 5% grade and 5 m/s.
– Answer for 200 lb wt (91 kg) is: 223 Watts
Conservation of Energy
• In some situations, total amount of mechanical energy
(potential + kinetic) does not change
– Stored elastic energy converted to kinetic energy
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diving board
bow (archery)
bending of pole in pole vault
landing on an elastic object (trampoline)
– Gravitational potential energy converted to kinetic energy
• Falling objects
• Videodisk on pole vault
Energy conservation – Case I : elastic potential (strain) and kinetic
Potential energy (FD) +
Kinetic energy (1/2mv2)
remains constant
Energy conservation – Case II : gravitational potential and kinetic
Potential energy
(Wh) + kinetic
energy (1/2mv2)
remains constant
Conservation of energy: gravitational potential and kinetic
Sample problem
#7, p 408
Falling objects and work-energy
relationship
• Problem:
– If a 2 kg object is dropped from a height of 1.5 meters, what will
be its velocity and kinetic energy when it hits the ground?
• Solution:
– Kinetic energy at impact (mgh)
equals the potential energy at drop
height (½ mv2)
• Potential energy at drop(mgh)
= 29.43 Nm
• Kinetic energy at impact
= 29.43 Nm; v = 5.42 m/s
Impacts and work-energy
relationship
From the previous problem, the kinetic energy at impact is 5.4 m/s. If the
object lands on a mat of .01 m in thickness, what force will the mat exert
on the ball while bringing KE to 0?
Solution: FD = (½ mv2); F = (½ mv2) ÷ D; F = 2,943 N
If the mat thickness is increased to .1 m in thickness (a factor of 10), then
F is reduced to 294 (a factor of 10).
Three ways to minimize impact
force of 2 colliding objects
• Force-time, or impulse-momentum relationship
(Ft = mv)
– Increase time through which force is applied
• Force-distance, or work-energy relationship (FD =
½ mv2)
– Increase distance through which force is applied
• Force-area, or pressure concept (P = F/a)
– Increase area over which force is applied
Revisiting the problem from week 1
regarding man falling from ledge
• A man fell from the railing of a walkway on a second-story
apartment building. He was found lying unconscious on
his back with his center of mass located 5 feet horizontally
from a second story walkway and railing. The top of the
railing was 21.5 ft above the ground. His blood alcohol
content was found to be .30 (inebriated) and he has no
memory of how he fell. In order to appraise liability for
the accident, we need to determine if the victim walked
into the railing or if he was sitting on the railing and fell
off. Can this be done from the information given? How?
(Hint: First, find time of flight, then find horizontal velocity, then try to
figure out what forces were required to obtain this velocity by using
Newton’s law of acceleration (F = ma)
Eleven steps in solving Formal
quantitative problems
• 1) Read the problem carefully.
• 2) List the given information.
• 3) List the desired (unknown) information for
which you are to solve.
• 4) Draw a diagram of the problem situation
showing the known and unknown information.
• 5) Write down formulas that may be of use.
• 6) Identify the formulae to use.
Solving Formal Quantitative Problems
• 7) If necessary, reread the problem statement to
determine whether any additional needed information
can be inferred
• 8) Carefully substitute the given information into the
formula.
• 9) Solve the equation to identify the unknown variable
(the desired information).
• 10) Check that the answer is both reasonable and
complete
• 11) Clearly box the answer.
• *Note: Be sure to provide the correct unit of
measurement with the answer.
Note: for impulse-momentum
Relationship, assume pushoff
time is .3 s and that, during
pushoff, peak force is 1.5 to 2
times average force. Also,
convert all distance units to
meters.
Linear Kinetics Formulae