UBC SCHOOL OF HUMAN KINETICS
HKIN 363
Mechanics 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...................................................................................................................................................12
RECORDING PHASE..........................................................................................................................................................12
Equipment Selection ..................................................................................................................................................12
Camera Positioning and Orientation........................................................................................................................12
Marker Placement .....................................................................................................................................................13
Light Placement .........................................................................................................................................................13
Film Calibration ........................................................................................................................................................13
DIGITIZING PHASE ...........................................................................................................................................................14
Linear Kinematics......................................................................................................................................................14
Angular Kinematics ...................................................................................................................................................15
KINETICS METHODS ........................................................................................................................................................17
FORCE PLATFORM KINETIC DATA ACQUISITION ...........................................................................................................17
IV. TIPS ON USING EXCEL ..........................................................................................................................................23
ABOUT EXCEL .................................................................................................................................................................24
BASIC EXCEL FUNCTIONS ...............................................................................................................................................24
Opening Excel...........................................................................................................................................................24
Creating a Worksheet ...............................................................................................................................................24
Opening a Worksheet ...............................................................................................................................................24
Selecting Cells ...........................................................................................................................................................25
Selecting Non-Adjacent Cells ..................................................................................................................................25
Calculations...............................................................................................................................................................25
Correcting Cells ........................................................................................................................................................26
Copy/Paste .................................................................................................................................................................26
Filling Cells ...............................................................................................................................................................27
Graphing....................................................................................................................................................................27
COPYING AN EXCEL CHART INTO A WORD WORKSHEET ..............................................................................................29
PRINTING .........................................................................................................................................................................30
DIFFERENTIATION IN EXCEL (ESTIMATING SLOPES OF LINES) ...............................................................30
V. TIPS ON USING MICROSOFT WORD ..................................................................................................................34
OPENING MICROSOFT WORD ..........................................................................................................................................35
CREATING A WORKSHEET ...............................................................................................................................................35
OPENING A WORKSHEET .................................................................................................................................................35
VI. PROBLEM SET QUESTIONS .................................................................................................................................36
PROBLEM SET #1: REVIEW PROBLEMS...........................................................................................................................37
PROBLEM SET #2: DIFFERENTIATION .............................................................................................................................39
PROBLEM SET #3: LINEAR KINEMATICS ........................................................................................................................41
PROBLEM SET #4: PROJECTILES .....................................................................................................................................42
PROBLEM SET #5: ANGULAR KINEMATICS ....................................................................................................................43
PROBLEM SET #6: LINEAR KINETICS..............................................................................................................................45
PROBLEM SET #7: ANGULAR KINETICS #1.....................................................................................................................47
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PROBLEM SET #8: ANGULAR KINETICS #2.....................................................................................................................49
PROBLEM SET #9: WORK, POWER, AND ENERGY ..........................................................................................................50
VII. PROBLEM SET ANSWERS ...................................................................................................................................52
SOLUTIONS FOR PROBLEM SET #1: REVIEW PROBLEMS ................................................................................................53
SOLUTIONS FOR PROBLEM SET #3: LINEAR KINEMATICS .............................................................................................57
SOLUTIONS FOR PROBLEM SET #4: PROJECTILES ..........................................................................................................60
SOLUTIONS FOR PROBLEM SET #5: ANGULAR KINEMATICS .........................................................................................64
SOLUTIONS FOR PROBLEM SET #6: LINEAR KINETICS ...................................................................................................67
SOLUTIONS FOR PROBLEM SET #7: ANGULAR KINETICS #1..........................................................................................71
SOLUTIONS FOR PROBLEM SET #8: ANGULAR KINETICS #2..........................................................................................76
SOLUTIONS 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 Columbia
School of Human Kinetics
Human Kinetics 363
Course Outline
Summary:
This objective of this course is to build on the principles of physics acquired in Human Kinetics
163, high-school Physics or entry-level university Physics and apply them to a quantitative
analysis 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 the
rehabilitation environment. The student should gain an understanding of the use of a quantitative
analysis to explain how mechanical principles govern human motion. At the completion of this
course it is desired that each student be able to: 1) to understand how 2D rigid body dynamics
can be used to quantify human motion, (2) understand the cause and effect relationship between
force and linear and angular motion, and (3) perform mathematical analysis of complex human
motion in two dimensions.
Dates:
Mall)
Lectures: Tuesday and Thursday, 9:30 am – 11:00 am MacMillan 166 (2357 Main
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 Gym
Office 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 Movement
Authors:
J. Hamill and K.M. Knutzen
Publisher:
Lippincott Williams and Wilkins
WebCT Resources:
There are considerable resources on the WebCT site for this course. You must have an WebCT
account 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 terms
6. Apply knowledge of biomechanical principles to the analysis of skilled and unskilled
performances. Have demonstrated personal and social responsibility towards class
participation.
7. Be able to facilitate active learning, critical thinking, and problem solving skills in the
qualitative 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 and
angular, 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 will
be important.
It is expected that students will demonstrate an ability to use vector analysis to solve problems in determining
velocities, 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 - Kinetics
Using 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 human
movements. Examination of data acquisition and reduction will be made and used to acquire
data for analysis.
It is expected that students will analyse forces acting on an object and predict their effects on it. Specifically, it
is 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 movement
This section will deal with those movements that occur in other mediums such as water, in
the case of swimming, and for those projectiles that exhibit aerodynamic movement patterns
such as discus and javelin throwing.
It is expected that students will analyse forces acting on an object that arise because of its movement through
air 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 movement
This section will build on both the current course and the preceding course and put the
student 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, it
is 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 the
lectures being didactic there will also be group discussion on topics presented by the
instructor. Class participation component will entail analysis of movements provided
primarily on video tape but also through photographs and Power Point presentations. There
will be time allocated within the lecture to address the issues of analysis and
coaching/teaching components. There will be an opportunity to compare each presentation
and form a class consensus. The instructor will facilitate discussion. The group work is
modelled after the discussion by McKay and Emmison, 1995)1.
Laboratory Format. There will be a two-hour laboratory component. There is a laboratory slot
for four days and maximum enrolment in the course will be restricted by the number of
students 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 computer
programs will be used during this course – Microsoft Excel, Microsoft Word, and HMA
Technology’s HU-M-AN. By now you will be familiar with Excel and Word. HU-M-AN is a
program that allows you to digitise video images. In the fashion you will be able to create
numeric data from video clips and use these data to compute joint, segment angles, in
addition to computing joint and segment angular velocity and acceleration. All labs will
include quantitative and qualitative analysis. During these sessions, students will acquire the
necessary 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 the
WEBCT page (http://www.elearning.ubc.ca/login) under resources. Essentially, the report
will be no longer than 4 pages and will be done in a format specific for a scientific
conference.
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
1
2
3
4
5
6
Jan-5
Jan-12
Jan-19
Jan-26
Feb-2
Feb-9
Feb-16
No labs
Lab 1: Lab tour and HU-M-AN practice
Lab 2:100 m Sprint Data Collection
Lab 2:100 m Sprint Data Analysis
Lab 3: Sprint Start Data Collection
Lab 3: Sprint Start Data Analysis
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8
9
10
11
12
13
Feb-23
Mar-1
Mar-8
Mar-15
Mar-22
Mar-29
Apr-5
Lab 4: Vertical Jump Data Collection
Lab 4: Vertical Jump Data Analysis
Lab 5: Muscle Torque Data Collection
Lab 5: Muscle Torque Data Analysis
Lab 6: Rowing Data Collection
Lab 6: Rowing Data Analysis
No Labs
Location
Problem
set
Computer Lab
Field Hockey pitch
Computer Lab
Osborne Gym A
Computer Lab
READING
BREAK
Osborne Gym A
Computer Lab
Osborne Gym A
Computer Lab
Osborne Gym A
Computer Lab
none
1
2
3
4
5
6
7
8
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Evaluation Profile
Learning objective
1, 2, 3, 4
Method
Written examinations
Mid-term
Final
5, 6, 7
Labs
5 labs each worth
Total
Value
60 marks
20 marks
40 marks
40 marks
8 marks
100 marks
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II. Virtual Biomechanics: The Biomechanics 363
Webpage
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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 download
either 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 “Human
Kinetics”, 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
Resources
Password
Quizzes
Bulletin Board
Links
Takes you to the labs and notes sections
This is where some of your labs are
Pops up in a new browser window
Contains helpful resources including background notes, problem sets
with answers, course readings, and exam hints
Pops up in a new browser window
Allows you to change your password from you student number (default)
to something else
Takes you to a variety of multiple choice questions for each of the 5
sections covered in this course
Useful for short answer exam questions and quick concepts
Answers are returned immediately
An interactive bulletin board that can be used to ask the instructor/TA
questions
Also enables students to interact with one another in order to solve
problems
Useful links to biomechanics sites worldwide
Check 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 using
our equipment and motion measurement system. There are two general stages to quantifying
movement: 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 positioning
and orientation, light placement, film calibration, and marker placement.
Equipment Selection
In this example, we will be analyzing the kinematics
of walking. We have chosen to use a treadmill to
simulate the walking motion. Although our
equipment enables us to measure motion in three
dimensions, for the sake of simplicity, this example
will only examine motion in the sagittal plane (two
dimensions). This plane divides the body into left
and right portions.
Camera Positioning and Orientation
In two dimensional analysis, it is imperative that the camera be oriented perpendicular to the
plane 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 Placement
The next stage of setup is marker placement. Highly
reflective markers are placed at various landmarks,
allowing accurate identification
of specific structures on the recorded film. In this case,
the marker landmarks approximate the centres of
rotation of the different segments being analyzed. These
include the greater trochanter (hip), lateral tibio-femoral
joint line (knee), lateral malleolus (ankle), talus (heel),
and fifth metatarsal (forefoot). When light is directed
from the vicinity of the camera, the markers are
illuminated as they reflect the light back towards the
camera lens.
Light Placement
The reflective markers demonstrate a narrow angle of
reflection, therefore light placement is important in
order to achieve maximal illumination of the
markers.
Film Calibration
The last portion of the setup phase is film
calibration. This involves filming an object of a
known length. The actual and filmed length of the
object, a metre stick in this case, will be inputted
to the computer in the next stage. This permits the
calculation 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 this
stage, the motion is analyzed by quantifying the changes in location of the joint markers. At the
UBC Biomechanics lab, we use the Peak Motus Motion Measurement System to measure
movement kinematics.
This process begins by manually indicating the locations of the markers in the initial frame. This
is achieved by positioning a "cross-hair" over the markers on the computer screen and clicking
the mouse to select the location. The computer then records the x and y coordinates of the crosshair as the marker location. This manual process is repeated for all of the markers in the first
frame. 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 complete
the Virtual Digitizing Session.
Once coordinate data have been obtained for each marker in each frame, the data can be fed into
a spreadsheet program such as Excel, as shown below, and analyzed from both linear and
angular perspectives. The data can also be represented graphically once this is done.
Linear Kinematics
The coordinate data can be examined in the context of linear kinematics, or in the context of
changes in the locations of the markers over time. In other words, each frame will provide a
vertical 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 the
graph below.
Similarly, both the x and y ankle marker coordinates can be graphed, giving the path of the
ankle during the walking cycle. This is shown in the picture below.
Angular Kinematics
These data can be further analyzed by joining the markers to form segments. By calculating the
orientation of the segments relative to one another, or relative to a fixed angle in space, the
angular kinematics of the walking stride can be quantified. For example, the angle between the
thigh 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 the
kinematic labs, both linear and angular, these techniques were used. Once the data had been
digitized they were processed to compute the respective data and stored in files that can be
loaded by Microsoft Excel. You should now be able to work with those files to complete your
assignments.
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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 the
biomechanics lab. This equipment uses 4 piezoelectric transducers, located at the corners of the
platform 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 the
lateral forces for a walker are negligible). The force platform is then connected to an amplifier
using electrical wires. The amplifier boosts the signal from the force platform so that the
computer can "hear" the data. The amplifier is also connected to the computer using electrical
wires. The Force platform is activated by powering up the amplifiers and connecting it to the
Peak Motus Motion Measurement System. It is then ready to measure movement kinetics. As an
athlete applies a force to the force platform, the data from the force platform is passed through
the amplifiers and fed into the computer.
If an person lunges forward and sideways from a force
platform, a 3 dimensional force is applied to the force
platform. This force is called a resultant vector and
can be broken into its three component vectors. These
lie in the vertical (Fz), AP (Fy) , and lateral (Fx)
directions. The Kistler force platform measures these
components and passes them on to the computer.
The force applied to the force platform, however, is not the force that we are
interested in. It is the reaction force of the platform on the athlete that is of interest. Newton's
third law states that for every force there is a reaction force that is equal in magnitude and
opposite in direction. For example, as the athlete's foot applies a force to the force platform, the
force platform exerts a ground reaction force on the foot that is equal in magnitude and
opposite in direction to the resultant force. This can also be broken into three vectors, which are
oriented in the vertical, AP, and lateral directions. These are also equal in magnitude and
opposite 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 is
applied to the force platform.
The Kistler force platform is also capable of measuring the coordinates of the point at which the
force is applied. This is measure using an x and y coordinate system. It is important to note that
a force platform analyses forces at single point, even though force is actually being distributed
about an area (pressure).
Lastly, the force platform measures moments
exerted on the platform. This is to say, if a force is
applied in such a way as to cause the platform to
rotate, the moment of this torque can be measured.
The kinetics of motion are analyzed by quantifying
the changes in the forces applied to the force
platform, and the resultant ground reaction forces.
This process begins by manually clicking a trigger to commence the recording of data. The
computer then records the x, y, and z resultant forces provided by the force platform. The
computer 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 the
output is a horizontal line of constant value. This value reflects the generated action/reaction
force. That is, the person applies a force to the platform by virtue of standing on it. In return, the
platform provides a reaction force to the person. During quiet standing, the acceleration of the
person is zero and thus the platform can be used to record the person's weight. There are two
forces acting on the individual: the green arrow (FG) represents the force of gravity (mg). The
red arrow (FR) represents the reaction force from the platform. We can sum these forces using
Newton's second law: -FG + FR = ma, where "m" is the person's mass and "a" is the acceleration
of the person's centre of mass. In quiet standing, "a" is zero and thus FR = FG and thus the
ground reaction force is equal to the force due to gravity, the person's weight. Note that in the
equation, 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 are
collected.
Once the data have been collected, it can be copied into a spreadsheet program (e.g.Excel). The
forces can be input with respect to time as shown in the Excel spreadsheet below.
The data can then be manipulated to
produce graphs which graphically
represent the data. A graph plotting
vertical force and AP (anterior-posterior)
force with respect to time is shown
below. The red line is Vertical GRF
(ground reaction force) plotted against
time, and the blue line is anteriorposterior GRF plotted versus time. It is
important to note that the computer
records the data for as long as it is told,
and not just during the time that the
person is in contact with the force
platform. Because of this, the computer
records forces of zero for the points
before and after the contact made by the
athlete. This can be seen in the graph
below, where the computer sampled the data for two seconds, yet the athlete only made contact
with 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 during
walking. As you can see, the vertical GRF increases sharply upon heel strike, then dips
somewhat during the stance phase. Finally, the forces rise again and fall off as the subject leaves
the 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 is
negative at this point (i.e. the athlete is slowing down). During the stance phase, the acceleration
becomes less and less negative as the athlete moves over the base of support, finally becoming
positive midway through the stance phase. The acceleration continues to increase in magnitude
because the subject pushes off the force platform which results in a GRF propelling the athlete
forward. The acceleration drops off as the subject leaves the force platform. 20
This picture shows the resultant GRF vector of the
force applied to the athlete upon heel-strike. This is
obtained by adding the horizontal and vertical force
vectors.
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IV. Tips on Using Excel
23
About Excel
Excel is a spreadsheet program that is useful for quick mathematical calculations and data
analysis. The spreadsheet has rows and columns that form an infinite number of cells. Each cell
has an address defined by the row number and column letter. Excel can be used on a Macintosh
or a PC in many different versions. Currently, the PCs in the Human Kinetics computer lab
have Microsoft Excel 97.
The biomechanics 363 laboratory activities will be easier if you understand a few of Excel’s
basic functions. The three most important functions to know are:
1. creating a new worksheet
2. opening a saved worksheet
3. saving a worksheet
Listed below are the three most important functions and a selected few of the most frequently
used 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 that
computer’s hard-drive; d: is the CD ROM; all other letters are other drives) by clicking
on the down arrow beside the box LOOK IN: and then selecting the file you wish to open
or another folder.
When you locate the file you want to open, double click on it or single click on it and
then press the OPEN button.
If you cannot find the file you want to open, click on the down arrow beside the box
FILES 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 still
cannot 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 the
cells 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 then
press the END key and then hold down the SHIFT key while you press the down arrow. The
entire 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 the
spreadsheet. 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 the
function wizard to guide you through the calculations. Click on the “fx” button near the top of
the screen to use the function wizard. It helps to know a few shortcuts to make your formulas
shorter and easier to understand.
Addition/Subtraction/Multiplication/Division
If you want to perform simple arithmetic, just enter the formula. For example, if you want to add
two numbers, enter: =cell+cell.
If you are adding a row of numbers, instead of adding each one separately, you can enter the
following equation: =SUM(cell1:cell5), where cell1 is the first cell and cell5 is the last cell in the
row you want to add together.
If the numbers you want to add are not in consecutive cells, you can combine the ideas above
and use =SUM(cell1:cell2,cell3,cell7,cell9:cell34).
Average
If you want to take the average of some numbers, use the word AVERAGE. For example, if
you wanted to take the average of a row of numbers, enter into the cell where you want the
value to be: =AVERAGE(cell1:cell2).
25
Min/Max
To find the minimum or maximum in a string of numbers, use the commands MIN and MAX.
For example: =MIN(cell1:cell22)
Standard Deviation
The STDEV command returns the standard deviation of a list of numbers. For Example:
=STDEV(cell1:cell22)
Square Root
The 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 the
values in the formula to stay the same while others will change as you fill the formula across.
When the ‘$’ is placed before the cell or value you DO NOT want to change, it will stay the
same across the column or row. For Example, if you want to add a row of values individually
to a value in cell A1, your first formula would be: =SUM($A$1:A2). If you copy the formula
across cells, the next cell’s formula will be ($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.
Undo
There 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 menu
2. Select the picture of two pieces of paper at the top of the screen
3. Click the left mouse button and select copy
4. 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 menu
2. Clicking on the clipboard from the standard toolbar at the top of the screen
3. Selecting ‘paste’ from the menu that pops up after clicking the left mouse button
4. Press CTRL and V on your keyboard
26
Filling Cells
If you have a function you would like to copy to many cells in a row or column, you could either
type 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 plotted
on 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 Colour
Double-click on the background of the graph.
Under the AREA section, click NONE and then OK.
Formatting the gridlines
Colour: To change the colour of a line, double-click on a line, and then click on the PATTERNS
tab in the new window. Click on the down arrow beside COLOR in the LINE box. Choose a
new colour.
Thickness: To change the line thickness click on the down arrow beside WEIGHT and select the
thickness you want.
Marker: The change the style of the marker click on the down arrow beside STYLE and select a
marker pattern. Marker colour can be changed by clicking on the arrow next to "Background"
Changing Axis Scales
1. Double click on the axis you want to change.
27
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 on
the axis. MINOR UNIT is the smallest interval between units and MAJOR UNIT is the largest
interval between units.
Adding/Changing Titles
To add or change a title to your graph or axis:
Click on the CHART tab at the top of the screen
Select 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 variable
selected.
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 Location
To 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 Options
There 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 AXIS
TITLES.
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 major
units 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” which
labels each point on the chart with the Y coordinate value or show label which labels each
point with the X-coordinate value.
Changing Graph Type
To change the type of graph you initially selected:
1. Open the menu under CHART
28
2.
3.
4.
5.
6.
Select ‘CHART TYPE.
Choose either the ‘STANDARD TYPE or CUSTOM TYPE tab from the new window.
Click on a ‘CHART TYPE from the list below choose the type of chart you want.
To view your new chart, click PRESS AND HOLD TO VIEW SAMPLE.
Click OK to finish.
Adding Information
To add data to your chart, enter the data in the worksheet with the original chart data and
highlight the new data.
You now have three options:
1. Select the copy command from the Edit menu, or
2. Select the picture of two pieces of paper at the top of the screen, or
3. 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 the
screen.
Click anywhere on the chart background.
Select PASTE from either:
the edit menu
the clipboard from the standard toolbar at the top of the screen
clicking the right mouse button
Your new data will now appear as part of the chart.
Having More Than One Graph Per Page
To 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 Worksheet
Begin 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 same
time, close Excel before going to Word.
29
Printing
Printing now has a fee!! Instructions will be available in the computer lab. You will be given a
print 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 of
a line by drawing tangents. What we need to do 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 velocity (shown in the graph below). Differentiation
of 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 time
interval.
To begin with, you have a series of data points representing
velocity at particular points in time which were used to plot this
graph. In your spreadsheet, the data points would typically be
arranged in a column where each row represents an equal
increment in time. In the spreadsheet, shown below, the
numbers representing time are located in column "A" and the
numbers representing velocity are located in column "B".
However, if these are located in different columns, (i.e. columns "G" and "H"), the formulas
explained below are modified accordingly. We can easily find the slope
between two points by simply calculating the change in velocity (V2 - V1) and divide that by the
change in time (t2-t1). In this example, you would take the velocity data (column "B"), and
calculate the difference between one data point and the point preceding it (ex. B2-B1). This
represents the change in velocity during a given interval V2-V1. This number is then divided by
the length of this interval (ex. A2-A1) or the time interval over which the change in velocity
occurs (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). As
explained, 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 be
adapted 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 would
you 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, we
must employ slightly different techniques to solve this problem.
30
We will simply calculate the slope between 3 points and place this value at the mid point. For
example, 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 = (B3B1)/(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 column
and have the differential of column B. This approximates the slope of the data in column B.
There are some costs for this estimation
and that is we loose data for the first
and last points. There are more
sophisticated methods of determining
these values but we will not need these
for this course.
Integration in Excel
THE INTEGRAL
By now you are familiar with integration by graphical means - that is, estimating the area
beneath a curve by counting the enclose squares or other geometrical shapes. What we need to
do 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). Integration
of the acceleration-time graph will provide information on velocity. The equation is V = Adt
where V is velocity and A is acceleration and dt is the time interval. This, of course, tells us the
change in velocity over some interval but not the actual velocity. You need to know the starting
velocity before you can determine that. So the final equation is V = Vi + Adt where Vi is the
initial velocity.
31
To begin with, you have a series of data points representing velocity at particular points in time
which were used to plot this graph. The time interval between points must be the same, a
constant. In your spreadsheet, the data points would typically be arranged in a column where
each row represents an equal increment in time. To find the area under this curve we will use the
Trapaziod 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). Because
dt 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 greater
the number of divisions the lower the error. This equation
can easily be put into a spreadsheet. Consider a column of
numbers with a constant time interval of 0.05. The values
are arranged in column B from row 1 through 11. The
integral of the curve would then be calculated by:
= (.5*B1 + SUM(B2:B10) + .5*B11) * 0.05
This formula will be useful in determining the impulse
during a vertical jump. In this example the answer
is...0.3543467
Integration and Plotting
There are times, like in the kinematic labs, when you are
asked to integrate acceleration and plot velocity (or
integrate velocity and plot displacement). In this case we
are 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 the
area 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 (0.05
increment) and in B are the acceleration values (shown here). To compute the instantaneous
integral enter the value 0 is column C1. In C2 enter the equation = C1 + B2*0.05
32
This will compute a simple change in velocity as a
result of that acceleration and add it to the initial
velocity (in C1). In this case we set the initial
velocity at 0 but of course it could be any value.
Now you can copy (or autofill) this to the bottom
of column C and you will have estimated the
integral of column B. These two columns (B and
C) can be plotted as a function of column A and
allow you to compare the velocity and
acceleration.
Of course, you can carry on in this fashion and
integrate column C, velocity, and compute a new
column in D which will contain the estimate of the
displacement.
33
V. TIPS ON USING MICROSOFT WORD
34
Microsoft word is a word processing program which will be useful for writing up your lab
assignments 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 of
your screen. A menu will pop up. Click on PROGRAMS -- MICROSOFT WORD. Microsoft
Word 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) by
clicking on the down arrow beside the box LOOK IN: and then selecting the file you wish to
open or another folder. When you locate the file you want to open, double click on it or single
click on it and then press the OPEN button.
To copy a graph from excel into a word document, follow the instructions above in Excel
Functions.
35
VI. Problem Set Questions
36
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
37
III GEOMETRY
Find the angles, in degrees, of Θ:
38
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.
39
7.
8.
9.
40
Problem Set #3: Linear Kinematics
1. A skater increases her speed with constant acceleration from 22 m/s to 30 m/s
over 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 a
meaningful measurement?
3. A cyclist increases her speed with constant acceleration from 20 m/s to 28 m/s
over 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?
41
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/s
upwards.
(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 5
m/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 became
airborne?
(c) What distance will she travel if she lands 1 meter lower than she took-off from? (Hint, find
the 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 his
vertical impulse and enabling a 5% increase in his vertical velocity, what will now be the rise of
his center of mass?
(c) How far will the jumper travel, horizontally, if he lands 1 m above the ground? Assume the
velocity used in part (a).
42
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.
What is the average angular velocity of: (i) its end?
(a)
(ii) a point 30 cm from the axis of rotation?
What is the linear velocity of: (i) its end?
(b)
(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 a
constant 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 m
above 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 she
do 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 her
angular velocity when performing piked somersaults?
(c)The diver is performing a piked dive. She realizes that she won't have enough time to finish
her last somersault and enter the water in a vertical position, so she moves into a tucked position
over 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 takes
0.75 seconds to complete 1 revolution before the thrower releases the hammer. Assume that the
thrower is at rest when t=0s and that the path of the arm is circular. Find the magnitude of the
angular acceleration immediately before release of the hammer and the magnitude of total linear
acceleration upon release.
43
7. A soccer player kicks a ball so that it leaves his foot traveling at an angle of 20° from the
ground. His leg length is 1.2 m, and his hip is rotating at 3π rad/s as he executes the kick. How
far will the ball travel before hitting the ground? 44
Problem Set #6: Linear Kinetics
1. Find the resultant force and angle for the following horizontal and vertical
forces:
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 kN
2. Find the components of a 400N force applied to a ball of mass 8 kg, if the force
is 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 a
rope (which passes 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 a mass of 20 kg, begins to accelerate
downwards 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
45
sloped at an angle of 30° to flat ground. The box has a mass of 5 kg. What is the
magnitude 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 headon. She weighs 47 kg and is traveling at 4.5 m/s; he weighs 75 kg and is traveling
at 4 m/s. They try to cover up their mistake by grabbing onto each other and
gliding 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 bat
moving 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 applied
by the bat?
7. A 90 kg hockey player travelling with a velocity of 10 m/s collides head-on with
a 75 kg hockey player travelling at 6 m/s. If the two players become entangled and
continue travelling together as a unit following the collision, what is their
combined velocity and direction of travel?
8. What is the initial resultant velocity of a projectile that has a vertical velocity of
25 m/s and was projected at a 30 degree angle? If it weighs 20N what is its linear
momentum?
46
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 the
following given elbow angles (θ), force directions (Φ) and forearm lengths (L). Assume the
contraction 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 each
diagram.
47
4. Dino Dude and his stud buddy decide to go to the park to play on the teeter totter. When the
two 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, how
much would Dino Stud have to weigh in order to balance out the teeter totter if he sat on the
broken end, .75 metres away from the fulcrum?
(b) What is the mechanical advantage of the lever?
48
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 her
feet in an erect postition and enter the water feet first in a vertical position. She misjudges badly
and does a belly flop after completing one and a quarter revolutions - 1/4 more than
intended...OUCH! There are several things that she might do to improve performance. List as
many of these as you can and explain in biomechanical terms how each of them might serve to
improve the performance.
2. Sketch a graph of the relative values of angular velocity and moment of inertia of a diver
completing a one and a half somersault dive.
3. A weightlifter performing an arm curl with a 30 kg barbell stops the lifting motion with his
forearms 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 the
magnitude 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 that
the 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 to
the axis of rotation was 3 cm, what force would the muscle have to exert to maintain the forearm
in 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 is
30 cm from the axis of rotation. However, this individual's biceps muscle inserts at a distance of
3.5 cm from the axis of rotation. Ignoring the weight of the forearm and hand, what force must
his 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 one
could 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 the
segment length. The mass of the thigh is 10.5% of total body mass, and the length of the thigh is
23.2% of total body height. What is the moment of inertia of the thigh with respect to the 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
6. Angular kinematic data show that in the swing phase the minimum knee angle was 117.7° for
walking and 84.5° for running. Why are the knee angles smaller in the swing phase of running?
49
Problem Set #9: Work, Power, and Energy
1. A person is performing a lift against a 400 N load. The load moves vertically through a
distance 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 total
energy, kinetic energy, and potential energy at this point (assume that the load started 50 cm
from the ground)?
3. How high will a 30 kg object travel after it is released if it is released with a vertical velocity
of 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 100
kg 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/h
and 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 2000
J 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 is
the 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 and
that 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?
50
51
VII. Problem Set Answers
52
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
(f)
a = b/d - c
d = b/(a+c) = 10 /(20+5)
d = 0.4
(g)
a2 = b2 - 2cd
d = (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) + 5/10
d = 80.5
(j)
d/a = b/c
d = ab/c = 20(10) / 5
d = 40.0
(d)
a = bd +3
d = (a - 3) / b = (20 - 3) / 10
d = 1.7
(e)
d = 25 + ab + b2 = 25 + (20)(10) + 102
d = 325.0
Question #2
recall: a / sin A = b / sin B = c / sin C
and, 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°
53
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°, so
d = sin 130° (4 cm / sin 20°) = 8.96 cm
(d)
c2 = a2 + b2 - 2ab cos θ
cosθ= (a2 + b2 - c2) / 2ab
assign c to the side opposite θ
θ = 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°,
so
θ = 180° - 25° - 110° = 45°
(c)
θ = 180° -α
α= 45°, so
θ = 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, so
θ = 180° - 40° - 20° = 120°
(h)
θ = 360° - 65° - α
due to parallel lines, α = 55°, so
θ = 360° - 65° - 55° = 240°
(i)
α = 180° - 110° = 70°
θ = 360° - 220° -α= 360° - 220° - 70°; = 70°
54
Solutions for Problem Set #2: Differentiation
1.
2.
3.
4.
5.
6.
7.
8.
55
9.
56
Solutions for Problem Set #3: Linear Kinematics
A general strategy useful in solving linear kinetics problems is as follows:
list all given values and unknowns
choose a formula which uses appropriate variables
isolate the unknown variable and write down the formula (this way if you make a calculation error you may still get
part marks for having the correct approach)
substitute givens and solve for unknowns
examine your answer: are the units correct? does the answer make sense? If you have time, you may wish to double
check your calculations.
1. A skater increases her speed with constant acceleration from 22 m/s to 30 m/s over 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?
(a)
vi = 22 m/s
vf = 30 m/s
t=4s
a=?
a=Δv / t = (vf - vi) / t = (30 m/s - 22 m/s) / 4s
a = 2 m/s2
(b)
vi = 22 m/s
vf = 30 m/s
t=4s
a = 2 m/s2
d=?
vf2 = vi2 + 2ad
d= (vf2 - vi2) / 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)/2
vavg= 26 m/s
(d)
vi = 22 m/s
a = 2 m/s2
d = 30 m
vf = ?
vf2 = 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/s
a = 2 m/s2
d = 50 m
t=?
vf2 = vi2 + 2ad
vf = √ (vi2 + 2ad)
vf = √ [(22 m/s)2 + 2(2 m/s2)(50 m)]
vf = 26.15 m/s
vf = vi + at
t = (vf - vi) / a = [(26.15 m/s) - (22 m/s)] / (2
m/s2)
t = 2.08 s
57
(f)
vf = vi + at
vi = 22 m/s
t = (vf - vi) / a = [(25 m/s) - (22 m/s)] / (2
vf = 25 m/s
m/s2)
a = 2 m/s2
t = 1.5 s
t=?
2. What is the average velocity of a sprinter who runs the 100 m in 9.8 s? Is this a meaningful
measurement?
v = d/t = 100 m / 9.8 s = 10.2 m/s
No, 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 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?
(a)
a =Δv /Δt = (28 m/s - 20 m/s) / 4s
a= 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)
vf2 = 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)] / 2A
t = [-20 ± √ (202 - 4(1)(-25))] / 2(1)
t = [-20 ±√500] / 2
t = 1.18 or -21.18, but t must be positive, so
t =1.18 s
Approach 2: First find vf
vf2 = vi2 + 2ad
vf = √ [(20 m/s)2 + 2(2 m/s2)(25 m)] = 22.36 m/s
vf = vi + at
t = (vf - vi) / a
t = (22.36 m/s - 20.00 m/s) / 2 m/s2
t = 1.18 s
(f)
vf = vi + at
t = (vf - vi) / a = (25 m/s - 20m/s) / 2 m/s2 = 2.5 s
59
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/s
upwards.
(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 + tdown
tup:
ay= -9.81 m/s2
vyf= 0 m/s
vyi= 2 m/s
tup= ?
tdown:
tup= 0.204 s
vyi= 2 m/s
vyf= 0 m/s
dup= ?
ddown= -10.204m
ay= -9.81 m/s2
vyi= 0 m/s
tdown= ?
vyf = vyi + atup
tup = (vyf - vyi) / a = [(0 m/s) - (2 m/s)] / (-9.81 m/s2)
tup = 0.204 s
dup= [(vyf + vyi) / 2] t
dup= [(0 m/s + 2 m/s) / 2] (0.204 s)
dup= 0.204 m
* ddown=10 m + 0.204m = 10.204 m
ddown= vyit + 1/2 (a tdown2)
ddown= (0 m/s) t + 1/2 (a tdown2)
= 1/2 (a tdown2)
tdown2 = 2 ddown / a
= [(2) (-10.204 m)] / (-9.81 m/s2)
tdown= 1.44 s
ttotal:
ttotal = tup + tdown = 0.204 s + 1.44 s
ttotal = 1.64 s
(b)
ddown= 10.204 m
vyi= 0 m/s
ay= -9.81 m/s2
v =?
yf
vyf2= vyi2 + 2 ay ddown
vyf2= (0 m/s)2 + (2)(-9.81 m/s2)(10.204 m)
vyf2= 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 cause
horizontal acceleration...in this case, the only external force is gravity which acts in the vertical
direction)
60
(d)
ax= 0 m/s
dx= vxi t + 1/2 (ax ttotal2)
ttotal= 1.65 s
dx= vxi t + 0 = vxi t
vxi= 3 m/s
dx= (3 m/s)(1.65 s)
dx= ?
dx= 4.95 m
2 An athlete leaves the ground with a horizontal velocity of 12 m/s and a vertical velocity of 5
m/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 became
airborne?
(c) What distance will she travel if she lands 1 meter lower than she took-off from? (Hint, find
the flight time.)
(a)
tan θ= 5/12
θ = tan-1(5/12)
θ = 22.62°
(b)
first find t total
ttotal = tup + tdown
tup:
ay= -9.81 m/s2
vyf = vyi + atup
vyf= 0 m/s
tup = (vyf - vyi) / a = [(0 m/s) - (5 m/s)] / (-9.81 m/s2)
vyi= 5 m/s
tup = 0.51 s
tup= ?
since initial and final heights are 0, tup = tdown, and ttotal = 2 tup
ttotal:
ttotal = 2 tup = 2 (0.51 s)
ttotal = 1.02 s
now find d horizontal
range = 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 m
dup = 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
61
so, ttotal = .68 + .51 = 1.19 s
range = (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 his
vertical impulse and enabling a 5% increase in his vertical velocity, what will now be the rise of
his center of mass?
(c) How far will the jumper travel, horizontally, if he lands 1 m above the ground? Assume the
velocity used in part (a).
(a)
initial vertical velocity = (10 m/s) sin 20° = 3.42 m/s
tup = (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 + .60
max 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 + at
t = (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 m
height = 1.25 + 0.66 = 1.91 m
(c)
from (a), tup = 0.35 s, and max height = 1.85 m
but 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
62
range = vhorizontalt = (10 cos 20°)(0.77) = 7.23 m
63
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°/s
ii) same as above
(b)
i) v =ωr = (1.05 rad/s)(1 m) = 1.05 m/s
ii) 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 a
constant 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/s
a ~α = 3 °/s2 = 0.0524 radians/s2
Δt = 5 s
d ~ 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 to
θ= 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 m
above 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 she
do 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 m
Δd = -9.8 m
a = -9.81 m/s2
vi = 0
t=?
d = vit + 1/2 at2
since vi = 0, then d = 1/2 at2
t2 = 2d/a
t = √[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 s
t = 1.4 s
ω= 5.7rad/s
θ= ?
in linear equivalents we have v and t, and are looking for d, so use v = d/t
ω=θ/t
θ=ωt = (5.7πrad/s)(1.4 s) = 8πrad/s
since 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 her
angular velocity when performing piked somersaults?
t = 1.4 s
θ = (3π rotations)(2πrad/rotation) = 6πrad
ω=?
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 finish
her last somersault and enter the water in a vertical position, so she moves into a tucked position
over 0.2 s. What is her angular acceleration?
think of a =Δv /Δt
α=Δω /Δt = (ωt - ωp) /Δt
α= (5.7πrad/s - 4.3πrad/s) /0.2 s
= 7πrad/s
65
6.
The hammer thrower's arm is 0.8m long and the hammer extends an
additional 1.2m. It takes 0.75 seconds to complete 1 revolution before the
thrower releases the hammer. Assume that the thrower is at rest when t=0s
and that the path of the arm is circular. Find the magnitude of the angular
acceleration immediately before release of the hammer and the magnitude
of total linear acceleration upon release.
List all known and unknown variables.
r = 0.8 m + 1.2 m = 2.0 m
Δt = 0.75 seconds
ω1 = 0 rad/s
θ= 2π
α= ?
a=?
First, let us solve for α.
α= Δω/Δt
ω1 = 0 rad/s
ω2 = Δθ/Δt = 2π/0.75 = 2.67π rad/s
α=Δω/Δt = (2.67 πrad/s - 0 rad/s)/0.75
α= 3.56 πrad/s = 11.18 rad/s2
Next, let's solve for a.
at = (v2 - v1) /Δt
we need v2 for this equation:
v2 = rω2 = (2.0 m) ( 2.67π rad/s ) = 16.78 m/s2
at = (16.78 m/s - 0 m/s) / 0.75 s = 22.37 m / s2
at = 22.37 m / s2
ar = (vt2 / r) = (16.78 m/s)2 / (2 m)
ar = 140.78 m / s2
a2 = ar2 + at2 = (140.78 m/s2)2 + (22.37 m/s2
a = √( 140.78 m/s2)2 + (22.37 m/s2) 2
a = 142.55 m/s2
66
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/h
θ = 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 a
40 degree angle to the horizontal? What is the ball's acceleration?
r = 400 N, θ= 40°
x = 400 cos 40° = 306.4 N
y = 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 (which
passes 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 a
mass of 20 kg, begins to accelerate downwards at a rate of 2 m/s2.
a) What is the net force acting on Dino Dude?
67
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 components
since 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 + Ftension
Ftension = 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 angle
of 30° to flat ground. The box has a mass of 5 kg. What is the magnitude and direction of the
force 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 solve
Fnet = ma = 5 kg x 1.5 m/s
68
Fnet = 7.5 N
Fnet x = Fnet cos 30° = 7.5 N cos 30° = 6.50 N
Fnet y = Fnet sin 30° = 7.5 N sin 30° = 3.75 N
Fnet x = Fgravity x + Fpush x
Therefore Fpush x = Fnet x - Fgravity x = 6.50 N - 0 N
Fpush x = 6.50 N
Fnet y = Fgravity y + Fpush y
Therefore 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] = √2830.10
Fpush = 53.20 N
From trig relations we know that tanθ = 52.8 N / 6.50 N
Therefore, 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. She
weighs 47 kg and is traveling at 4.5 m/s; he weighs 75 kg and is traveling at 4 m/s. They try to
cover up their mistake by grabbing onto each other and gliding together. What is their resultant
velocity?
P i = Pf
m1v1i + m2v2i = m1v1f + m2v2f
m1v1i + m2v2i = (m1 + m2)vf
vf = (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 in
the 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 = FΔt
F = 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 kg
hockey player travelling at 6 m/s. If the two players become entangled and continue travelling
together 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 collision
equals the total momentum after impact.
m1v1 + m2v2 = (m1 + m2)vf
(90 kg)(10 m/s) + (75 kg)(-6 m/s) = (90 kg + 75 kg)vf
vf = 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 and
was projected at a 30 degree angle? If it weighs 20N what is its linear momentum?
Vi = ?
Vv = 25 m/s
θ= 30°
sin 30° = 25 m/s / Vi
Vi = 25 m/s / sin 30°
Vi = 50 m/s
M=mv
M= (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/s
y = 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 N
y = 10 sin 30° = 5.0 N
(d)
x = 25 cos 160° = -23.5 N
y = 25 sin 160° = 8.6 N
(e)
x = 100 cos (-45°) = 70.7 m/s
y = 100 sin (-45°) = -70.7 m/s
(f)
x = 25 cos (π/4 radians) = 17.7 m/s
y = 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 of
application of the force and the axis of rotation.
M = Fd = (1000 N)(0.30 m)
M = Fd = (1000 N)(0.30 m)
M = 300 Nm
M = 300 Nm
(b) M = Fd = (1000 N)(0.35 m)
M = Fd = (1000 N)(0.35 m)
M = 350 Nm
M = 350 Nm
(c) Fx passes through the axis of rotation
xs = 0.3 sin 45° = 0.212 m
of the elbow, so
ys = 0.3 sin 45° = 0.212 m
Mxe = 0 Nm
dxs = 0.3 + 0.212 m = 0.512 m
dxe = 0.3 m
dys = y = 0.212 m
Fy = 1000 cos 45° = 707.1 N
Fx = 1000 sin 45° = 707.1 N
Fy = 1000 cos 45° = 707.1 N
Mye = Fydxe
= (707.1 N)(0.3 m) = 212.1 Nm
Mxs = Fxdys
= (707.1 N)(0.212 m) = 149.9 Nm
Mys = Fydxs
Me = Mxe + Mye
= (707.1 N)(0.512 m) = 362.0 Nm
= 0 + 212.1 = 212.1 Nm
moments are in same direction, so
Ms = Mxs + Mys
= 149.9 +362 = 511.9 Nm
(d) Fx passes through the axis of rotation of the elbow, xs = 0.3 cos 45° = 0.212 m
so
ys = 0.3 sin 45° = 0.212 m
Mxe = 0 Nm
dxs = 0.25 - 0.212 = 0.038 m
dxe = 0.25 m
dys = y = 0.212 m
Fy = 1000 sin 45° = 707.1 N
Fx = 1000 cos 45° = -707.1 N
Fy = 1000 sin 45° = 707.1 N
Mye = Fydxe
= (707.1 N)(0.25 m) = 176.8 Nm
Mxs = Fxdys
= (-707.1 N)(0.212 m) = -149.9 Nm
Me = Mxe + Mye
Mys = Fydxs
= 0 + 176.8 = 176.8 Nm
= (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° , Same as above, but replace 45° angles with 75° ,
and 0.25 m forearm length with 0.30 m.
and 0.25 m forearm length with 0.30 m.
72
Fx passes through the axis of rotation of the elbow,
so
Mxe = 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 m
ys = 0.3 sin 75° = 0.290 m
dxs = 0.3 - 0.078 m = 0.222 m
dys = ys = 0.290 m
Fx = 1000 cos 75° = -258.8 N
Fy = 1000 sin 75° = 965.9 N
Mxs = Fxdys
= (-258.8 N)(0.290 m) = -75.1 Nm
Mys = 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 with Same as d, but replace 0.25 m forearm length
0.30 m.
with 0.30 m.
Fx passes through the axis of rotation of the elbow, xs = 0.3 cos 45° = 0.212 m
so
ys = 0.3 sin 45° = 0.212 m
Mxe = 0 Nm
dxs = 0.30 - 0.212 = 0.088 m
dxe = 0.30 m
dys = y = 0.212 m
Fy = 1000 sin 45° = 707.1 N
Fx = 1000 cos 45° = -707.1 N
Fy = 1000 sin 45° = 707.1 N
Mye = Fydxe
= (707.1 N)(0.30 m) = 212.1 Nm
Mxs = Fxdys
= (-707.1 N)(0.212 m) = -149.9 Nm
Mys = Fydxs
Me = Mxe + Mye
= (707.1 N)(0.088 m) = 62.2 Nm
= 0 + 212.1 = 212.1 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
The force acts through axis B, so
d = 0.20 m
d = 0, and conseqently,
M = 0 Nm
M = Fd = 433.0 N (0.20 m)
= 86.6 Nm
(b) note: l = parallel, and p = perpendicular
note: l = parallel, and p = perpendicular
BCl = 20 cos 60° = 10 cm
BCl = 20 cos 60° = 10 cm
73
BCp = 20 sin 60° = 17.3 cm
dl = AB + BCl = 20 + 10 = 30 cm
dp = BCp = 17.3 cm
Fl = -500 cos 60° = -250.0 N
Fp = -500 sin 60° = -433.0 N
Ml = Fldp = -250.0 N (0.173 m)
= -43.3 Nm
Mp = Fpdl = -433.0 N (0.30 m)
= -129.9 Nm
M = Ml + Mp
M = -173.2 Nm
(c) Fa passes through axis of rotation, so Ma = 0
dAB = 0.50 m
dAC = 0.25 m
Mb = FbdAB = -1000 N (0.50 m)
= -500 Nm
Mc = FcdAC = 500 N (0.25 m)
= 125 Nm
M = Ma + Mb + Mc
= 0 Nm + -500 Nm + 125 Nm
M = -375 Nm
(d) note: l = parallel, and p = perpendicular
BCl = 30 cos 40° = 23.0 cm
BCp = 30 sin 40° = 19.3 cm
dl = AB + BCl = 60 + 23
= 83 cm
dp = BCp = 19.3 cm
Fl = -300 N
Fp = 0 N
Ml = Fldp = -300.0 N (0.193 m)
= -57.9 Nm
Mp = Fpdl = 0 N (0.83 m)
= 0 Nm
M = Ml + Mp
BCp = 20 sin 60° = 17.3 cm
dl = BCl = 10 cm
dp = BCp = 17.3 cm
Fl = -500 cos 60° = -250.0 N
Fp = -500 sin 60° = -433.0 N
Ml = Fldp = -250.0 N (0.173 m)
= -43.3 Nm
Mp = Fpdl = -433.0 N (0.10 m)
= -43.3 Nm
M = Ml + Mp
M = -86.6 Nm
Fb passes through axis of rotation, so Mb = 0
dAB = 0.50 m
dBC = 0.75 m
Ma = FadAB = -1500 N (0.50 m)
= -750 Nm
Mc = FcdBC = 500 N (0.75 m)
= 375 Nm
M = Ma + Mb + Mc
= -750 Nm + 0 Nm + 375 Nm
M = -375 Nm
note: l = parallel, and p = perpendicular
BCl = 23.0 cm
BCp = 19.3 cm
dl = BCl = 23 cm
dp = BCp = 19.3 cm
Fl = -300 N
Fp = 0 N
Ml = Fldp = -300.0 N (0.193 m)
= -57.9 Nm
Mp = Fpdl = 0 N (0.23 m)
= 0 Nm
M = Ml + Mp
M = -57.9 Nm
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M = -57.9 Nm
(e) dx = 0.40 m
dx = 0 m
dy = 0.30 m
dy = 0.30 m
Fx = 200 N sin 53° = -159.7 N
Fx = 200 N sin 53° = -159.7 N
Fy = 200 N cos 53° = 120.4 N
Fy = 200 N cos 53° = 120.4 N
Mx = Fxdy
Mx = Fxdy
= (-159.7 N)(0.30 m) = -47.9 Nm
= (-159.7 N)(0.30 m) = -47.9 Nm
My = Fydx
My = Fydx
= (120.4 N)(0.40 m) = 48.2 Nm
= (120.4 N)(0 m) = 0 Nm
M = 0.3 Nm
M = -47.9 Nm
Question #3
Dino Dude and his stud buddy decide to go to the park to play on the teeter totter. When the two
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, how
much would Dino Stud have to weigh in order to balance out the teeter totter if he sat on the
broken 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 her
feet in an erect postition and enter the water feet first in a vertical position. She misjudges badly
and does a belly flop after completing one and a quarter revolutions - 1/4 more than
intended...OUCH! There are several things that she might do to improve performance. List as
many of these as you can and explain in biomechanical terms how each of them might serve to
improve the performance.
τ= I αI = mk2 I = Σmr2
τ= torque - produces angular acceleration
I = moment of inertia - inertial property of rotating bodies that increases with both mass and the
distance the mass is distributed from the axis of rotation the more closely concentrated the mass
to the axis of rotation, the easier it is to rotate the object the more mass positioned away from the
axis of rotation, the more difficult it is to initiate or stop angular motion.
r = radius of rotation
k = 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 theoretically
be concentrated without altering the inertial characteristics of the rotating body
Suggestions:
come out of tuck earlier: ↑ k which ↑ I which in turn, ↓ α (τ remains constant due to
Newton's #1)
"looser tuck": ↑ k which ↑ I which in turn, ↓ α
move into tuck slower: IΔt = mΔv ∴ decreases angular momentum
lean 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 diver
completing a one and a half somersault dive.
3. A weightlifter performing an arm curl with a 30 kg barbell stops the lifting motion with his
forearms 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 the
magnitude 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 = Fd
τ = (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 that
the 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 to
the axis of rotation was 3 cm, what force would the muscle have to exert to maintain the forearm
in 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 = Fd
Mbb = (-294 N) (.3 m)
Mbb = -88.29 Nm
∴Mb = 88.29 Nm
Mb = Fd
88.29 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 is
30 cm from the axis of rotation. However, this individual's biceps muscle inserts at a distance of
3.5 cm from the axis of rotation. Ignoring the weight of the forearm and hand, what force must
his muscles exert in order to maintain the forearm in it's horizontal position?
Mb = Fd
88.29 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 one
could 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 ↓ τ
concentrate mass as close to the axis of rotation as possible
b) Harder
extend arms behind head so body (abs) must generate more τ to oppose the resistance ( ↑ k
which 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
77
thigh is 23.2% of total body height. What is the moment of inertia of the thigh with respect to
the 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 m2
b) I = (60 kg)(0.105) [(1.8 m)(0.232)(0.54)]2 = 0.32 kg m2
c) I = (70 kg)(0.105) [(1.6 m)(0.232)(0.54)]2 = 0.30 kg m2
d) 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° for
walking 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 preparation
for next heelstrike. This decreases radius of gyration which decreases I.
78
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 a
distance 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 total
energy, kinetic energy, and potential energy at this point (assume that the load started 50 cm
from the ground)?
F = 400 N = mg
vf = 7.0 m/s
df = 0.6 m
di = 0.5 m
vi = 0 m/s
m = 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 velocity
of 2.0 m/s?
m = 30 kg
vv = 2.0 m/s
dto apex = ?
g = -9.81 m/s
According to the law of conservation of energy, KE + PE = constant. Initially PE = 0, and at the
apex KE = 0; so,
79
KEi = PEapex
1/2 mv2 = mgh
1/2 v2 = gh
h = 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 100
kg that is moving at 6 m/s?
F = 500 N
v = 6 m/s
m = 100 kg
P=?
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 m
m = 700 kg
W=?
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/h
and is then stopped by the catcher in .01 s?
m = 0.360 kg
vi = 100 km/h = (100 km/h)(3600 s/h) /(1000 m/km) = 360 m/s
vf = 0 m/s
t = 0.01 s
I=?
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 2000
J when it hit the ground?
F = 80 N = mg
KE = 2000 J
h=?
due to law of conservation of energy, KE at ground = PE at apex
KEi = PEapex
1/2 mv2 = mgh
2000 J = (80 N) h
h = (2000 J) /(80 N) = 25 m
80
8. A 55 kg person moves at the constant speed of 7 m/s along a straight stretch of track. What is
the person's:
(a) acceleration
m = 55 kg
v = 7 m/s
if velocity is constant, then a = 0
(b) momentum
M = mv = (55kg)(7 m/s) = 385 kg m/s
(c) kinetic energy
KE = 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 kg
vi = 8.5 m/s
vf = 0 m/s
W=?
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 and
that moves the object a displacement of 12 m?
F = 500 N at 25°
d = 12 m
W=?
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?
81
The principle of conservation of mechanical energy must be used to solve the first part of this
problem. The total mechanical energy Dino Dude possesses at a height of 10 m is his potential
energy. Immediately before impact with the water, his gravitational potential energy can be
assumed to be 0 and 100% of his mechanical energy will then be kinetic.
Find the total energy possessed by Dino Dude
PE + KE = Constant
magh + 1/2 mv2 = Constant
(20 kg)(9.81 m/s2)(10 m) + 1/2 (20 kg)(0)2 = Constant
1962 J = Constant
Find Dino Dude's velocity just before impact
PE + KE = 1962 J
magh + 1/2 mv2 = 1962 J
(20 kg)(9.81 m/s2)(0 m) + 1/2 (20 kg)(v2) = 1962 J
v2 = 196.2 m2/s2
v = 14.0 m/s
82
VIII. Biomechanics Terminology
83
Acceleration
Angle of Attack
Angular Acceleration (α)
SI: rad/s2
Angular Displacement (θ)
SI: radians or °
Angular Impulse
SI: Nms
Angular Momentum
SI: kg*m2/s
Angular Velocity (ω)
SI: rad/s
Archimedes Principle
Bernoulli Principle
Biomechanics
Center of Gravity
Center of Volume
Differentiation
Displacement
Equilibrium
First Class Lever
Force (F)
SI: Newton
Ground Reaction Force
(GRF)
A vector quantity involving positive, negative, or zero
values.
The angle between the longitudinal axis of an object and
the flow of the fluid.
The rate of change of angular velocity with respect to
time. The first derivative of angular velocity
The rotational change in position of a body. Each point on
an object is displaced around an axis.
An impulse that produces a change in angular momentum
Angular Impulse = torque x time interval over which the
torque acts.
A vector quantity of the amout of angular motion
possessed by a body.
Angular Momentum= angular velocity x moment of
inertia
The rate at which a body rotates. The first derivative of
angular displasement.
The buoyant force acting on a body is equal in magnitude
to the weight of the fluid displaced by the body.
The inverse of the relationship between relative velocity
and relative pressure in a fluid flow.
The application of mechanical priniciples in the study of
living organisms.
The point at which the weight of the body or system can
be considered to act.
The point around which a body’s volume is equally
balanced. The point at which the bouyant force acts.
The rate of change or slope of a line. For more
information on differentiation, click on the link in the lab
notes or look under mathematical calculation in the course
manual.
A vector quantity involving positive, negative, or zero
values. It may be translational, rotational or both.
All forces and moments acting on a body are at zero. The
object may be at rest or in motion.
Effort force and resistance force are on opposite sides of
the axis.
A vector quantity that desribes the action of one body on
another.
The forces acting on the body as a result of interaction
with the ground. These forces are equal but opposite in
direction to the forces the body is applying to the ground.
84
Impulse
SI: Ns
Integration
Kinematics
Kinetics
Lever
Lift
Linear Acceleration (a)
SI: m/s2
Linear Displacement (s or
d or x)
SI: m
Linear Momentum
SI: kg*m/s
Linear Velocity (v)
SI: m/s
Magnus Effect
Magnus Force
Mechanical Advantage
(MA)
Moment (M)
SI: Nm
Moment Arm
Moment of Inertia (I)
Relative Velocity
Resultant Force
Second Class Lever
The effect of a force acting over a period of time. It is
the integral of the force –time curve
Impulse = force x time interval over which the force acts
The area under a curve. For more information on
integration, click on the link in the lab notes or look under
mathematical calculation in the course manual.
The study of the description of motion including
considerations of space and time.
The study of the forces associated with linear motion
A simple machine consisting of two forces (an effort force
and a resistance force) and a barlike body that rotates
about an axis. The lever arm is the perpendicular distance
of each force from the axis of rotation.
A force acting perpendicular to the fluid flow on a body in
fluid
The rate of change of linear velocity with respect to time.
The first deriviative of linear velocity
The translational change in position of a body. Each point
on an object is displaced along parallel lines.
A vector quantity calculated by the mass of an object x the
linear velocity of the object
The rate at which a body moves in a strainght line. The
first derivative of linear displacement.
The deviation in the trajectory of a spinning object toward
the direction of a spin resulting from the magnus force.
The lift force created by a spin
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’ MA
will increase
The turning effect produced by a force.
Moment= force x perpendicular distance between the
point of application of the force and the axis of rotation.
The shortest distance between a forces line of action and
the axis of rotation
The inertial property of rotating bodies. It increases with
both mass and the distance that the mass is located from
the axis of rotation.
The velocity of a body with respect to another object
around it (ie. fluid).
The resultant force when all forces acting on an object are
added together and expressed as a single force
Effort force and resistance force are on the same side of
the axis with the resistance force being closer to the axis.
85
Third Class Lever
Velocity
Effort force and resistance force are on the same side of
the axis with the effort force being closer to the axis.
A vector quantity involving positive, negative, or zero
values.
86
IX. References
87
Hall, Susan J. 1995. In: V. Malinee, C. Wells (Eds.) Basic Biomechanics 2nd Edition. MosbyYear 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.
88