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

• 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)

• 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: Impulse-Momentum Relationship (pp 295-399)

• 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: areaunder force-time curve

Net impulse (Ft) produces a change in momentum (mV)

Sample problem #4, p 397

Vertical impulse While Running: Area underForce-timecurve

Anterioposterior(frictional) component of GRF: impulseIs area under Force-time curvePositive andNegative impulseAre equal ifHorizontal compOf velocity isconstant

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 (2r 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 (2r 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

• 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) + kineticenergy (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-momentumRelationship, 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