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UBC SCHOOL OF HUMAN KINETICS

HKIN 363Mechanics and Kinetics

RESOURCE NOTES

2004

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I. COURSE INFORMATION............................................................................................................................................4

COURSE OUTLINE..............................................................................................................................................................5

DATE.....................................................................................................................................................................................8

LABORATORY EXERCISE.............................................................................................................................................8

II. VIRTUAL BIOMECHANICS: THE BIOMECHANICS 363 WEBPAGE............................................................9

SIGNING ONTO THE VIRTUAL LAB..................................................................................................................................10

III. BIOMECHANICS METHODS ................................................................................................................................11

KINEMATICS METHODS...................................................................................................................................................12RECORDING PHASE..........................................................................................................................................................12

Equipment Selection ..................................................................................................................................................12Camera Positioning and Orientation........................................................................................................................12Marker Placement .....................................................................................................................................................13Light Placement .........................................................................................................................................................13Film Calibration ........................................................................................................................................................13

DIGITIZING PHASE...........................................................................................................................................................14Linear Kinematics......................................................................................................................................................14Angular Kinematics ...................................................................................................................................................15

KINETICS METHODS........................................................................................................................................................17FORCE PLATFORM KINETIC DATA ACQUISITION ...........................................................................................................17

IV. TIPS ON USING EXCEL..........................................................................................................................................23

ABOUT EXCEL .................................................................................................................................................................24BASIC EXCEL FUNCTIONS ...............................................................................................................................................24

Opening Excel ...........................................................................................................................................................24Creating a Worksheet ...............................................................................................................................................24Opening a Worksheet ...............................................................................................................................................24Selecting Cells ...........................................................................................................................................................25Selecting Non-Adjacent Cells ..................................................................................................................................25Calculations...............................................................................................................................................................25Correcting Cells ........................................................................................................................................................26Copy/Paste .................................................................................................................................................................26Filling Cells ...............................................................................................................................................................27Graphing....................................................................................................................................................................27

COPYING AN EXCEL CHART INTO A WORD WORKSHEET ..............................................................................................29PRINTING .........................................................................................................................................................................30

DIFFERENTIATION IN EXCEL (ESTIMATING SLOPES OF LINES) ...............................................................30

V. TIPS ON USING MICROSOFT WORD ..................................................................................................................34

OPENING MICROSOFT WORD..........................................................................................................................................35CREATING A WORKSHEET...............................................................................................................................................35OPENING A WORKSHEET.................................................................................................................................................35

VI. PROBLEM SET QUESTIONS.................................................................................................................................36

PROBLEM SET #1: REVIEW PROBLEMS...........................................................................................................................37PROBLEM SET #2: DIFFERENTIATION.............................................................................................................................39PROBLEM SET #3: LINEAR KINEMATICS ........................................................................................................................41PROBLEM SET #4: PROJECTILES .....................................................................................................................................42PROBLEM SET #5: ANGULAR KINEMATICS ....................................................................................................................43PROBLEM SET #6: LINEAR KINETICS..............................................................................................................................45PROBLEM SET #7: ANGULAR KINETICS #1.....................................................................................................................47

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PROBLEM SET #8: ANGULAR KINETICS #2.....................................................................................................................49PROBLEM SET #9: WORK, POWER, AND ENERGY..........................................................................................................50

VII. PROBLEM SET ANSWERS...................................................................................................................................52

SOLUTIONS FOR PROBLEM SET #1: REVIEW PROBLEMS................................................................................................53SOLUTIONS FOR PROBLEM SET #3: LINEAR KINEMATICS .............................................................................................57SOLUTIONS FOR PROBLEM SET #4: PROJECTILES ..........................................................................................................60SOLUTIONS FOR PROBLEM SET #5: ANGULAR KINEMATICS .........................................................................................64SOLUTIONS FOR PROBLEM SET #6: LINEAR KINETICS...................................................................................................67SOLUTIONS FOR PROBLEM SET #7: ANGULAR KINETICS #1..........................................................................................71SOLUTIONS FOR PROBLEM SET #8: ANGULAR KINETICS #2..........................................................................................76SOLUTIONS FOR PROBLEM SET #9: WORK, POWER AND ENERGY ................................................................................79

VIII. BIOMECHANICS TERMINOLOGY ..................................................................................................................83

IX. REFERENCES............................................................................................................................................................87

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I. Course Information

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The University of British ColumbiaSchool of Human Kinetics

Human Kinetics 363Course Outline

Summary:This objective of this course is to build on the principles of physics acquired in Human Kinetics163, high-school Physics or entry-level university Physics and apply them to a quantitativeanalysis of human movement. Examples of movement will include those pertaining to exercise,sport, and physical activity in addition to more general activities such as walking and in therehabilitation environment. The student should gain an understanding of the use of a quantitativeanalysis to explain how mechanical principles govern human motion. At the completion of thiscourse it is desired that each student be able to: 1) to understand how 2D rigid body dynamicscan be used to quantify human motion, (2) understand the cause and effect relationship betweenforce and linear and angular motion, and (3) perform mathematical analysis of complex humanmotion in two dimensions.

Dates: Lectures: Tuesday and Thursday, 9:30 am – 11:00 am MacMillan 166 (2357 MainMall)

Laboratories: L1A Monday 8:30 am,L1B Tuesday 3:30 pm,L1C Wednesday 4:00 pm,L1E Friday 8:30 am.

Each lab session is two hours long. We will meet at either Osborne or WMG room 120.

Instructor: Dr. David Sanderson, room 29, War Memorial GymOffice hours: Tuesday and Thursday 8:30 am or by appointment

Teaching Assistants: Tamika Heiden (tamika@interchange.ubc.ca)Natalie Vanicek (nvanicek@interchange.ubc.ca)

Office hours: By appointment

Prerequisites:Human Kinetics 163, either all of HKIN 290, 291 or (b) all of ANAT 390,391.

Textbook (Required):Title: Biomechanical Basis of Human MovementAuthors: J. Hamill and K.M. KnutzenPublisher: Lippincott Williams and Wilkins

WebCT Resources:There are considerable resources on the WebCT site for this course. You must have an WebCTaccount access this course. You can go through the elearning site (http://www.elearning.ubc.ca)or from the “My UBC” portal (http://my.ubc.ca).

Course Learning Objectives:1. Understand and use the concept of a free-body diagram as it applies to human movement.

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2. Be able to derive and solve equations of human motion in 2 dimensions.3. Demonstrate and ability to interpret graphs and simple models which are used to explain

human movement.4. Describe which tools are used to acquire human movement data and show an understanding

of their efficacy.5. Apply knowledge of applied anatomy to describe human movement and motor skills in both

anatomical and mechanical terms6. Apply knowledge of biomechanical principles to the analysis of skilled and unskilled

performances. Have demonstrated personal and social responsibility towards classparticipation.

7. Be able to facilitate active learning, critical thinking, and problem solving skills in thequalitative analysis of human movement.

Course Content (Specific Learning Objectives):1. Quantification of human movement description - Kinematics

The interrelationship between displacement, velocity and acceleration, both linear andangular, will be used to quantify human movement. Multi-segment motions will be the focus,including running and jumping. Attention to methods of data acquisition and reduction willbe important.It is expected that students will demonstrate an ability to use vector analysis to solve problems in determiningvelocities, displacement, and time of travel of objects. Specifically, it is expected that students will:• gather and organize data, produce and interpret graphs, and determine relationships between kinematic

variables in 2D• identify situations involving the use of kinematics and use classical mechanics to quantify the kinematics• solve problems involving: displacement, initial velocity, final velocity, average velocity, acceleration, time

2. Forces that change motion - KineticsUsing Newton's Laws, the effect of external forces on human motion will be explored.Impulse and momentum, in linear and angular terms, will be used to explore complex humanmovements. Examination of data acquisition and reduction will be made and used to acquiredata for analysis.It is expected that students will analyse forces acting on an object and predict their effects on it. Specifically, itis expected that students will:• solve problems involving: force, mass, acceleration, friction, coefficient of friction, normal and shear

forces, impulse, momentum, angular impulse, angular momentum, moment of inertia• construct free-body diagrams for objects in various situations and use them to solve problems involving

balanced or unbalanced forces, objects on inclined surfaces• define work and power, identify where these variables are useful in assessing human motion and solve

problems involving: force, displacement, work, power, and efficiency• differentiate between kinetic energy and gravitational potential energy and give examples of each and solve

problems involving: rotational and translational kinetic energy, potential energy and spring energy• state the law of conservation of momentum and determine whether a collision is elastic or inelastic• analyse conservation of momentum in two dimensions

3. Aerodynamic and fluid forces and their effects on human movementThis section will deal with those movements that occur in other mediums such as water, inthe case of swimming, and for those projectiles that exhibit aerodynamic movement patternssuch as discus and javelin throwing.It is expected that students will analyse forces acting on an object that arise because of its movement throughair or water. Specifically, it is expected that students will:

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• identify drag (surface, skin, form), lift, thrust and how they affect motion.• identify how Magnus force is created• explain how objects can be made to change their direction as they move through space• explain buoyancy and solve problems using buoyancy, gravity, thrust and lift• explain how swimmers can use drag forces to move through the water

4. Analysis of human movementThis section will build on both the current course and the preceding course and put thestudent into a position to complete a sophisticated description/analysis of human movement.The student will be able to explore movements in many domains, including sports,rehabilitation, and ergonomics.It is expected that students will analyse forces acting on an object and predict their effects on it. Specifically, itis expected that students will:• identify workplace and community situations involving Newton's laws• Apply mechanical principles and anatomical knowledge to describe motion and to describe control of that

motion

Course Structure:

Lecture Format. The lecture component will be two 90-minute lectures. Rather than all thelectures being didactic there will also be group discussion on topics presented by theinstructor. Class participation component will entail analysis of movements providedprimarily on video tape but also through photographs and Power Point presentations. Therewill be time allocated within the lecture to address the issues of analysis andcoaching/teaching components. There will be an opportunity to compare each presentationand form a class consensus. The instructor will facilitate discussion. The group work ismodelled after the discussion by McKay and Emmison, 1995)1.

Laboratory Format. There will be a two-hour laboratory component. There is a laboratory slotfor four days and maximum enrolment in the course will be restricted by the number ofstudents who can fit into the laboratory room. The lab work will comprise a series of 6 labs.Some of these will be done on computer in the War Memorial Gym microcomputer lab(room 120). Others will involve participation in a gym or out-of-doors. Three computerprograms will be used during this course – Microsoft Excel, Microsoft Word, and HMATechnology’s HU-M-AN. By now you will be familiar with Excel and Word. HU-M-AN is aprogram that allows you to digitise video images. In the fashion you will be able to createnumeric data from video clips and use these data to compute joint, segment angles, inaddition to computing joint and segment angular velocity and acceleration. All labs willinclude quantitative and qualitative analysis. During these sessions, students will acquire thenecessary skills that will enable them to perform quantitative analysis of human movement.Each lab requires a written report. The format for the written report can be found on theWEBCT page (http://www.elearning.ubc.ca/login) under resources. Essentially, the reportwill be no longer than 4 pages and will be done in a format specific for a scientificconference.

1 McKay, J. and Emmison, M. (1995). Using learner-centred learning (LCL) in undergraduate sociology courses.

Australian and New Zealand Journal of Sociology 31(3)102.

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Week Date Laboratory Exercise Location Problemset

1 Jan-5 No labs none2 Jan-12 Lab 1: Lab tour and HU-M-AN practice Computer Lab 13 Jan-19 Lab 2:100 m Sprint Data Collection Field Hockey pitch 24 Jan-26 Lab 2:100 m Sprint Data Analysis Computer Lab 35 Feb-2 Lab 3: Sprint Start Data Collection Osborne Gym A 46 Feb-9 Lab 3: Sprint Start Data Analysis Computer Lab

Feb-16 READINGBREAK

7 Feb-23 Lab 4: Vertical Jump Data Collection Osborne Gym A 58 Mar-1 Lab 4: Vertical Jump Data Analysis Computer Lab 69 Mar-8 Lab 5: Muscle Torque Data Collection Osborne Gym A 710 Mar-15 Lab 5: Muscle Torque Data Analysis Computer Lab 811 Mar-22 Lab 6: Rowing Data Collection Osborne Gym A 912 Mar-29 Lab 6: Rowing Data Analysis Computer Lab13 Apr-5 No Labs

Evaluation Profile

Learning objective Method Value

1, 2, 3, 4 Written examinations 60 marksMid-term 20 marksFinal 40 marks

5, 6, 7 Labs 40 marks5 labs each worth 8 marks

Total 100 marks

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II. Virtual Biomechanics: The Biomechanics 363Webpage

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To go to the Virtual Biomechanics Lab, use either Netscape (3.0 or newer) or Internet Explorer.You must go to the respective site and then follow the instructions if you need to downloadeither browser.

Signing onto the Virtual Lab

Go to Remote Address: http://www.elearning.ubc.ca and logon (or select the logon option).You will need an interchange account to access this site. Select “Course Listing”, select “HumanKinetics”, Select “HK363” and sign in with your UserID and password.

363 HOMEPAGE: VIRTUAL BIOMECHANICS

Once you have logged onto the 363 Webpage, there are 6 options for you to choose from.

Virtual Labs Takes you to the labs and notes sectionsThis is where some of your labs arePops up in a new browser window

Resources Contains helpful resources including background notes, problem setswith answers, course readings, and exam hintsPops up in a new browser window

Password Allows you to change your password from you student number (default)to something else

Quizzes Takes you to a variety of multiple choice questions for each of the 5sections covered in this courseUseful for short answer exam questions and quick conceptsAnswers are returned immediately

Bulletin Board An interactive bulletin board that can be used to ask the instructor/TAquestionsAlso enables students to interact with one another in order to solveproblems

Links Useful links to biomechanics sites worldwideCheck it out…you may be surprised!

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III. Biomechanics Methods

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Kinematics Methods

At the UBC Biomechanics Lab, we can analyze virtually any body movement sequence usingour equipment and motion measurement system. There are two general stages to quantifyingmovement: recording and digitizing.

Recording Phase

In the recording phase, the movement is filmed by a video camera. Before recording can begin,the subject and equipment must be prepared. This includes activity location, camera positioningand orientation, light placement, film calibration, and marker placement.

Equipment SelectionIn this example, we will be analyzing the kinematicsof walking. We have chosen to use a treadmill tosimulate the walking motion. Although ourequipment enables us to measure motion in threedimensions, for the sake of simplicity, this examplewill only examine motion in the sagittal plane (twodimensions). This plane divides the body into leftand right portions.

Camera Positioning and OrientationIn two dimensional analysis, it is imperative that the camera be oriented perpendicular to theplane of motion. It is also important that the camera be level front to back, as well as side to side.This is achieved by placing a small level on the appropriate axes of the camera.

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Marker PlacementThe next stage of setup is marker placement. Highlyreflective markers are placed at various landmarks,allowing accurate identificationof specific structures on the recorded film. In this case,the marker landmarks approximate the centres ofrotation of the different segments being analyzed. Theseinclude the greater trochanter (hip), lateral tibio-femoraljoint line (knee), lateral malleolus (ankle), talus (heel),and fifth metatarsal (forefoot). When light is directedfrom the vicinity of the camera, the markers areilluminated as they reflect the light back towards thecamera lens.

Light Placement

The reflective markers demonstrate a narrow angle ofreflection, therefore light placement is important inorder to achieve maximal illumination of themarkers. 

Film CalibrationThe last portion of the setup phase is filmcalibration. This involves filming an object of aknown length. The actual and filmed length of theobject, a metre stick in this case, will be inputtedto the computer in the next stage. This permits thecalculation of a screen pixel: actual size ratio.

 

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Digitizing Phase

Once the setup is complete, the movement is filmed, and the digitizing phase can begin. In thisstage, the motion is analyzed by quantifying the changes in location of the joint markers. At theUBC Biomechanics lab, we use the Peak Motus Motion Measurement System to measuremovement kinematics.

This process begins by manually indicating the locations of the markers in the initial frame. Thisis achieved by positioning a "cross-hair" over the markers on the computer screen and clickingthe mouse to select the location. The computer then records the x and y coordinates of the cross-hair as the marker location. This manual process is repeated for all of the markers in the firstframe. Then computer then automatically follows the markers as the movement progresses,recording the x and y coordinates of each marker in each frame. Now you are ready to completethe Virtual Digitizing Session. Once coordinate data have been obtained for each marker in each frame, the data can be fed intoa spreadsheet program such as Excel, as shown below, and analyzed from both linear andangular perspectives. The data can also be represented graphically once this is done.

Linear KinematicsThe coordinate data can be examined in the context of linear kinematics, or in the context ofchanges in the locations of the markers over time. In other words, each frame will provide avertical position which can then be plotted against percent stride on a graph. For instance, the

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vertical position of the 5th metatarsal marker can be plotted against percent stride as seen in thegraph below.

 

Similarly, both the x and y ankle marker coordinates can be graphed, giving the path of theankle  during the walking cycle. This is shown in the picture below.

Angular KinematicsThese data can be further analyzed by joining the markers to form segments. By calculating theorientation of the segments relative to one another, or relative to a fixed angle in space, theangular kinematics of the walking stride can be quantified. For example, the angle between thethigh and shank segments, or the knee angle, can be plotted over time as demonstrated below.The frame shown corresponds to the angles at 68% stride.

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 This section summarized the methods used to collect and process kinematic data. For each of thekinematic labs, both linear and angular, these techniques were used. Once the data had beendigitized they were processed to compute the respective data and stored in files that can beloaded by Microsoft Excel. You should now be able to work with those files to complete yourassignments.

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Kinetics Methods

Force Platform Kinetic Data Acquisition

A Kistler force platform and Kistler force pedals are used to acquire kinetic data in thebiomechanics lab. This equipment uses 4 piezoelectric transducers, located at the corners of theplatform to measure the applied forces. Forces are measured in three planes: vertical,anteroposterior (AP), and in the lateral direction. (For our purposes, we will assume that thelateral forces for a walker are negligible). The force platform is then connected to an amplifierusing electrical wires. The amplifier boosts the signal from the force platform so that thecomputer can "hear" the data. The amplifier is also connected to the computer using electricalwires. The Force platform is activated by powering up the amplifiers and connecting it to thePeak Motus Motion Measurement System. It is then ready to measure movement kinetics. As anathlete applies a force to the force platform, the data from the force platform is passed throughthe amplifiers and fed into the computer.

If an person lunges forward and sideways from a forceplatform, a 3 dimensional force is applied to the forceplatform. This force is called a resultant vector andcan be broken into its three component vectors. Theselie in the vertical (Fz), AP (Fy) , and lateral (Fx)directions. The Kistler force platform measures thesecomponents and passes them on to the computer.

The force applied to the force platform, however, is not the force that we areinterested in. It is the reaction force of the platform on the athlete that is of interest. Newton'sthird law states that for every force there is a reaction force that is equal in magnitude andopposite in direction. For example, as the athlete's foot applies a force to the force platform, theforce platform exerts a ground reaction force on the foot that is equal in magnitude andopposite in direction to the resultant force. This can also be broken into three vectors, which areoriented in the vertical, AP, and lateral directions. These are also equal in magnitude andopposite in direction to the corresponding force components applied to the force platform. The

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ground reaction force components can be calculated from the components of the force that isapplied to the force platform.

The Kistler force platform is also capable of measuring the coordinates of the point at which theforce is applied. This is measure using an x and y coordinate system. It is important to note thata force platform analyses forces at single point, even though force is actually being distributedabout an area (pressure).

Lastly, the force platform measures momentsexerted on the platform. This is to say, if a force isapplied in such a way as to cause the platform torotate, the moment of this torque can be measured.

The kinetics of motion are analyzed by quantifyingthe changes in the forces applied to the forceplatform, and the resultant ground reaction forces.

This process begins by manually clicking a trigger to commence the recording of data. Thecomputer then records the x, y, and z resultant forces provided by the force platform. Thecomputer automatically stops recording after a preset amount of time (usually 2 seconds). 

The Kistler force platform is precalibrated. When a subject stands stationary on the platform theoutput is a horizontal line of constant value. This value reflects the generated action/reactionforce. That is, the person applies a force to the platform by virtue of standing on it. In return, theplatform provides a reaction force to the person. During quiet standing, the acceleration of theperson is zero and thus the platform can be used to record the person's weight. There are twoforces acting on the individual: the green arrow (FG) represents the force of gravity (mg). Thered arrow (FR) represents the reaction force from the platform. We can sum these forces usingNewton's second law: -FG + FR = ma, where "m" is the person's mass and "a" is the accelerationof the person's centre of mass. In quiet standing, "a" is zero and thus FR = FG and thus theground reaction force is equal to the force due to gravity, the person's weight. Note that in theequation, FG is negative because the direction of the gravitational force is down.

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 Now you are ready to go to the Virtual Kinetics Data Acquisition page to see how the data arecollected.Once the data have been collected, it can be copied into a spreadsheet program (e.g.Excel). Theforces can be input with respect to time as shown in the Excel spreadsheet below.

The data can then be manipulated toproduce graphs which graphicallyrepresent the data. A graph plottingvertical force and AP (anterior-posterior)force with respect to time is shownbelow. The red line is Vertical GRF(ground reaction force) plotted againsttime, and the blue line is anterior-posterior GRF plotted versus time. It isimportant to note that the computerrecords the data for as long as it is told,and not just during the time that theperson is in contact with the forceplatform. Because of this, the computerrecords forces of zero for the pointsbefore and after the contact made by theathlete. This can be seen in the graph

below, where the computer sampled the data for two seconds, yet the athlete only made contactwith the force platform for  0.6 seconds.

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The graph below shows the vertical ground reaction forces plotted with respect to time duringwalking. As you can see, the vertical GRF increases sharply upon heel strike, then dipssomewhat during the stance phase. Finally, the forces rise again and fall off as the subject leavesthe platform.

The next graph shows the analysis of the AP (anterior-posterior) GRF forces during walking.Upon heel-strike, the reaction force is in a negative direction. This is because the acceleration isnegative at this point (i.e. the athlete is slowing down). During the stance phase, the accelerationbecomes less and less negative as the athlete moves over the base of support, finally becomingpositive midway through the stance phase. The acceleration continues to increase in magnitudebecause the subject pushes off the force platform which results in a GRF propelling the athleteforward. The acceleration drops off as the subject leaves the force platform.  

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This picture shows the resultant GRF vector of theforce applied to the athlete upon heel-strike. This isobtained by adding the horizontal and vertical forcevectors.

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IV. Tips on Using Excel

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About Excel

Excel is a spreadsheet program that is useful for quick mathematical calculations and dataanalysis. The spreadsheet has rows and columns that form an infinite number of cells. Each cellhas an address defined by the row number and column letter. Excel can be used on a Macintoshor a PC in many different versions. Currently, the PCs in the Human Kinetics computer labhave Microsoft Excel 97.

The biomechanics 363 laboratory activities will be easier if you understand a few of Excel’sbasic functions. The three most important functions to know are:

1. creating a new worksheet2. opening a saved worksheet3. saving a worksheet

Listed below are the three most important functions and a selected few of the most frequentlyused functions. Excel instructions are also available from the UBC Bookstore for $5.95.

Basic Excel Functions

Opening Excel

Excel may be opened by clicking on of MICROSOFT EXCEL 97 on the task bar on your screen.An Excel worksheet will open on the screen.

Creating a Worksheet

To create a new Excel worksheet select FILE--NEW from the pop down menu that appears.

Opening a Worksheet

All Excel files have the suffix XLS. Excel will open files from other applications.

1. There are three ways to open an excel file.a) Click on FILE and a pop-down menu will appear. Click on open.b) Click on the yellow folder near the top of the screen.c) On the keyboard, simultaneously press CTRL and O.

2. A window will open.Select appropriate drive/folder (a: drive is your disk if you put one in; c: drive is thatcomputer’s hard-drive; d: is the CD ROM; all other letters are other drives) by clickingon the down arrow beside the box LOOK IN: and then selecting the file you wish to openor another folder.When you locate the file you want to open, double click on it or single click on it andthen press the OPEN button.If you cannot find the file you want to open, click on the down arrow beside the boxFILES OF TYPE and select ALL FILES (*.*). More files should appear. If your file still

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is not there, double-check to make sure you are looking in the correct place. If you stillcannot find it, it is not there!

Selecting Cells

Click on the cell you wish to select/highlight.

If you want to select multiple cells, click on one cell and drag the mouse to include all of thecells you want to select.

If you would like to select a row(s) of data, select the first (few) cells in that row(s) and thenpress the END key and then hold down the SHIFT key while you press the down arrow. Theentire list of numbers will be selected.

If you want to select the entire row or column, you can click on the letter or number around thespreadsheet. The entire row(s)/column(s) will be selected. That’s right, the entire 10,000+ cells!

Selecting Non-Adjacent Cells

Select the first group of cells.Hold down the CTRL key and click on the other group(s) of cells you want to select.If you are using a Mac, use the apple button instead of CTRL.

Calculations

To calculate using excel, enter the equal sign (=) and then your formula. You can also use thefunction wizard to guide you through the calculations. Click on the “fx” button near the top ofthe screen to use the function wizard. It helps to know a few shortcuts to make your formulasshorter and easier to understand.

Addition/Subtraction/Multiplication/DivisionIf you want to perform simple arithmetic, just enter the formula. For example, if you want to addtwo numbers, enter: =cell+cell.If you are adding a row of numbers, instead of adding each one separately, you can enter thefollowing equation: =SUM(cell1:cell5), where cell1 is the first cell and cell5 is the last cell in therow you want to add together.If the numbers you want to add are not in consecutive cells, you can combine the ideas aboveand use =SUM(cell1:cell2,cell3,cell7,cell9:cell34).

AverageIf you want to take the average of some numbers, use the word AVERAGE. For example, ifyou wanted to take the average of a row of numbers, enter into the cell where you want thevalue to be: =AVERAGE(cell1:cell2).

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Min/MaxTo find the minimum or maximum in a string of numbers, use the commands MIN and MAX.For example: =MIN(cell1:cell22)

Standard DeviationThe STDEV command returns the standard deviation of a list of numbers. For Example:=STDEV(cell1:cell22)

Square RootThe function for square root is either "SQRT" or "^0.5".

'$' Function (how to keep one cell the same when filling in a formula)When you are copying a formula across a column or row of cells, you may want some of thevalues in the formula to stay the same while others will change as you fill the formula across.When the ‘$’ is placed before the cell or value you DO NOT want to change, it will stay thesame across the column or row. For Example, if you want to add a row of values individuallyto a value in cell A1, your first formula would be: =SUM($A$1:A2). If you copy the formulaacross cells, the next cell’s formula will be ($A$1:A3).

Correcting Cells

If you want to correct a formula or valve in the cell, you have two options:1. Click on the cell and retype the information.2. Click on the cell you want to correct.

The cell contents will appear at the top of the spreadsheet.Then click the cursor on these cell contents and correct them.

UndoThere are two ways to undo mistakes.

1. Press CTRL and Z to undo your last function.2. Click on the curved arrow buttons on the top of your screen to undo or redo functions.

Copy/Paste

To Copy : Select the cells that you want to copy.There are four ways of copying information:1. Select the copy command from the Edit menu2. Select the picture of two pieces of paper at the top of the screen3. Click the left mouse button and select copy4. On the keyboard, press CTRL and C

To Paste: Move the cursor to the cell where you want to paste the data.You can paste by:1. Selecting the ‘paste’ command from the Edit menu2. Clicking on the clipboard from the standard toolbar at the top of the screen3. Selecting ‘paste’ from the menu that pops up after clicking the left mouse button4. Press CTRL and V on your keyboard

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Filling Cells

If you have a function you would like to copy to many cells in a row or column, you could eithertype the same formula in each cell or can FILL cells.

1. Select the cell that you want to copy.2. Put the cursor on the bottom right corner of that cell. The cursor will change from a fat

white plus sigh to a skinny black plus sign.3. When the cursor is black, hold down the mouse button and drag the cursor until the

desired number of cells are highlighted. The formula will fill in these cells.

Graphing

1. Select the cells you want to graph.2. Click on INSERT and then CHART and the chart wizard will pop up.3. Click on the desired graph format (choose XY SCATTER if you want the first column to be

plotted as x-axis data; choose LINE GRAPH if there are no columns you want to be plottedon the x-axis).

4. On the right side of that window, choose the type of line that you want.5. Click the Preview Graph button if you want to preview your graph.6. Click on NEXT of you are happy with your graph.7. Label your chart and then click NEXT.8. Choose where you want your graph to be plotted, either as a new sheet or on different sheet.9. To plot the graph on a different sheet, clock on the down arrow and select a sheet.10. Click FINISH to plot your graph.

Changing Background ColourDouble-click on the background of the graph.Under the AREA section, click NONE and then OK.

Formatting the gridlinesColour: To change the colour of a line, double-click on a line, and then click on the PATTERNStab in the new window. Click on the down arrow beside COLOR in the LINE box. Choose anew colour.

Thickness: To change the line thickness click on the down arrow beside WEIGHT and select thethickness you want.

Marker: The change the style of the marker click on the down arrow beside STYLE and select amarker pattern. Marker colour can be changed by clicking on the arrow next to "Background"

Changing Axis Scales1. Double click on the axis you want to change.

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2. Click on the SCALE tab in the format axis window that opens up.3. Choose the variable you want to change and click on the box to the right of it.

The MINIMUM UNIT is the smallest unit on the axis. MAXIMUM UNIT is the largest unit onthe axis. MINOR UNIT is the smallest interval between units and MAJOR UNIT is the largestinterval between units.

Adding/Changing TitlesTo add or change a title to your graph or axis:Click on the CHART tab at the top of the screenSelect chart options.Select the TITLES tab from the new window.Enter or change the Title, X-axis or Y-axis by clicking on the box to the right of the variableselected.

To change or edit a title already on the graph:1. Click once on the title. A grey box should appear around the title.2. Click once inside the box so a flashing cursor is on the title. Edit the title.3. Click anywhere on the background of the graph to finish.

Changing LocationTo change the location of your chart:

1. Click on the CHART heading at the top of the page.2. Select LOCATION from the pop up menu.3. To move the graph to a different sheet, click on the down arrow beside AS OBJECT

IN and select the sheet you want to move it to.4. To move a graph to a new sheet, select AS A NEW SHEET.

Changing OptionsThere are a number of options available to adjust your chart.Click the CHART tab at the top of the screen and select chart options.

The TITLES option has previously been discussed under ADDING/CHANGING AXISTITLES.The AXIS Option will allow you to view or hide the units in the X and Y axes.The GRIDLINES option allows you to place lines on your chart along either the majorunits of minor units of the X and Y axes.The LEGEND option allows you to place the legend at different spots on your chart.The DATA LABELS option to inserted labels on your graph. as “show value” whichlabels each point on the chart with the Y coordinate value or show label which labels eachpoint with the X-coordinate value.

Changing Graph TypeTo change the type of graph you initially selected:

1. Open the menu under CHART

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2. Select ‘CHART TYPE.3. Choose either the ‘STANDARD TYPE or CUSTOM TYPE tab from the new window.4. Click on a ‘CHART TYPE from the list below choose the type of chart you want.5. To view your new chart, click PRESS AND HOLD TO VIEW SAMPLE.6. Click OK to finish.

Adding InformationTo add data to your chart, enter the data in the worksheet with the original chart data andhighlight the new data.You now have three options:

1. Select the copy command from the Edit menu, or2. Select the picture of two pieces of paper at the top of the screen, or3. Click the right mouse button and select copy.Your selected data should now have a moving dashed line surrounding it.Choose the chart where you want the data to appear by using the tab at the bottom of thescreen.Click anywhere on the chart background.Select PASTE from either:the edit menuthe clipboard from the standard toolbar at the top of the screenclicking the right mouse buttonYour new data will now appear as part of the chart.

Having More Than One Graph Per PageTo place two charts on one page:1. Begin by opening the chart that you want to move.2. Click on the chart tab at the top and select LOCATION.3. Click on the down arrow beside AS OBJECT IN and select the sheet or chart you want to

place it in.4. Select OK to place the chart as an embeded object in another chart or worksheet.5. The chart will now appear on the screen with the chart you have placed it in, or in the

worksheet.6. To move the chart around the page:7. Click on the chart’s corner once so a black square appears in the corner.8. Click again on the outline so a thick black line becomes the border of the chart.9. Position the mouse on the corner so the thick white arrow changes to a narrow black two

headed arrow.10. Hold the mouse key down and drag the chart to the size you want.

Copying an Excel Chart into a Word WorksheetBegin in Excel by using the mouse to select the chart you want to copy.Follow the commands under COPY/PASTE at the beginning of the Excel Functions document.Open Word using the start button and then select Programs—Microsoft Word.Follow the COPY/PASTE commands in the Microsoft Word instructions.If your computer does not have enough memory to have both Excel and Word open at the sametime, close Excel before going to Word.

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PrintingPrinting now has a fee!! Instructions will be available in the computer lab. You will be given aprint card at the first lab for you to print out your labs as you see fit.

Advanced Excel Functions

Differentiation in Excel (estimating slopes of lines)

By now you are familiar with differentiation by graphical means - that is, estimating the slope ofa line by drawing tangents. What we need to do now is develop a means to integrate numericallywithin a computer program such as Excel (these comments would apply to any spreadsheetprogram).Consider a graph of kinematic data, such as velocity (shown in the graph below). Differentiationof the velocity-time graph will provide information on acceleration. The equation is A = dV/dt

where V is velocity and A is acceleration and dt is the timeinterval.To begin with, you have a series of data points representingvelocity at particular points in time which were used to plot thisgraph. In your spreadsheet, the data points would typically bearranged in a column where each row represents an equalincrement in time. In the spreadsheet, shown below, thenumbers representing time are located in column "A" and thenumbers representing velocity are located in column "B".

However, if these are located in different columns, (i.e. columns "G" and "H"), the formulasexplained below are modified accordingly. We can easily find the slopebetween two points by simply calculating the change in velocity (V2 - V1) and divide that by thechange in time (t2-t1). In this example, you would take the velocity data (column "B"), andcalculate the difference between one data point and the point preceding it (ex. B2-B1). Thisrepresents the change in velocity during a given interval V2-V1. This number is then divided bythe length of this interval (ex. A2-A1) or the time interval over which the change in velocityoccurs (t2-t1). You are now ready to put these formulas together:You want to calculate the slope of the graph over a certain time interval: (V2-V1)/ (t2-t1). Asexplained, this can be calculated in Excel using a formula such as (B2 - B1)/(A2-A1).Remember, B represents the velocity, and A represents the time. These letters will have to beadapted to the columns in which you have placed the velocity and time information.With this equation, you can calculate the average slope between these points. But where wouldyou enter them in the spreadsheet? Not at C1 nor at C2.

You have calculated the average slope between these points but not AT these points. Thus, wemust employ slightly different techniques to solve this problem.

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We will simply calculate the slope between 3 points and place this value at the mid point. Forexample, to calculate the slope at point 2, we take the value at point 3 minus the value at point 1,divide by twice the time interval and place the new value at point 2. That is, in cell C2 = (B3-B1)/(A3-A1) or, if you are only given the time interval (not successive time values), in cell C2 =(B3-B1) / (2* time interval). Now you can copy (or autofill) this to the bottom of the columnand have the differential of column B. This approximates the slope of the data in column B.There are some costs for this estimationand that is we loose data for the firstand last points. There are moresophisticated methods of determiningthese values but we will not need thesefor this course.

Integration in Excel

THE INTEGRALBy now you are familiar with integration by graphical means - that is, estimating the areabeneath a curve by counting the enclose squares or other geometrical shapes. What we need todo now is develop a means to integrate numerically within a computer program such as Excel(these comments would apply to any spreadsheet program).Consider a graph of kinematic data, such as acceleration (shown in the graph below). Integrationof the acceleration-time graph will provide information on velocity. The equation is V = Adtwhere V is velocity and A is acceleration and dt is the time interval. This, of course, tells us thechange in velocity over some interval but not the actual velocity. You need to know the startingvelocity before you can determine that. So the final equation is V = Vi + Adt where Vi is theinitial velocity.

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To begin with, you have a series of data points representing velocity at particular points in timewhich were used to plot this graph. The time interval between points must be the same, aconstant. In your spreadsheet, the data points would typically be arranged in a column whereeach row represents an equal increment in time. To find the area under this curve we will use theTrapaziod Approximation. The formula is quite straight forward:

Integral = (.5*A1*dt + A2*dt + A3*dt + .. + An-1*dt + .5*An*dt)where dt is the time interval and A is the value of acceleration at some time (1, 2, 3 ...n). Becausedt is a constant it can be factored out and the equation now looks like this:Integral = (.5*A1 + A2+ A3 + .. + An-1+ .5*An)*dt

You can see that this is an approximation but the greaterthe number of divisions the lower the error. This equationcan easily be put into a spreadsheet. Consider a column ofnumbers with a constant time interval of 0.05. The valuesare arranged in column B from row 1 through 11. Theintegral of the curve would then be calculated by:

= (.5*B1 + SUM(B2:B10) + .5*B11) * 0.05This formula will be useful in determining the impulseduring a vertical jump. In this example the answeris...0.3543467

Integration and PlottingThere are times, like in the kinematic labs, when you areasked to integrate acceleration and plot velocity (orintegrate velocity and plot displacement). In this case weare not so interested in the total integral but rather in an

estimate of the instantaneous integral over the time interval. In this case, you can estimate thearea under each point by simply multiplying the magnitude of the value and the time interval.For each successive point you must add this computed value to the previous value. For example,you have a series of values in a spreadsheet. In column A are the incremental time values (0.05increment) and in B are the acceleration values (shown here). To compute the instantaneousintegral enter the value 0 is column C1. In C2 enter the equation = C1 + B2*0.05

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This will compute a simple change in velocity as aresult of that acceleration and add it to the initialvelocity (in C1). In this case we set the initialvelocity at 0 but of course it could be any value.Now you can copy (or autofill) this to the bottomof column C and you will have estimated theintegral of column B. These two columns (B andC) can be plotted as a function of column A andallow you to compare the velocity andacceleration.Of course, you can carry on in this fashion andintegrate column C, velocity, and compute a newcolumn in D which will contain the estimate of thedisplacement.

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V. TIPS ON USING MICROSOFT WORD

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Microsoft word is a word processing program which will be useful for writing up your labassignments and other documents. The most frequently used functions are described below.

Opening Microsoft Word

In Windows 95, Microsoft word may be opened by clicking on the Start button at the bottom ofyour screen. A menu will pop up. Click on PROGRAMS -- MICROSOFT WORD. MicrosoftWord will open on the screen.

Creating a Worksheet

To create a new Word document select FILE--NEW from the pop down menu that appears.

Opening a Worksheet

All Word files have the suffix .doc.

There are three ways to open an word file.Click on FILE and a pop-down menu will appear. Click on open.Click on the yellow folder near the top of the screen.On the keyboard, simultaneously press CTRL and O.

A window will open. Select appropriate drive/folder (a: drive is your disk if you put one in; c:drive is that computer’s hard-drive; d: is the CD ROM; all other letters are other drives) byclicking on the down arrow beside the box LOOK IN: and then selecting the file you wish toopen or another folder. When you locate the file you want to open, double click on it or singleclick on it and then press the OPEN button.

To copy a graph from excel into a word document, follow the instructions above in ExcelFunctions.

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VI. Problem Set Questions

37

Problem Set #1: Review Problems

I ALGEBRA

Given a = 20, b = 10, c=5, find d below:

II TRIGONOMETRY

Find the side d and/or the angle Θ

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III GEOMETRY

Find the angles, in degrees, of Θ:

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Problem Set #2: Differentiation

1. Differentiate the graphs below by finding the slope(s) of the line.

2. Differentiate the graph you just obtained from #1 above.

 

3. 4.

5. 6.

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7. 8.

9.

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Problem Set #3: Linear Kinematics 1. A skater increases her speed with constant acceleration from 22 m/s to 30 m/sover a 4 second duration.(a) What is her acceleration?(b) What distance will she travel in the 4 seconds?(c) What is her average velocity?(d) What will her velocity be after skating 30 m?(e) How long will it take her to cover the first 50 m?(f) How long will it take her to reach a speed of 25 m/s?

2. What is the average velocity of a sprinter who runs the 100 m in 9.8 s? Is this ameaningful measurement?

3. A cyclist increases her speed with constant acceleration from 20 m/s to 28 m/sover a 4s period.(a) What was her acceleration?(b) What distance did she cover in the 4 s period?(c) What was her average velocity?(d) What was her velocity 30 m after starting to accelerate?(e) How long would it take her to cover the first 25 m?(f) How long would it take her to reach 25 m/s? 

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Problem Set #4: Projectiles

 1. A ball is thrown off a 10 m tower with an initial velocity of 3 m/s forwards and 2 m/supwards.(a) How long will it take the ball to strike the ground?(b) What will be the ball's downward (vertical) velocity immediately before it strikes the ground?(c) What will be the ball's forward (horizontal) velocity immediately before it strikes the ground?(d) How far will the ball travel forwards?

2. An athlete leaves the ground with a horizontal velocity of 12 m/s and a vertical velocity of 5m/s, forwards and upwards, respectively.(a) What is her take-off angle?(b) What distance will she travel forwards if she lands at the same height that she becameairborne?(c) What distance will she travel if she lands 1 meter lower than she took-off from? (Hint, findthe flight time.)

3. A high jumper takes off with a velocity of 10 m/s at an angle of 20 degrees to the horizontal.His center of mass (calculated from a still photograph) at take-off was 1.25 m above the ground.(a) What will be the maximum height his center of mass will rise above the ground? (Hint, v = 0,upwards).(b) If the athlete reduces his horizontal velocity slightly and thereby increases the duration of hisvertical impulse and enabling a 5% increase in his vertical velocity, what will now be the rise ofhis center of mass?(c) How far will the jumper travel, horizontally, if he lands 1 m above the ground? Assume thevelocity used in part (a).

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Problem Set #5: Angular Kinematics

 1. Convert the following to degrees:(a) 1.5 revolutions(b) 1.5 π radians(c) 1.5 radians(d) 0.75 revolution(e) π/2 radians

2. Convert the following to radians:(a) one half revolution(b) 45 degrees(c) 110 degrees(d) -30 degrees(e) 210 degrees

3. A limb which is 1 m long moves through an angle of 30 degrees in 0.5 seconds.(a) What is the average angular velocity of: (i) its end?

(ii) a point 30 cm from the axis of rotation?(b) What is the linear velocity of: (i) its end?

(ii) a point 30 cm from the axis of rotation? 4. Consider a horse on a spinning carousel. At t=0 s it is moving at 2°/s and accelerating at aconstant rate of 3°/s2. What is the displacement of the horse after 5 seconds?

5. An athlete performs a dive from a handstand off a 10 m tower. Her center of mass is 0.8 mabove the tower as she falls into the dive. (a) If she can rotate at 5.7π rad/s in a tucked position, how many complete somersaults can shedo in her dive? Assume that she must stop rotating 1 m above the water to achieve a clean entry. (b) If she can perform 3 somersaults in a piked position in the same amount of time, what is herangular velocity when performing piked somersaults? (c)The diver is performing a piked dive. She realizes that she won't have enough time to finishher last somersault and enter the water in a vertical position, so she moves into a tucked positionover 0.2 s. What is her angular acceleration? 6. The hammer thrower's arm is 0.8m long and the hammer extends an additional 1.2m. It takes0.75 seconds to complete 1 revolution before the thrower releases the hammer. Assume that thethrower is at rest when t=0s and that the path of the arm is circular. Find the magnitude of theangular acceleration immediately before release of the hammer and the magnitude of total linearacceleration upon release. 

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7. A soccer player kicks a ball so that it leaves his foot traveling at an angle of 20° from theground. His leg length is 1.2 m, and his hip is rotating at 3π rad/s as he executes the kick. Howfar will the ball travel before hitting the ground?  

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Problem Set #6: Linear Kinetics 1. Find the resultant force and angle for the following horizontal and verticalforces:

Horizontal Vertical(a) 250 N 350 N(b) 1000 N 2000 N(c) 2 kN 1 KN(d) 25 lbs -35 lbs(e) -10 lbs -20 lbs(f) -5 kN 15 kN2. Find the components of a 400N force applied to a ball of mass 8 kg, if the forceis applied at a 40 degree angle to the horizontal? What is the ball's acceleration?

3. Dino Dude is rock climbing and falls off the rock. Luckily, he is attatched by arope (which passes over a frictionless pulley) to his buddy who has already made itto the top of the cliff. But, Dino Buddy is not quite strong enough to support DinoDude's weight. Dino Dude, who has a mass of 20 kg, begins to acceleratedownwards at a rate of 2 m/s2.

a) What is the net force acting on Dino Dude?b) What is the tension through the rope?c) How much force is being exerted by Dino Buddy?

4. A box is being pushed up a hill, accelerating at a rate of 1.5 m/s2. The hill is

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sloped at an angle of 30° to flat ground. The box has a mass of 5 kg. What is themagnitude and direction of the force applied by the person pushing the box?

5. A pair of novice figure skaters make a mistake in their routine and collide head-on. She weighs 47 kg and is traveling at 4.5 m/s; he weighs 75 kg and is travelingat 4 m/s. They try to cover up their mistake by grabbing onto each other andgliding together. What is their resultant velocity? 6. A baseball (mass 0.8 kg) reaches a bat traveling at a velocity of 30 m/s.(a) How much momentum does the ball have?(b) How much impulse must be applied in order for the ball to leave the batmoving at 35 m/s in the opposite direction?(c) If the ball is in contact with the bat for 0.3 s, how much average force is appliedby the bat? 7. A 90 kg hockey player travelling with a velocity of 10 m/s collides head-on witha 75 kg hockey player travelling at 6 m/s. If the two players become entangled andcontinue travelling together as a unit following the collision, what is theircombined velocity and direction of travel?8. What is the initial resultant velocity of a projectile that has a vertical velocity of25 m/s and was projected at a 30 degree angle? If it weighs 20N what is its linearmomentum?

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Problem Set #7: Angular Kinetics #1

1. Find the X and Y components for the following parameters:(a) 25 m/s at 25 degrees to the horizontal(b) 5 m at π/2 radians to the vertical(c) 0 N at 30 degrees to the horizontal(d) 25 N at 160 degrees to the horizontal(e) 100 m/s2 at -45 degrees to the vertical(f) 25 m/s at π/4 radians to the horizontal 2. Calculate the isometric moments of force (M) at the elbow and shoulder joints for thefollowing given elbow angles (θ), force directions (Φ) and forearm lengths (L). Assume thecontraction is in the horizontal plane and that the isometric force is 1000 N.

    Length (L)  Φ  θ (a)  30 cm 90°  90° (b)  35 cm  90°  90° (c)  30 cm  135°  135° (d)  25 cm  45°  45° (e)  30 cm  75°  75° (f)  30 cm  45°  45°

  3. Calculate the moments of force about the axes through point A and through point B for eachdiagram.

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 4. Dino Dude and his stud buddy decide to go to the park to play on the teeter totter. When thetwo arrive at the park, they discover that the teeter totter has been broken.(a) If Dino Dude sits on the long end (1.55 metre end) of the teeter totter and weighs 20kg, howmuch would Dino Stud have to weigh in order to balance out the teeter totter if he sat on thebroken end, .75 metres away from the fulcrum?(b) What is the mechanical advantage of the lever?

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Problem Set #8: Angular Kinetics #2

1. A diver sets out to perform a front somersault off a 5 m tower. She intends to take off from herfeet in an erect postition and enter the water feet first in a vertical position. She misjudges badlyand does a belly flop after completing one and a quarter revolutions - 1/4 more thanintended...OUCH! There are several things that she might do to improve performance. List asmany of these as you can and explain in biomechanical terms how each of them might serve toimprove the performance.2. Sketch a graph of the relative values of angular velocity and moment of inertia of a divercompleting a one and a half somersault dive.

3. A weightlifter performing an arm curl with a 30 kg barbell stops the lifting motion with hisforearms in a horizontal position.

a) If the line of action of the weight of the barbell is 30 cm from the axis of rotation, what is themagnitude of the moment that the elbow flexors must exert in order to maintain the position?(Ignore the weight of the lifter's forearms and hands in the calculations.)b) If the weight of the lifter's forearms and hands were taken into account would the moment thatthe elbow flexors have to exert be more or less than that in (a)? Explain.c) If the only elbow flexor muscles active were the biceps, and the distance from its insertion tothe axis of rotation was 3 cm, what force would the muscle have to exert to maintain the forearmin it's horizontal position? (Again, ignore the weight of the forearms and hands.)d) A second individual performs an arm curl. He, too, lifts a 30 kg barbell whose line of action is30 cm from the axis of rotation. However, this individual's biceps muscle inserts at a distance of3.5 cm from the axis of rotation. Ignoring the weight of the forearm and hand, what force musthis muscles exert in order to maintain the forearm in it's horizontal position?e) What type of lever is the forearm acting as?

4. There are several techniques for doing situps. Using the concept of torque, explain how onecould change the technique to make a situp easier, and more difficult.

5. The radius of gyration of the thigh with respect to the transverse at the hip is 54% of thesegment length. The mass of the thigh is 10.5% of total body mass, and the length of the thigh is23.2% of total body height. What is the moment of inertia of the thigh with respect to the hip forpeople of the following body masses and heights?

   MASS (kg)  HEIGHT (m)

 A  60  1.6 B  60  1.8 C  70  1.6 D  70  1.8

6. Angular kinematic data show that in the swing phase the minimum knee angle was 117.7° forwalking and 84.5° for running. Why are the knee angles smaller in the swing phase of running?

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Problem Set #9: Work, Power, and Energy   1. A person is performing a lift against a 400 N load. The load moves vertically through adistance of 20 cm in a time of 0.25 s.(a) How much work was done on the load?(b) What is the person's average power during the lift?(c) What is the load's increased potential energy after the lift?

2. If the load mentioned above was moving at 7.0 m/s after moving 10 cm, what was its totalenergy, kinetic energy, and potential energy at this point (assume that the load started 50 cmfrom the ground)?

3. How high will a 30 kg object travel after it is released if it is released with a vertical velocityof 2.0 m/s?

4. How much power is present when a force of 500 N is applied to an object with a mass of 100kg that is moving at 6 m/s?

5. What work is done when an object is lowered 1.80 m if the object has a mass of 700 kg?

6. What is the impulse delivered to a catcher's mitt by a 360 g ball that is traveling at 100 km/hand is then stopped by the catcher in .01 s?

7. How high would you have to drop an 80 N object so that it will have a kinetic energy of 2000J when it hit the ground?

8. A 55 kg person moves at the constant speed of 7 m/s along a straight stretch of track. What isthe person's:(a) acceleration(b) momentum(c) kinetic energy

9. How much work is done to stop an 90 kg football player who is running at 8.5 m/s,horizontally, and you do not have to knock him down?

10. What is the work done by a force of 500 N that acts at an angle of 25 degrees to an object andthat moves the object a displacement of 12 m?

11. Dino Dude is peacefully floating over the water when a big blue bird pops his balloons,causing Dino Dude to fall 10 m before crashing into the water. If Dino Dude has a mass of 20 kg,what is his velocity immediately before impact with the water?

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52

VII. Problem Set Answers

53

Solutions for Problem Set #1: Review Problems

Question #1(a)d = a2 + b2

d = 202 + 102

d = 500.0

(b)d2 = a2 + b2

d = √(a2 + b2) = √ (202 + 102)d = √500 = 22.4 (c)a2 = b2 + d2

d = √ (a2 - b2) = √ (202 - 102)d = √ 300 = 17.3

(d)a = bd +3d = (a - 3) / b = (20 - 3) / 10d = 1.7 (e)d = 25 + ab + b2 = 25 + (20)(10) + 102

d = 325.0 

(f)a = b/d - cd = b/(a+c) = 10 /(20+5)d = 0.4 (g)a2 = b2 - 2cdd = (b2 - a2)/2c = (102 - 202) / 2(5)d = -30.0 (h)d = a [(b - 10) / c] = 20 [(10 - 10) / 5)d = 0 (i)d = 2a + 4b + c/10 = 2(20) + 4(10) + 5/10d = 80.5 (j)d/a = b/cd = ab/c = 20(10) / 5d = 40.0

   Question #2 recall: a / sin A = b / sin B = c / sin Cand, c2 = a2 + b2 - 2ab cos C (a)d / sin 25° = 25 m / sin 120°d = sin 25° (25 m / sin 120°) = 12.20 m (b)d / sin D = 10 m / sin 30°d = sin D (10 m / sin 30°)since all angles in a triangle add to 180°, D = 180° - 70° - 30° = 80°

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d = sin 80° (10 m / sin 30°) = 19.70 m  (c)d / sin D = 4 cm / sin 20°d = sin D (4 cm / sin 20°)D = 180° - 50° = 130°, sod = sin 130° (4 cm / sin 20°) = 8.96 cm  (d)c2 = a2 + b2 - 2ab cos θcosθ= (a2 + b2 - c2) / 2abassign c to the side opposite θθ = cos-1 [(152 + 102 - 102) / 2(15)(10)]θ = 41.41° Question #3 (a)θ + 50° +30° = 180°θ = 180° - 50° - 30°; = 100°   (b)θ + α + 25° = 180°θ = 180°- 25°-α and α = 180°-70°; = 110°,soθ = 180° - 25° - 110° = 45° (c)θ = 180° -αα= 45°, soθ = 180° - 45° = 135° (d)eqilateral triangle, so θ+θ+20° = 180°2θ = 180° - 20°θ = 80°

 

(e)  θ +α+ 90° = 180°α = 30°θ = 180° - 90° - 30° = 60° (f)θ + 90° +α= 180°parallel lines so α= 180° - 120° = 60°θ = 180° - 90°; - 60° = 30° (g)symmetric triangles, soθ = 180° - 40° - 20° = 120° (h)θ = 360° - 65° - αdue to parallel lines, α = 55°, soθ = 360° - 65° - 55° = 240° (i)α = 180° - 110° = 70°θ = 360° - 220° -α= 360° - 220° - 70°; = 70°

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Solutions for Problem Set #2: Differentiation

1. 2.

3. 4.

5. 6.

7. 8.

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9.

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Solutions for Problem Set #3: Linear Kinematics

 A general strategy useful in solving linear kinetics problems is as follows:list all given values and unknownschoose a formula which uses appropriate variablesisolate the unknown variable and write down the formula (this way if you make a calculation error you may still getpart marks for having the correct approach)substitute givens and solve for unknownsexamine your answer: are the units correct? does the answer make sense? If you have time, you may wish to doublecheck your calculations.

1. A skater increases her speed with constant acceleration from 22 m/s to 30 m/s over a 4 secondduration.(a) What is her acceleration?(b) What distance will she travel in the 4 seconds?(c) What is her average velocity?(d) What will her velocity be after skating 30 m?(e) How long will it take her to cover the first 50 m?(f) How long will it take her to reach a speed of 25 m/s?

(a)vi = 22 m/svf = 30 m/st = 4 sa = ? a=Δv / t = (vf - vi) / t = (30 m/s - 22 m/s) / 4sa = 2 m/s2

 (b)vi = 22 m/svf = 30 m/st = 4 sa = 2 m/s2

d = ? vf

2 = vi2 + 2ad

d= (vf2 - vi

2) / 2a= [(30 m/s)2 - (22 m/s)2] / 2(2 m/s2)d= 104 m (c)vavg= (vi + vf)/2 = (30 m/s + 22 m/s)/2vavg= 26 m/s 

(d)vi = 22 m/sa = 2 m/s2

d = 30 mvf = ? vf

2 = vi2 + 2ad

vf =√(vi2 + 2ad)

vf = √ [(22 m/s)2 + 2(2 m/s2)(30 m)]vf = 24.58 m/s (e)vi = 22 m/sa = 2 m/s2

d = 50 mt = ? vf

2 = vi2 + 2ad

vf = √ (vi2 + 2ad)

vf = √ [(22 m/s)2 + 2(2 m/s2)(50 m)]vf = 26.15 m/s vf = vi + att = (vf - vi) / a = [(26.15 m/s) - (22 m/s)] / (2m/s2)t = 2.08 s

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 (f)vi = 22 m/svf = 25 m/sa = 2 m/s2

t = ?

 vf = vi + att = (vf - vi) / a = [(25 m/s) - (22 m/s)] / (2m/s2)t = 1.5 s

  2. What is the average velocity of a sprinter who runs the 100 m in 9.8 s? Is this a meaningfulmeasurement? v = d/t = 100 m / 9.8 s = 10.2 m/sNo, it doesn't tell us how the sprinter ran the run. Velocities over smaller time intervals (i.e.instantaneous velocities) would be more informative. 3. A cyclist increases her speed with constant acceleration from 20 m/s to 28 m/s over a 4speriod.(a) What was her acceleration?(b) What distance did she cover in the 4 s period?(c) What was her average velocity?(d) What was her velocity 30 m after starting to accelerate?(e) How long would it take her to cover the first 25 m?(f) How long would it take her to reach 25 m/s?

(a)a =Δv /Δt = (28 m/s - 20 m/s) / 4sa= 2 m/s2

 (b)d = vit + 1/2 at2 = 20 m/s (4s) + 1/2 (2 m/s2)(4s)2 = 96 m (c)v = d/t = 96 m / 4 s = 24 m/s (d)vf

2 = vi2 + 2ad

vf = √[(20 m/s) + 2(2 m/s2)(30 m)] = 22.8 m/s (e)d = 25m, a = 2 m/s2, vi = 20m/s, t = ? Approach 1: Use quadratic equation d = vit + 1/2 at2

25 m = (20 m/s)t + 1/2 (2m/s2)t2

25 = 20t + t2

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t2 + 20t - 25 = 0 quadratic equation: when At2 +Bt + C = 0, then t = [-B ±√(B2 - 4AC)] / 2At = [-20 ± √ (202 - 4(1)(-25))] / 2(1)t = [-20 ±√500] / 2t = 1.18 or -21.18, but t must be positive, sot =1.18 s Approach 2: First find vf vf

2 = vi2 + 2ad

vf = √ [(20 m/s)2 + 2(2 m/s2)(25 m)] = 22.36 m/s vf = vi + att = (vf - vi) / at = (22.36 m/s - 20.00 m/s) / 2 m/s2

t = 1.18 s (f)vf = vi + att = (vf - vi) / a = (25 m/s - 20m/s) / 2 m/s2 = 2.5 s

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Solutions for Problem Set #4: Projectiles

 1. A ball is thrown off a 10 m tower with an initial velocity of 3 m/s forwards and 2 m/supwards.(a) How long will it take the ball to strike the ground?(b) What will be the ball's downward (vertical) velocity immediately before it strikes the ground?(c) What will be the ball's forward (horizontal) velocity immediately before it strikes the ground?(d) How far will the ball travel forwards?

(a)ttotal = tup + tdowntup:

ay= -9.81 m/s2

vyf= 0 m/svyi= 2 m/stup= ?

vyf = vyi + atuptup = (vyf - vyi) / a = [(0 m/s) - (2 m/s)] / (-9.81 m/s2)tup = 0.204 s

tdown:tup= 0.204 svyi= 2 m/svyf= 0 m/sdup= ?

dup= [(vyf + vyi) / 2] tdup= [(0 m/s + 2 m/s) / 2] (0.204 s)dup= 0.204 m* ddown=10 m + 0.204m = 10.204 m

ddown= -10.204may= -9.81 m/s2

vyi= 0 m/stdown= ?

ddown= vyit + 1/2 (a tdown2)

ddown= (0 m/s) t + 1/2 (a tdown2)

= 1/2 (a tdown2)

 tdown

2 = 2 ddown / a= [(2) (-10.204 m)] / (-9.81 m/s2)tdown= 1.44 s

ttotal:ttotal = tup + tdown = 0.204 s + 1.44 sttotal = 1.64 s

(b)ddown= 10.204 mvyi= 0 m/say= -9.81 m/s2

vyf

= ?

vyf2= vyi

2 + 2 ay ddownvyf

2= (0 m/s)2 + (2)(-9.81 m/s2)(10.204 m)vyf

2= 200.20 m2/s2

vyf = -14.15 m/s

(c)vxf= 3 m/s (horizontal velocity only changes if there are horizontal forces acting to causehorizontal acceleration...in this case, the only external force is gravity which acts in the verticaldirection)

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(d)ax= 0 m/sttotal= 1.65 svxi= 3 m/sdx= ?

dx= vxi t + 1/2 (ax ttotal2)

dx= vxi t + 0 = vxi tdx= (3 m/s)(1.65 s)dx= 4.95 m

 2 An athlete leaves the ground with a horizontal velocity of 12 m/s and a vertical velocity of 5m/s, forwards and upwards, respectively.(a) What is her take-off angle?(b) What distance will she travel forwards if she lands at the same height that she becameairborne?(c) What distance will she travel if she lands 1 meter lower than she took-off from? (Hint, findthe flight time.)

(a)tan θ= 5/12θ = tan-1(5/12)θ = 22.62° (b)first find t totalttotal = tup + tdowntup:ay= -9.81 m/s2

vyf= 0 m/svyi= 5 m/stup= ?

vyf = vyi + atuptup = (vyf - vyi) / a = [(0 m/s) - (5 m/s)] / (-9.81 m/s2)tup = 0.51 s

since initial and final heights are 0, tup = tdown, and ttotal = 2 tupttotal:ttotal = 2 tup = 2 (0.51 s)ttotal = 1.02 s now find d horizontalrange = vht = (12 m/s)(1.02 s)range = 12.24 m (c)dup = vit + 1/2 at2

dup = (5 m/s)(0.51s) + 1/2 (-9.81 m/s2)(0.51)2 = 1.274 mdup = 1.274 + 1 = 2.274 m ddown = vit + 1/2 at2

1/2 atdown2 = ddown - vit

tdown = √[2(2.274 m)/9.81] = 0.68 s 

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so, ttotal = .68 + .51 = 1.19 srange = (12 m/s)(1.19 s) = 14.28 m 3. A high jumper takes off with a velocity of 10 m/s at an angle of 20 degrees to the horizontal.His center of mass (calculated from a still photograph) at take-off was 1.25 m above the ground.(a) What will be the maximum height his center of mass will rise above the ground? (Hint, v = 0,upwards).(b) If the athlete reduces his horizontal velocity slightly and thereby increases the duration of hisvertical impulse and enabling a 5% increase in his vertical velocity, what will now be the rise ofhis center of mass?(c) How far will the jumper travel, horizontally, if he lands 1 m above the ground? Assume thevelocity used in part (a).

(a)initial vertical velocity = (10 m/s) sin 20° = 3.42 m/stup = (vf - vi) / a = (0 - 3.42 ) / -9.81 = 0.35 s dup = vit + 1/2 at2

= (3.42 m/s)(0.35 s) +1/2(-9.81 m/s2)(0.35 s)2 = 0.60 m max height = height at take off + dup = 1.25 + .60max height = 1.85 m (b)The key here is the 5% increase in vertical velocity.vi = (10 m/s) sin 20° + (0.05)(10 m/s) sin 20° = 3.59 m/s vf = vi + att = (vf - vi)/a = (0 - 3.59) /(- 9.81)t = 0.37 s dup = vit + 1/2 at2 = 3.59(0.37) + 1/2 (-9.81)(0.37)2

dup = 0.66 mheight = 1.25 + 0.66 = 1.91 m (c)from (a), tup = 0.35 s, and max height = 1.85 mbut at landing c of m will be 1.00 m above ground,so ddown = 1.85 m - 1.00 m = 0.85 m ddown = vit + 1/2 at2

t2 = (0.85 m)(2) / (9.81 m/s2)t = 0.42 s ttotal = 0.35 s + 0.42 s = 0.77 s 

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range = vhorizontalt = (10 cos 20°)(0.77) = 7.23 m

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Solutions for Problem Set #5: Angular Kinematics

 1. 1 rev = 360° = 2πradians(a) (1.5 rev)(360°/rev) = 540°(b) (1.5 radians)(180°/π radians) = 270° or -90°(c) (1.5 radians)(180°/π radians) = 85.9°(d) (0.75 rev)(360°/ rev) = 270°(e) (π/2 radians)(180°/π radians) = 90° 2.(a) (0.5 rev)(2πradians/rev) = π or 3.14 radians(b) (45°)(π radians / 180°) = π / 4 or 0.79 radians(c) (110°)(π radians / 180°) = 1.92 radians(d) (-30°)(π radians / 180°) = -0.52 radians(e) (210°)(π radians / 180°) = 3.67 radians 3.(a)i) ω=θ/ t = (30°)(π radians / 180°) / 0.5 seconds = 1.05 rad/s or 60°/sii) same as above (b)i) v =ωr = (1.05 rad/s)(1 m) = 1.05 m/sii) v = ωr = (1.05 rad/s)(0.30m) = 0.315 m/s 4. Consider a horse on a spinning carousel. At t=0 s it is moving at 2 °/s and accelerating at aconstant rate of 3 °/s2. What is the displacement of the horse after 5 seconds?

List the variables and their linear equivalents, and summarize the question.v ~ ω1 = 2 °/s = 0.0349 radians/sa ~α = 3 °/s2 = 0.0524 radians/s2

Δt = 5 sd ~ θ= ?Select linear equation that you would use.In this case, d = vit + 1/2 at2 would be most appropriate.Convert to angular and solve.From d = vit + 1/2 at2 we get θ= ωit + 1/2αt2.

Substituting our given values leads toθ= ωit + 1/2αt2

θ= (0.0349 radians/s)(5 s) + 1/2(0.0524 radians/s2)(5 s)2

θ= 0.8295 radians

Converting back to degrees, we get:θ= 0.8295 x 180/π = 47.5°

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 5. An athlete performs a dive from a handstand off a 10 m tower. Her center of mass is 0.8 mabove the tower as she falls into the dive. (a) If she can rotate at 5.7πrad/s in a tucked position, how many complete somersaults can shedo in her dive? Assume that she must stop rotating 1 m above the water to achieve a clean entry. first find out how much time she has to do these rotations:d = 10 m + 0.8 m - 1 m = 9.8 mΔd = -9.8 ma = -9.81 m/s2

vi = 0t = ? d = vit + 1/2 at2

since vi = 0, then d = 1/2 at2

t2 = 2d/at = √[2d/a] = √ [2(-9.8 m) /(-9.81 m/s2)]t = 1.4 s now determine how many rotations she can do in 1.4 st = 1.4 sω= 5.7rad/sθ= ? in linear equivalents we have v and t, and are looking for d, so use v = d/tω=θ/tθ=ωt = (5.7πrad/s)(1.4 s) = 8πrad/ssince 2πrad = 1 rotation, then she can do 8π/ 2π= 4 rotations (b) If she can perform 3 somersaults in a piked position in the same amount of time, what is herangular velocity when performing piked somersaults? t = 1.4 sθ = (3π rotations)(2πrad/rotation) = 6πradω = ?ω=θ/t = 6πrad / 1.4 s = 4.3πrad/s (c)The diver is performing a piked dive. She realizes that she won't have enough time to finishher last somersault and enter the water in a vertical position, so she moves into a tucked positionover 0.2 s. What is her angular acceleration? think of a =Δv /Δtα=Δω /Δt = (ωt - ωp) /Δtα= (5.7πrad/s - 4.3πrad/s) /0.2 s= 7πrad/s

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   6.

The hammer thrower's arm is 0.8m long and the hammer extends anadditional 1.2m. It takes 0.75 seconds to complete 1 revolution before thethrower releases the hammer. Assume that the thrower is at rest when t=0sand that the path of the arm is circular. Find the magnitude of the angularacceleration immediately before release of the hammer and the magnitudeof total linear acceleration upon release.

List all known and unknown variables.r = 0.8 m + 1.2 m = 2.0 mΔt = 0.75 secondsω1 = 0 rad/sθ= 2πα= ?a = ?First, let us solve for α.α= Δω/Δtω1 = 0 rad/sω2 = Δθ/Δt = 2π/0.75 = 2.67π rad/sα=Δω/Δt = (2.67 πrad/s - 0 rad/s)/0.75α= 3.56 πrad/s = 11.18 rad/s2

Next, let's solve for a.at = (v2 - v1) /Δtwe need v2 for this equation:v2 = rω2 = (2.0 m) ( 2.67π rad/s ) = 16.78 m/s2at = (16.78 m/s - 0 m/s) / 0.75 s = 22.37 m / s2

at = 22.37 m / s2

 ar = (vt

2 / r) = (16.78 m/s)2 / (2 m)ar = 140.78 m / s2

 a2 = ar

2 + at2 = (140.78 m/s2)2 + (22.37 m/s2

a = √( 140.78 m/s2)2 + (22.37 m/s2) 2a = 142.55 m/s2

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Solutions for Problem Set #6: Linear Kinetics

 1. Find the resultant force and angle for the following horizontal and vertical forces: For force, use r2 = h2 + v2, and tanθ = v/h, so θ = tan-1 (v/h) (a)r2 = h2 + v2

r = √(h2 + v2) = √ (2502 + 3502) = 430.12 tanθ = v/hθ = tan-1(v/h) = tan-1(350 / 250) = 54.5° or 0.91 rad (b)r = 2236.1 N at 63.4° or 1.11 rad (c)r = 2.2 N at 26.6° or 0.46 rad  (d)r = 43.0 lbs. at -54.5° or -0.95 rad (e)r = 22.4 lbs. at -116.6° or -2.03 rad  (f)r = 15.8 kN at 108.4° or 1.89 rad 2. Find the components of a 400N force applied to a ball of mass 8 kg, if the force is applied at a40 degree angle to the horizontal? What is the ball's acceleration? r = 400 N, θ= 40° x = 400 cos 40° = 306.4 Ny = 400 sin 40° = 257.1 N F = ma, so:a = F/m = 400 N / 8 kg = 50 m/s2 at 40° to the horizontal

3. Dino Dude is rock climbing and falls off the rock. Luckily, he is attatched by a rope (whichpasses over a frictionless pulley) to his buddy who has already made it to the top of the cliff. But,Dino Buddy is not quite strong enough to support Dino Dude's weight. Dino Dude, who has amass of 20 kg, begins to accelerate downwards at a rate of 2 m/s2.a) What is the net force acting on Dino Dude?

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b) What is the tension through the rope?c) How much force is being exerted by Dino Buddy?

a) i) Free body diagram

ii) Add axes and resolve vector into componentssince all forces act in the same plane, there is no need for vector resolution

iii) Apply Newton's Laws and solve.ΣF = ma = 20 kg x 2 m/s2

ΣF = - 40 N

b)Fnet = Fgravity + FtensionFtension = Fnet - Fgravity = -40 N - (-196.2 N)Ftension = 156.2 N

c)Since the pulley is frictionless,FDino Buddy = Ftension = 156.2 N 4. A box is being pushed up a hill, accelerating at a rate of 1.5 m/s2. The hill is sloped at an angleof 30° to flat ground. The box has a mass of 5 kg. What is the magnitude and direction of theforce applied by the person pushing the box?

i) Free-Body Diagram (see below, left)

ii) Axes (see above, right)

iii) Apply Newton's Laws and solveFnet = ma = 5 kg x 1.5 m/s

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Fnet = 7.5 N

Fnet x = Fnet cos 30° = 7.5 N cos 30° = 6.50 NFnet y = Fnet sin 30° = 7.5 N sin 30° = 3.75 N

Fnet x = Fgravity x + Fpush xTherefore Fpush x = Fnet x - Fgravity x = 6.50 N - 0 NFpush x = 6.50 N

Fnet y = Fgravity y + Fpush yTherefore Fpush y = Fnet y - Fgravity y = 3.75 N - (-49.05 N)Fpush x = 52.80 N

Using Pythagoras' Theorem, we can calculate Fpush:(Fpush)2 = (Fpush x)2 + (Fpush y)2

Therefore Fpush = √[(Fpush x)2 + (Fpush y)2]Fpush = √ [(6.50 N)2 + (52.80 N)2] = √2830.10Fpush = 53.20 N

From trig relations we know that tanθ = 52.8 N / 6.50 NTherefore, θ= 83° So, the applied force is 53 N at 83° 5. A pair of novice figure skaters make a mistake in their routine and collide head-on. Sheweighs 47 kg and is traveling at 4.5 m/s; he weighs 75 kg and is traveling at 4 m/s. They try tocover up their mistake by grabbing onto each other and gliding together. What is their resultantvelocity? Pi = Pfm1v1i + m2v2i = m1v1f + m2v2fm1v1i + m2v2i = (m1 + m2)vfvf = (m1v1i + m2v2i) / (m1 + m2)vf = [(47 kg)(4.5 m/s) + (75 kg)(4 m/s)] / (47 kg + 75 kg)vf = 4.2 m/s  6. A base ball (mass 0.8 kg) reaches a bat traveling at a velocity of 30 m/s.(a) How much momentum does the ball have?(b) How much impulse must be applied in order for the ball to leave the bat moving at 35 m/s inthe opposite direction?(c) If the ball is in contact with the bat for 0.3 s, how much average force is applied by the bat? (a)Pi = mv = (0.8 kg)(30 m/s) = 24 kg m/s

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 (b)I = FΔt = mΔv = (0.8 kg)(-35 - 30 m/s) = -52 Ns 

(c)I = FΔtF = I /Δt = (-52 N/s) / 0.3 s = -173.3 N 7. A 90 kg hockey player travelling with a velocity of 10 m/s collides head-on with a 75 kghockey player travelling at 6 m/s. If the two players become entangled and continue travellingtogether as a unit following the collision, what is their combined velocity and direction of travel?

According to the Law of Conservation of Momentum, the total momentum before the collisionequals the total momentum after impact.m1v1 + m2v2 = (m1 + m2)vf(90 kg)(10 m/s) + (75 kg)(-6 m/s) = (90 kg + 75 kg)vfvf = 2.72 m/s in the direction of the 90 kg hockey player 8. What is the initial resultant velocity of a projectile that has a vertical velocity of 25 m/s andwas projected at a 30 degree angle? If it weighs 20N what is its linear momentum? Vi = ?Vv = 25 m/sθ= 30° sin 30° = 25 m/s / ViVi = 25 m/s / sin 30°Vi = 50 m/s M=mvM= (20N / 9.81 m/s2) (50 m/s)M = 101.94 kg m/s 

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Solutions for Problem Set #7: Angular Kinetics #1 Recall that moment of force = torque = Iα, and is the angular equivalent of F = ma.   Question #1

(a)assign θ= 25°, r = 25 m/s, and we wish to find x(= h) and y(= v).Cos 25° = x /(25 m/s) and sin 25° = y /(25 m/s), therefore,x = 25 m/s cos 25° = 22.7 m/sy = 25 sin 25° = 10.6 m/s   (b)sinceπ/ 2 radians (= 90 degrees) to the vertical is directed horizontally,x = 5.0 m, and y = 0.0 m (c)x = 10 cos 30° = 8.7 Ny = 10 sin 30° = 5.0 N (d)x = 25 cos 160° = -23.5 Ny = 25 sin 160° = 8.6 N (e)x = 100 cos (-45°) = 70.7 m/sy = 100 sin (-45°) = -70.7 m/s (f)x = 25 cos (π/4 radians) = 17.7 m/sy = 25 sin (π/4 radians) = 17.7 m/s 

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Question #2

  ELBOW SHOULDER(a) Since the force acts perpendicular to the forearm, and does not pass through the axis of rotation, we

can simply multiply the magnitude of the force by the perpendicular distance between the point ofapplication of the force and the axis of rotation.

  M = Fd = (1000 N)(0.30 m)M = 300 Nm

M = Fd = (1000 N)(0.30 m)M = 300 Nm

(b) M = Fd = (1000 N)(0.35 m)M = 350 Nm

M = Fd = (1000 N)(0.35 m)M = 350 Nm

(c) Fx passes through the axis of rotationof the elbow, soMxe = 0 Nm dxe = 0.3 m Fy = 1000 cos 45° = 707.1 N Mye = Fydxe= (707.1 N)(0.3 m) = 212.1 Nm  Me = Mxe + Mye= 0 + 212.1 = 212.1 Nm

xs = 0.3 sin 45° = 0.212 mys = 0.3 sin 45° = 0.212 m dxs = 0.3 + 0.212 m = 0.512 mdys = y = 0.212 m Fx = 1000 sin 45° = 707.1 NFy = 1000 cos 45° = 707.1 N Mxs = Fxdys= (707.1 N)(0.212 m) = 149.9 NmMys = Fydxs= (707.1 N)(0.512 m) = 362.0 Nm moments are in same direction, soMs = Mxs + Mys= 149.9 +362 = 511.9 Nm

(d) Fx passes through the axis of rotation of the elbow,soMxe = 0 Nm dxe = 0.25 m Fy = 1000 sin 45° = 707.1 N Mye = Fydxe= (707.1 N)(0.25 m) = 176.8 Nm Me = Mxe + Mye= 0 + 176.8 = 176.8 Nm

xs = 0.3 cos 45° = 0.212 mys = 0.3 sin 45° = 0.212 m dxs = 0.25 - 0.212 = 0.038 mdys = y = 0.212 m Fx = 1000 cos 45° = -707.1 NFy = 1000 sin 45° = 707.1 N Mxs = Fxdys= (-707.1 N)(0.212 m) = -149.9 NmMys = Fydxs= (707.1 N)(0.038 m) = 26.9 Nm   Ms = Mxs + Mys= -149.9 + 26.9 = -123.0 Nm

(e) Same as above, but replace 45° angles with 75° ,and 0.25 m forearm length with 0.30 m. 

Same as above, but replace 45° angles with 75° ,and 0.25 m forearm length with 0.30 m. 

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Fx passes through the axis of rotation of the elbow,soMxe = 0 Nm dxe = 0.3 m Fy = 1000 sin 75° = 965.9 N Mye = Fydxe= (965.9 N)(0.3 m) = 289.8 Nm  Me = Mxe + Mye= 0 + 176.8 = 289.8 Nm

xs = 0.3 cos 75° = 0.078 mys = 0.3 sin 75° = 0.290 m dxs = 0.3 - 0.078 m = 0.222 mdys = ys = 0.290 m Fx = 1000 cos 75° = -258.8 NFy = 1000 sin 75° = 965.9 N Mxs = Fxdys= (-258.8 N)(0.290 m) = -75.1 NmMys = Fydxs= (965.9 N)(0.222 m) = 214.4 Nm Ms = Mxs + Mys= -20.2 + 214.4 = 139.3 Nm

(f) Same as d, but replace 0.25 m forearm length with0.30 m. Fx passes through the axis of rotation of the elbow,soMxe = 0 Nm dxe = 0.30 m Fy = 1000 sin 45° = 707.1 N Mye = Fydxe= (707.1 N)(0.30 m) = 212.1 Nm  Me = Mxe + Mye= 0 + 212.1 = 212.1 Nm

Same as d, but replace 0.25 m forearm lengthwith 0.30 m. xs = 0.3 cos 45° = 0.212 mys = 0.3 sin 45° = 0.212 m dxs = 0.30 - 0.212 = 0.088 mdys = y = 0.212 m Fx = 1000 cos 45° = -707.1 NFy = 1000 sin 45° = 707.1 N Mxs = Fxdys= (-707.1 N)(0.212 m) = -149.9 NmMys = Fydxs= (707.1 N)(0.088 m) = 62.2 Nm Ms = Mxs + Mys= -149.9 + 26.9 = -87.7 Nm

 Question #3  THROUGH A THROUGH B(a) Fp = 500 N cos 30° = 433.0 N

d = 0.20 m M = Fd = 433.0 N (0.20 m)= 86.6 Nm

The force acts through axis B, sod = 0, and conseqently,M = 0 Nm

(b) note: l = parallel, and p = perpendicular BCl = 20 cos 60° = 10 cm

note: l = parallel, and p = perpendicular BCl = 20 cos 60° = 10 cm

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BCp = 20 sin 60° = 17.3 cm dl = AB + BCl = 20 + 10 = 30 cmdp = BCp = 17.3 cm Fl = -500 cos 60° = -250.0 NFp = -500 sin 60° = -433.0 N Ml = Fldp = -250.0 N (0.173 m)= -43.3 NmMp = Fpdl = -433.0 N (0.30 m)= -129.9 Nm M = Ml + MpM = -173.2 Nm

BCp = 20 sin 60° = 17.3 cm dl = BCl = 10 cmdp = BCp = 17.3 cm Fl = -500 cos 60° = -250.0 NFp = -500 sin 60° = -433.0 N Ml = Fldp = -250.0 N (0.173 m)= -43.3 NmMp = Fpdl = -433.0 N (0.10 m)= -43.3 Nm M = Ml + MpM = -86.6 Nm

(c) Fa passes through axis of rotation, so Ma = 0 dAB = 0.50 mdAC = 0.25 m Mb = FbdAB = -1000 N (0.50 m)= -500 NmMc = FcdAC = 500 N (0.25 m)= 125 Nm M = Ma + Mb + Mc= 0 Nm + -500 Nm + 125 NmM = -375 Nm

 Fb passes through axis of rotation, so Mb = 0 dAB = 0.50 mdBC = 0.75 m Ma = FadAB = -1500 N (0.50 m)= -750 NmMc = FcdBC = 500 N (0.75 m)= 375 Nm M = Ma + Mb + Mc= -750 Nm + 0 Nm + 375 NmM = -375 Nm

(d) note: l = parallel, and p = perpendicular BCl = 30 cos 40° = 23.0 cmBCp = 30 sin 40° = 19.3 cm dl = AB + BCl = 60 + 23= 83 cmdp = BCp = 19.3 cm Fl = -300 NFp = 0 N Ml = Fldp = -300.0 N (0.193 m)= -57.9 NmMp = Fpdl = 0 N (0.83 m)= 0 Nm M = Ml + Mp

note: l = parallel, and p = perpendicular BCl = 23.0 cmBCp = 19.3 cm dl = BCl = 23 cmdp = BCp = 19.3 cm Fl = -300 NFp = 0 N Ml = Fldp = -300.0 N (0.193 m)= -57.9 NmMp = Fpdl = 0 N (0.23 m)= 0 Nm M = Ml + MpM = -57.9 Nm

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M = -57.9 Nm(e) dx = 0.40 m

dy = 0.30 m Fx = 200 N sin 53° = -159.7 NFy = 200 N cos 53° = 120.4 N Mx = Fxdy= (-159.7 N)(0.30 m) = -47.9 NmMy = Fydx= (120.4 N)(0.40 m) = 48.2 Nm M = 0.3 Nm

dx = 0 mdy = 0.30 m Fx = 200 N sin 53° = -159.7 NFy = 200 N cos 53° = 120.4 N Mx = Fxdy= (-159.7 N)(0.30 m) = -47.9 NmMy = Fydx= (120.4 N)(0 m) = 0 Nm M = -47.9 Nm

 Question #3Dino Dude and his stud buddy decide to go to the park to play on the teeter totter. When the twoarrive at the park, they discover that the teeter totter has been broken.

(a) If Dino Dude sits on the long end (1.55 metre end) of the teeter totter and weighs 20kg, howmuch would Dino Stud have to weigh in order to balance out the teeter totter if he sat on thebroken end, .75 metres away from the fulcrum?

We must determine how much mass is required to balance the lever.F x FMA = R x RMA

Dino Stud must have a mass of 41.3 kg to balance the teeter totter. (b) What is the mechanical advantage of the lever?

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Solutions for Problem Set #8: Angular Kinetics #2

1. A diver sets out to perform a front somersault off a 5 m tower. She intends to take off from herfeet in an erect postition and enter the water feet first in a vertical position. She misjudges badlyand does a belly flop after completing one and a quarter revolutions - 1/4 more thanintended...OUCH! There are several things that she might do to improve performance. List asmany of these as you can and explain in biomechanical terms how each of them might serve toimprove the performance.τ= I αI = mk2 I = Σmr2

τ= torque - produces angular accelerationI = moment of inertia - inertial property of rotating bodies that increases with both mass and thedistance the mass is distributed from the axis of rotation the more closely concentrated the massto the axis of rotation, the easier it is to rotate the object the more mass positioned away from theaxis of rotation, the more difficult it is to initiate or stop angular motion.r = radius of rotationk = radius of gyration - the objects mass distribution woth respect to a given axis of rotation;the distance from the axis of rotation to a point at which the mass of the body can theoreticallybe concentrated without altering the inertial characteristics of the rotating bodySuggestions:come out of tuck earlier: ↑ k which ↑ I which in turn, ↓ α (τ remains constant due toNewton's #1)"looser tuck": ↑ k which ↑ I which in turn, ↓ αmove into tuck slower: IΔt = mΔv ∴ decreases angular momentumlean forward less at take-off: gravity contributes to angular momentum

2. Sketch a graph of the relative values of angular velocity and moment of inertia of a divercompleting a one and a half somersault dive.      3. A weightlifter performing an arm curl with a 30 kg barbell stops the lifting motion with hisforearms in a horizontal position.

a) If the line of action of the weight of the barbell is 30 cm from the axis of rotation, what is themagnitude of the moment that the elbow flexors must exert in order to maintain the position?(Ignore the weight of the lifter's forearms and hands in the calculations.)τ = Fdτ = (30 kg) (-9.81 m/s2) (-0.3 m)τ = 88.29 Nm

b) If the weight of the lifter's forearms and hands were taken into account would the moment thatthe elbow flexors have to exert be more or less than that in (a)? Explain.

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The moment would be more than above because there would be an additional force in the"downward" direction to take into account.

c) If the only elbow flexor muscles active were the biceps, and the distance from its insertion tothe axis of rotation was 3 cm, what force would the muscle have to exert to maintain the forearmin it's horizontal position? (Again, ignore the weight of the forearms and hands.)Fbb = (30 kg) (-9.81 m/s2) = -294 N∴Fb = + 294 N Mbb = FdMbb = (-294 N) (.3 m)Mbb = -88.29 Nm∴Mb = 88.29 Nm Mb = Fd88.29 Nm = Fb (0.03m)Fb = 2943 N

d) A second individual performs an arm curl. He, too, lifts a 30 kg barbell whose line of action is30 cm from the axis of rotation. However, this individual's biceps muscle inserts at a distance of3.5 cm from the axis of rotation. Ignoring the weight of the forearm and hand, what force musthis muscles exert in order to maintain the forearm in it's horizontal position?Mb = Fd88.29 Nm = Fb (0.035m)Fb = 2522.6 N

e) What type of lever is the forearm acting as?Third Class Lever

4. There are several techniques for doing situps. Using the concept of torque, explain how onecould change the technique to make a situp easier, and more difficult. τ= IαI = mk2 I = Σmr2

τ= Fd

a) Easier:↓ k by bringing arms in which ↓ I which in turn ↓ τconcentrate mass as close to the axis of rotation as possible

b) Harderextend arms behind head so body (abs) must generate more τ to oppose the resistance ( ↑ kwhich also ↑ I which decreases τ) 5. The radius of gyration of the thigh with respect to the transverse at the hip is 54% of the

segment length. The mass of the thigh is 10.5% of total body mass, and the length of the

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thigh is 23.2% of total body height. What is the moment of inertia of the thigh with respect tothe hip for people of the following body masses and heights?

   MASS (kg)  HEIGHT (m)

 A  60  1.6 B  60  1.8 C  70  1.6 D  70  1.8 a) I = (60 kg)(0.105) [(1.6 m)(0.232)(0.54)]2 = 0.25 kg m2b) I = (60 kg)(0.105) [(1.8 m)(0.232)(0.54)]2 = 0.32 kg m2c) I = (70 kg)(0.105) [(1.6 m)(0.232)(0.54)]2 = 0.30 kg m2d) I = (70 kg)(0.105) [(1.8 m)(0.232)(0.54)]2 = 0.37 kg m2 6. Angular kinematic data show that in the swing phase the minimum knee angle was 117.7° forwalking and 84.5° for running. Why are the knee angles smaller in the swing phase of running?Knee angles are smaller in running because leg must swing through faster in preparationfor next heelstrike. This decreases radius of gyration which decreases I.

 

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Solutions for Problem Set #9: Work, Power and Energy

   1. A person is performing a lift against a 400 N load. The load moves vertically through adistance of 20 cm in a time of 0.25 s. F = 400 N; d = 0.2 m; t = 0.25s

(a) How much work was done on the load?W = Fd = (400 N)(0.2 m) = 80 Joules

(b) What is the person's average power during the lift?P = W /Δt = 80 J / 0.25s = 320 Watts

(c) What is the load's increased potential energy after the lift?PE = mgh = Fd = (400 N)(0.2 m) = 80 Joules  2. If the load mentioned above was moving at 7.0 m/s after moving 10 cm, what was its totalenergy, kinetic energy, and potential energy at this point (assume that the load started 50 cmfrom the ground)? F = 400 N = mgvf = 7.0 m/sdf = 0.6 mdi = 0.5 mvi = 0 m/sm = F/g = 400 N / 9.81 N/kg = 40.8 kg KE = 1/2 mv2 = 1/2 (40.8 kg)(7.0 m/s)2

= 1000 J PE = mgh = Fd = (400 N)(0.6 m)= 240 J Et = KE + PE = 1240 J

3. How high will a 30 kg object travel after it is released if it is released with a vertical velocityof 2.0 m/s?m = 30 kgvv = 2.0 m/sdto apex = ?g = -9.81 m/s According to the law of conservation of energy, KE + PE = constant. Initially PE = 0, and at theapex KE = 0; so,

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KEi = PEapex1/2 mv2 = mgh1/2 v2 = ghh = v2/2g = (2.0 m/s)2 / 2(9.81 m/s2)h = 0.2 m 4. How much power is present when a force of 500 N is applied to an object with a mass of 100kg that is moving at 6 m/s?F = 500 Nv = 6 m/sm = 100 kgP = ? P = W/t = Fd/t = Fv = (500 N)(6 m/s) = 3000 Watts 5. What work is done when an object is lowered 1.80 m if the object has a mass of 700 kg?h = 1.8 mm = 700 kgW = ? W = Fd = mgd = (700 kg)(-9.81 N)(-1.8 m) = 12361 Joules  

6. What is the impulse delivered to a catcher's mitt by a 360 g ball that is traveling at 100 km/hand is then stopped by the catcher in .01 s?m = 0.360 kgvi = 100 km/h = (100 km/h)(3600 s/h) /(1000 m/km) = 360 m/svf = 0 m/st = 0.01 sI = ? I = Ft = mΔv = 0.360 kg (0 - 360 m/s) = -130 Ns  7. How high would you have to drop an 80 N object so that it will have a kinetic energy of 2000J when it hit the ground?F = 80 N = mgKE = 2000 Jh = ? due to law of conservation of energy, KE at ground = PE at apexKEi = PEapex1/2 mv2 = mgh2000 J = (80 N) hh = (2000 J) /(80 N) = 25 m 

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8. A 55 kg person moves at the constant speed of 7 m/s along a straight stretch of track. What isthe person's:(a) accelerationm = 55 kgv = 7 m/s if velocity is constant, then a = 0

(b) momentumM = mv = (55kg)(7 m/s) = 385 kg m/s

(c) kinetic energyKE = 1/2 mv2 = 1/2(55 kg)(7.0 m/s)2 = 1348 J

9. How much work is done to stop an 90 kg football player who is running at 8.5 m/s,horizontally, and you do not have to knock him down?m = 90 kgvi = 8.5 m/svf = 0 m/sW = ? The work you do must be equal in magnitude and opposite in direction to his kinetic energy.W = -KE = -1/2 mv2 = -1/2(90 kg)(8.5 m/s)2 = -3251 J 10. What is the work done by a force of 500 N that acts at an angle of 25 degrees to an object andthat moves the object a displacement of 12 m?F = 500 N at 25°d = 12 mW = ? W = Fd = (500 N)(cos 25°)(12 m) = 5438 J 11.Dino Dude is peacefully floating over the water when a big blue bird pops his balloons,causing Dino Dude to fall 10 m before crashing into the water. If Dino Dude has a mass of 20 kg,what is his velocity immediately before impact with the water?

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The principle of conservation of mechanical energy must be used to solve the first part of thisproblem. The total mechanical energy Dino Dude possesses at a height of 10 m is his potentialenergy. Immediately before impact with the water, his gravitational potential energy can beassumed to be 0 and 100% of his mechanical energy will then be kinetic.

Find the total energy possessed by Dino DudePE + KE = Constantmagh + 1/2 mv2 = Constant(20 kg)(9.81 m/s2)(10 m) + 1/2 (20 kg)(0)2 = Constant1962 J = Constant Find Dino Dude's velocity just before impactPE + KE = 1962 Jmagh + 1/2 mv2 = 1962 J(20 kg)(9.81 m/s2)(0 m) + 1/2 (20 kg)(v2) = 1962 Jv2 = 196.2 m2/s2

v = 14.0 m/s

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VIII. Biomechanics Terminology

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Acceleration A vector quantity involving positive, negative, or zerovalues.

Angle of Attack The angle between the longitudinal axis of an object andthe flow of the fluid.

Angular Acceleration (α)SI: rad/s2

The rate of change of angular velocity with respect totime. The first derivative of angular velocity

Angular Displacement (θ)SI: radians or °

The rotational change in position of a body. Each point onan object is displaced around an axis.

Angular ImpulseSI: Nms

An impulse that produces a change in angular momentumAngular Impulse = torque x time interval over which thetorque acts.

Angular MomentumSI: kg*m2/s

A vector quantity of the amout of angular motionpossessed by a body.Angular Momentum= angular velocity x moment ofinertia

Angular Velocity (ω)SI: rad/s

The rate at which a body rotates. The first derivative ofangular displasement.

Archimedes Principle The buoyant force acting on a body is equal in magnitudeto the weight of the fluid displaced by the body.

Bernoulli Principle The inverse of the relationship between relative velocityand relative pressure in a fluid flow.

Biomechanics The application of mechanical priniciples in the study ofliving organisms.

Center of Gravity The point at which the weight of the body or system canbe considered to act.

Center of Volume The point around which a body’s volume is equallybalanced. The point at which the bouyant force acts.

Differentiation The rate of change or slope of a line. For moreinformation on differentiation, click on the link in the labnotes or look under mathematical calculation in the coursemanual.

Displacement A vector quantity involving positive, negative, or zerovalues. It may be translational, rotational or both.

Equilibrium All forces and moments acting on a body are at zero. Theobject may be at rest or in motion.

First Class Lever Effort force and resistance force are on opposite sides ofthe axis.

Force (F)SI: Newton

A vector quantity that desribes the action of one body onanother.

Ground Reaction Force(GRF)

The forces acting on the body as a result of interactionwith the ground. These forces are equal but opposite indirection to the forces the body is applying to the ground.

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ImpulseSI: Ns

The effect of a force acting over a period of time. It isthe integral of the force –time curveImpulse = force x time interval over which the force acts

Integration The area under a curve. For more information onintegration, click on the link in the lab notes or look undermathematical calculation in the course manual.

Kinematics The study of the description of motion includingconsiderations of space and time.

Kinetics The study of the forces associated with linear motionLever A simple machine consisting of two forces (an effort force

and a resistance force) and a barlike body that rotatesabout an axis. The lever arm is the perpendicular distanceof each force from the axis of rotation.

Lift A force acting perpendicular to the fluid flow on a body influid

Linear Acceleration (a)SI: m/s2

The rate of change of linear velocity with respect to time.The first deriviative of linear velocity

Linear Displacement (s ord or x)SI: m

The translational change in position of a body. Each pointon an object is displaced along parallel lines.

Linear MomentumSI: kg*m/s

A vector quantity calculated by the mass of an object x thelinear velocity of the object

Linear Velocity (v)SI: m/s

The rate at which a body moves in a strainght line. Thefirst derivative of linear displacement.

Magnus Effect The deviation in the trajectory of a spinning object towardthe direction of a spin resulting from the magnus force.

Magnus Force The lift force created by a spinMechanical Advantage

(MA)The ratio of effort arm/ resistance arm. If the ratio is ‘1’there will be no change in MA, if the ratio is less than ‘1’MA will decrease, and if the ration is greater than ‘1’ MAwill increase

Moment (M)SI: Nm

The turning effect produced by a force.Moment= force x perpendicular distance between thepoint of application of the force and the axis of rotation.

Moment Arm The shortest distance between a forces line of action andthe axis of rotation

Moment of Inertia (I) The inertial property of rotating bodies. It increases withboth mass and the distance that the mass is located fromthe axis of rotation.

Relative Velocity The velocity of a body with respect to another objectaround it (ie. fluid).

Resultant Force The resultant force when all forces acting on an object areadded together and expressed as a single force

Second Class Lever Effort force and resistance force are on the same side ofthe axis with the resistance force being closer to the axis.

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Third Class Lever Effort force and resistance force are on the same side ofthe axis with the effort force being closer to the axis.

Velocity A vector quantity involving positive, negative, or zerovalues.

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IX. References

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Hall, Susan J. 1995. In: V. Malinee, C. Wells (Eds.) Basic Biomechanics 2nd Edition. Mosby-Year Book, Inc..Rodgers, Mary M., Cavanagh, Peter R.. December 1984. Glossary of biomechancial terms,concepts, and units. Physical Therapy. Volume 64, No. 12. P. 1886-1902.