Mathcad Intro

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Solving Dynamics Problems

in Mathcad

Brian D. HarperMechanical Engineering

The Ohio State University

A supplement to accompany

Engineering Mechanics: Dynamics, 6thEdition

by J.L. Meriam and L.G. Kraige

JOHN WILEY & SONS, INC.

New York Chichester Brisbane Toronto Singapore


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CONTENTS

Introduction 5

Chapter 1 An Introduction to Mathcad 7

Numerical Calculations 7Variables and Functions 9

Graphics 11Symbolic Math 17

Vector Algebra 20Differentiation and Integration 23

Solving Equations 26

Chapter 2 Kinematics of Particles 33

2.1 Sample Problem 2/4 (Rectilinear Motion) 34

2.2 Problem 2/87 (Rectangular Coordinates) 372.3 Problem 2/120 (n-t Coordinates) 422.4 Sample Problem 2/9 (Polar Coordinates) 44

2.5 Sample Problem 2/10 (Polar Coordinates) 482.6 Problem 2/183 (Space Curvilinear Motion) 512.7 Sample Problem 2/16 (Constrained Motion

of Connected Particles) 54

Chapter 3 Kinetics of Particles 57

3.1 Sample Problem 3/3 (Rectilinear Motion) 583.2 Problem 3/98 (Curvilinear Motion) 613.3 Sample Problem 3/17 (Potential Energy) 64

3.4 Problem 3/218 (Linear Impulse/Momentum) 66

3.5 Problem 3/250 (Angular Impulse/Momentum) 68

3.6 Problem 3/365 (Curvilinear Motion) 70


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4 CONTENTS

Chapter 4 Kinetics of Systems of Particles 73

4.1 Problem 4/26 (Conservation of Momentum) 74

4.2 Problem 4/62 (Steady Mass Flow) 774.3 Problem 4/86 (Variable Mass) 79

Chapter 5 Plane Kinematics of Rigid Bodies 83

5.1 Problem 5/3 (Rotation) 845.2 Problem 5/44 (Absolute Motion) 89

5.3 Sample Problem 5/9 (Relative Velocity) 925.4 Problem 5/108 (Instantaneous Center) 95

5.5 Problem 5/123 (Relative Acceleration) 97

5.6 Sample Problem 5/15 (Absolute Motion) 99

Chapter 6 Plane Kinetics of Rigid Bodies 105

6.1 Sample Problem 6/2 (Translation) 106

6.2 Sample Problem 6/4 (Fixed-Axis Rotation) 1126.3 Problem 6/98 (General Plane Motion) 1146.4 Problem 6/104 (General Plane Motion) 117

6.5 Sample Problem 6/10 (Work and Energy) 119

6.6 Problem 6/206 (Impulse/Momentum) 124

Chapter 7 Introduction to Three-Dimensional

Dynamics of Rigid Bodies 127

7.1 Sample Problem 7/3 (General Motion) 1287.2 Sample Problem 7/6 (Kinetic Energy) 131

Chapter 8 Vibration and Time Response 135

8.1 Sample Problem 8/2 (Free Vibration of Particles) 1368.2 Problem 8/139 (Damped Free Vibrations) 138

8.3 Sample Problem 8/6 (Forced Vibration of Particles) 141


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INTRODUCTION

Computers and software have had a tremendous impact upon engineering

education over the past several years and most engineering schools nowincorporate computational software such as Mathcad in their curriculum. Since

you have this supplement the chances are pretty good that you are already aware

of this and will have to learn to use Mathcad as part of a Dynamics course. Thepurpose of this supplement is to help you do just that.

There seems to be some disagreement among engineering educators regarding

how computers should be used in an engineering course such as Dynamics. I willuse this as an opportunity to give my own philosophy along with a little advice.In trying to master the fundamentals of Dynamics there is no substitute for hard

work. The old fashioned taking of pencil to paper, drawing free body and massacceleration diagrams, struggling with equations of motion and kinematic

relations, etc. is still essential to grasping the fundamentals of Dynamics. A

sophisticated computational program is not going to help you to understand thefundamentals. For this reason, my advice is to use the computer only whenrequired to do so. Most of your homework can and should be done without a

computer. A possible exception might be using Mathcads symbolic algebra

capabilities to check some messy calculations.

The problems in this booklet are based upon problems taken from your text. Theproblems are slightly modified since most of the problems in your book do notrequire a computer for the reasons discussed in the last paragraph. One of the

most important uses of the computer in studying Mechanics is the convenience

and relative simplicity of conducting parametric studies. A parametric studyseeks to understand the effect of one or more variables (parameters) upon ageneral solution. This is in contrast to a typical homework problem where you

generally want to find one solution to a problem under some specified conditions.

For example, in a typical homework problem you might be asked somethingabout the trajectory of a particle launched at an angle of 30 degrees from the

horizontal with an initial speed of 30 ft/sec. In a parametric study of the sameproblem you might typically find the trajectory as a function of two parameters,

the launch angle and initial speed v. You might then be asked to plot thetrajectory for different launch angles and speeds. A plot of this type is very

beneficial in visualizing the general solution to a problem over a broad range of

variables as opposed to a single case.


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6 INTRODUCTION

As you will see, it is not uncommon to find Mechanics problems that yield

equations that cannot be solved exactly. These problems require a numericalapproach that is greatly simplified by computational software such as Mathcad.

Although numerical solutions are extremely easy to obtain in Mathcad this is stillthe method of last resort. Chapter 1 will illustrate several methods for obtainingsymbolic (exact) solutions to problems. These methods should always be triedfirst. Only when these fail should you generate a numerical approximation.

Many students encounter some difficulties the first time they try to use acomputer as an aid to solving a problem. In many cases they are expecting thatthey have to do something fundamentally different. It is very important to

understand that there is no fundamental difference in the way that you wouldformulate computer problems as opposed to a regular homework problem. Each

problem in this booklet has a problem formulation section prior to the solution.

As you work through the problems be sure to note that there is nothing peculiar

about the way the problems are formulated. You will see free-body and massacceleration diagrams, kinematic equations etc. just like you would normally

write. The main difference is that most of the problems will be parametric studies

as discussed above. In a parametric study you will have at least one and possiblymore parameters or variables that are left undefined during the formulation. For

example, you might have a general angle as opposed to a specific angle of 20.If it helps, you can pretend that the variable is some specific number while youare formulating a problem.

This supplement has eight chapters. The first chapter contains a brief introduction

to Mathcad. If you already have some familiarity with Mathcad you can skip thischapter. Although the first chapter is relatively brief it does introduce all the

methods that will be used later in the book and assumes no prior knowledge ofMathcad. Chapters 2 through 8 contain computer problems taken from chapters 2

through 8 of your textbook. Thus, if you would like to see some computer

problems involving the kinetics of particles you can look at the problems inchapter 3 of this supplement. Each chapter will have a short introduction thatsummarizes the types of problems and computational methods used. This would

be the ideal place to look if you are interested in finding examples of how to usespecific functions, operations etc.

This supplement uses Mathcad 13. Mathcad is a registered trademark ofMathSoft, Inc., 101 Main Street, Cambridge, Massachusetts, 02142.


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AN INTRODUCTIONTO MATHCAD

This chapter provides an introduction to the Mathcad programming language.Although the chapter is introductory in nature it will cover everything needed tosolve the computer problems in this booklet.

1.1 Numerical Calculations

Mathcad has four different equals signs. The most important of these are the

evaluation equals sign (=) and the assignment equals sign (:=). Numerical

calculations use the evaluation equals sign. As a simple example, type thefollowing expression into a Mathcad worksheet: "(2+6^3)*4/5=". After pressing

the "=" key, Mathcad will immediately evaluate the expression. It should looklike the following.

2 63

+( ) 45

174.4=

Note that the result looks very much like what you would write on a sheet of

paper. Now try typing "10+12/3-6*2^4=" into the worksheet. You should get thefollowing.

1012

3 6 24

+ 9.871=

At first it may surprising that the 6*2^4 remains in the denominator. Now try

entering the same keystrokes but press the space bar immediately after typing"3". Note how the blue placeholder changes when the space bar is pressed. With

a little practice, you shouldn't have too much trouble getting the expression youwant. The main thing is to pay attention to the placeholder. The arrow keys can

also be used to move the placeholder.

Numerical calculations can also include standard functions. The most commonly

used functions can be found in the calculator toolbar. The calculator toolbar canbe opened with View...Toolbars or by pressing shortcut button that looks like a

calculator. Mathcad has many built in functions besides those shown in the

1


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8 CH. 1 AN INTRODUCTION TO MATHCAD

Calculator toolbar. If you already know the name of the function you can simply

type it in or select from a list by using the shortcut Cntrl+E or by choosingInsert...Function in the menu bar. Here are a few examples. Explanations are

given to the right when appropriate.

64

18 1.155=

sinh 0.5( ) 0.521=

sin 10( ) 0.544=

Mathcad, like most mathematical software packages, assumes that angles are

given in radians. Thus the last line calculates the sine of 10 radians (573 degrees).Use one of the following to methods to obtain the sine of 10 degrees.

sin 10

180

!

#

$

&0.174= sin 10 deg( ) 0.174=

Of course, inverse trig functions also return results in radians and similarmethods can be used to obtain results in degrees. The following calculates an

inverse sine (asin in Mathcad) and converts the result to degrees.

180

asin

3

2

!

#

$

& 60=

asin3

2

!

#

$

&deg

60=

Press the square root button in the Calculatotoolbar then type "6*4/18="

Type "sin(10)=" or select sin from the

Calculator toolbar.

The hyperbolic sine. Either type "sinh(0.5)="

or selectInsert...Function...Hyperbolic...sinhin the menu bar.


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INTRODUCTION TO MATHCAD 9

1.2Variables and Functions

A variable is a name or alias which can be defined as a number or an expressionusing the assignmentequals sign ":=" (type ":" in Mathcad). Mathcad has many

built in variables. A good example is the variable deg (an alias for the number

/180) used in the previous examples. To see this, type "deg=" in a Mathcadworksheet. Of course, you can also define your own variables and functions in

Mathcad. The following example assigns a number to the variable x and an

expression (a function of x) to the variable f. Technically, both x and f arevariables though it is customary to refer to f as a function of x. Following the twoassignments we also use the evaluation equals sign (=) in order to illustrate the

difference between these two equals signs. As the names suggest, one is used forassigning (giving names to) numbers or expressions while the other is used for

evaluating (calculating) names or expressions.

x 5:=

f 3 x 5 x2 2 x3+:=

f 140=

Assigning expressions to names is very useful when you want to calculate thevalues of a function for several different values of a parameter. Note, however,

that x must be assigned a numerical value before assigning the expression above

to the name f. It is also possible to define functions explicitly in terms of one or

more parameters. In this way you can define functions that work just like built infunctions such as sin, cos, log etc. When functions are defined in this way it isnot necessary to specify beforehand the values of the parameters in the equation.

Here are a few examples.

f y( ) 3 y 5 y2 2 y3+:=

f 5( ) 140= f 2( ) 2=

g x y,( ) x2 y2+:=

Type "x:5"

to enter a function type "f(y):" followed by

the expression

note that the function f operates like a built in

function

note that it is okay to use x as a parameter in

a function definition even though it has been

previously defined a value


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g x x,( ) 7.071=

g 4 f 2( ),( ) 4.472=

Range Variables

As the name implies, a range variable is a variable which has been assigned a

range of values. Assigning a range to a variable is accomplished by typingsomething like "x:a,b;c" where a, b and c are numbers or variables previouslyassigned a numerical value. The first value in the range for the variable x is a, thesecond value is b while the last value in the range is c. Note that b is the second

value, not the increment. Mathcad will automatically determine the increment

from a and b. Let's try it out. Type "x:1,1.5;3" followed by "x=". You should seethe following.

x 1 1.5, 3..:= x

1

1.5

2

2.5

3

=

Notice that two dots (..) are displayed when you type the semicolon (;). The twodots is Mathcad's range variable operator. A shortcut (m..n) is available on the

Calculator toolbar. Once a range variable has been defined it can be used likeany other variable.

z 0 1, 6..:=

f y( ) 2 y y3+:=

note the evaluation equals sign. NowMathcad substitutes the value previously

assigned to x (5) into the function g, resultingin the square root of 5^2+5^2

One function can be used as the argument foranother. Mathcad first evaluates f(2) and then

substitutes this for y in the function g(x,y).

Type "z:0,-1;-6". Notice that the range can

either increase or decrease!

Type "f(y):2*y+y^3"


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f 3( ) 33=

f z( )

0

-3

-12

-33

-72

-135

-228

=

If the second value in the range is omitted, Mathcad will assume an increment of

1. To illustrate, type "x:6;9" followed by "x=". The result is,

x 6 9..:= x

6

7

8

9

=

1.3 Graphics

One of the most useful things about a computational software package such asMathcad is the ability to easily create graphs of functions. As we will see, these

graphs allow one to gain a lot of insight into a problem by observing how a

solution changes as some parameter (the magnitude of a load, an angle, adimension etc.) is varied. This is so important that practically every problem inthis supplement will contain at least one plot. By the time you have finished

reading this supplement you should be very proficient at plotting in Mathcad.

This section will introduce you to the basics of plotting in Mathcad.

Mathcad has the capability of creating a number of different types of graphs.Here we will consider only the X-Y plot. The most common and easiest way to

generate a plot of a function is to use range variables. The following examplewill guide you through the basic procedure.

Type "f(z)=". Notice that the function f canoperate either on a single value or a range of

values.


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First define the function to be plotted. Type "f(x):x*exp(-x^2)"

f x( ) x exp x2( ):=

Now define a range variable covering the range over which you would like toplot the function.

x 3 2.9, 3..:=

Now click at the desired location on the worksheet and insert an X-Y plot by (a)selectingInsert...Graph...X-YPlotfrom the main menu, (b) using the shortcut key

"@" or (c) selecting the X-Y Plot icon from the graph toolbar. You should see an

empty graph like the following.

You should see two empty placeholders on the x and y axes. By default, the

insertion point should already be on the x placeholder. If not, click on thatplaceholder and type "x". Now click on the y placeholder and type "f(x)". Afterclicking away you should see the following graph.


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3 2 1 0 1 2 30.6

0.4

0.2

0

0.2

0.4

0.6

f x( )

x

Parametric Studies

One of the most important uses of the computer in studying Mechanics is the

convenience and relative simplicity of conducting parametric studies (not to beconfused with parametric plotting discussed below). A parametric study seeks tounderstand the effect of one or more variables (parameters) upon a general

solution. This is in contrast to a typical homework problem where you generallywant to find one solution to a problem under some specified conditions. For

example, in a typical homework problem you might be asked to find the reactions

at the supports of a structure with a concentrated force of magnitude 200 lb thatis oriented at an angle of 30 degrees from the horizontal. In a parametric study ofthe same problem you might typically find the reactions as a function of two

parameters, the magnitude of the force and its orientation. You might then be

asked to plot the reactions as a function of the magnitude of the force for severaldifferent orientations. A plot of this type is very beneficial in visualizing thegeneral solution to a problem over a broad range of variables as opposed to a

single case.

Parametric studies generally require making multiple plots of the same function

with different values of a particular parameter in the function. Following is a verysimple example.

f a x,( ) 5 x+ 5 x2 a x3+:=

What we would like to do is gain some understanding of how f varies with both xand a. We will illustrate this by plotting f as a function of x for a = -1, 0, and 1.

As before, we first define a range variable.


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x 5 4.9, 5..:=

Now bring up an empty X-Y plot by typing "@". Type "x" into the placeholderon the x axis and then click on the y axis placeholder. Now type "f(-1,x),f(0,x),f(1,x)". Note that each time you type a comma, a new placeholder

appears. When you click away you should see something like the following.

6 4 2 0 2 4 6250

200

150

100

50

0

50

f 1 x,( )

f 0 x,( )

f 1 x,( )

x

Parametric Plots

It often happens that one needs to plot some function y versus x but y is notknown explicitly as a function of x. For example, suppose you know the x and y

coordinates of a particle as a function of time but want to plot the trajectory ofthe particle, i.e. you want to plot the y coordinate of the particle versus the x

coordinate. A plot of this type is generally called a parametric plot. Parametric

plots are easy to obtain in Mathcad. You start by defining the two functions interms of the common parameter and then define the common parameter as arange variable. Next, open an empty X-Y plot and type the two functions into the

x and y axis placeholders. The following example illustrates this procedure.

f a( ) 10 a 2 a( ):=

g a( ) sin 3 a( ):=

a 1 .95, 3.5..:=

In this example the parameter is a.


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1 0.5 0 0.5 160

40

20

0

20

f a( )

g a( )

The range selected for the parameter can have a big, and sometimes surprisingeffect on the resulting graph. To illustrate, try increasing the upper limit on the

range on a a few times and see how the graph changes.

You can, of course, also plot g as a function of f.

a 1 .95, 6..:=

250 200 150 100 50 0 501

0.5

0

0.5

1

g a( )

f a( )


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In the above examples we have more or less just accepted whatever graph

Mathcad produced. This is the easiest approach and is certainly acceptable formany situations. You should be aware, though, that it is possible to change the

appearance of a graph in several ways. To do this, first prepare your graph in theusual manner and then double click on it. You will get a pop-up menu that youcan use to reformat the graph. At the top of the menu there are four tabs that youcan select to alter different aspects of the graph's appearance. The figure below

shows a menu where the "Traces" tab has been selected.

Here you can modify the line style, color and thickness (weight) of each curve.You can also plot symbols instead of lines. You should spend some time

familiarizing yourself with the various graph formatting possibilities available inMathcad.


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1.4 Symbolic Math

Up to this point we have been using Mathcad essentially as a calculator. Well,obviously, a very sophisticated calculator, but a calculator nevertheless. Thereare times where it is very useful to have Mathcad perform mathematical

calculations with symbols rather than numbers. This is very much like what you

might do when deriving or manipulating equations in a homework problem.Except, Mathcad is less prone to making algebra mistakes.

Three of the most important applications of symbolic math will be discussed in

the next three sections, namely symbolic vector algebra, symbolic calculus(integration and differentiation) and symbolic solution of one or more equations.

The purpose of the present section is to introduce you to the basic procedures ofsymbolic math as well as to give a few other useful applications.

There are several approaches that can be used to perform symbolic mathematics.

Here we will use just one primary approach and a slight modification of thatapproach. Start by opening the symbolic toolbar. This can be done either by

selecting View...Toolbars...Symbolic from the main menu or by clicking thesymbolic icon on the math toolbar. It looks like a graduation cap. Here's what

you should see (note that the appearance might be slightly different in differentversions of Mathcad).

Let's start by illustrating symbolic simplification. First enter the expression youwish to simplify on the worksheet (no equals signs). Now click anywhere on thisexpression and then click on the simplify tab on the Symbolic tool bar. Finally,click anywhere on the worksheet and the simplified expression will appear.

Here's a simple example.


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a x b x2+( )2

x4

Here's the expression we want to simplify. If you need help, type

"(a*x+b*x^2)^2[space bar]/x^4". Now click anywhere on the expression and

then press thesimplifytab. After clicking away you should see the following.

a x b x2+( )2

x4

simplify1

x2

a b x+( )2

You can also simplify expressions containing previously defined functions.Here's another way to obtain the simplification above. See if you can reproduce it

on your worksheet.

f a b, x,( ) a x b x2+:= g x( ) x4:=

f a b, x,( )2

g x( )simplify

1

x2

a b x+( )2

Another useful symbolic operation is substitution. The substitution operator

allows you to substitute an expression for a variable in another expression. Startwith the expression you would like to substitute into. Click anywhere on this

expression and then click the "substitute" tab on the Symbolic tool bar. You willget a bold equal sign with placeholders on either side. Fill in the placeholders so

that you have variable1=variable2, where variable2 is to be substituted forvariable1. The following example illustrates the substitute operator.

a x 4 x2+( )2

x2

a+

Start with the expression into which you would like to substitute. Click anywhere

on this expression and then click the "substitute" tab on the Symbolic toolbar.

You should see something like the following.


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Click in the left placeholder and type "a". Now click in the right placeholder and

type "1+x^2". You should see the following result.

a x 4 x2+( )2

x2

a+substitute a 1 x

2+,1 x

2+( ) x 4 x2+ 2

2 x2 1+( )

It is also possible to substitute previously defined functions.

f x( ) x2

1+:=

2 x 4 x2+( )2

x

2

2+

substitute x f x( ),2 2 x

2+ 4 1 x2+( )2

+2

1 x2

+( )2

2+

The results following a substitution are often rather messy. To simplify, onecould always copy the final result into the clipboard, paste it onto the worksheet,and then follow the procedure above to simplify. It is also possible to do several

symbolic operations at once. The following example shows a substitution

followed by a simplification. The procedure is the same as for substitution withone difference. After filling out the before and after placeholders, click on the

"simplify" tab beforeclicking away.

2 x 4 x2+( )2

x2 2+

substitute x f x( ),

simplify

43 5 x

2+ 2 x4+( )2

3 2 x2+ x4+( )

Finally, here is an example with two substitutions followed by simplification.

a x b x2+( )2

x5

substitute a x3

tan x( )+,

substitute b x2,

simplify

1

x3

tan x( )2


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1.5 Vector Algebra

The main application of vector algebra is in three dimensional problems wherethe geometry is difficult to visualize. Some of these difficulties include finding

the x, y, and z components of a vector, moment arms for a force, projections of a

force onto a line etc. The most useful vector operations are finding the magnitude

of a vector and finding the dot or cross product of two vectors.

First we need to learn how to represent a vector in Mathcad. Start by opening theVector and Matrix Toolbar. You can do this by selecting

View...Toolbars...Matrix or by clicking the matrix icon on the Math Toolbar (itlooks like a 3x3 matrix). Cartesian vectors are represented by three element

column matrices. The following example shows how to create a vector.

Start by typing "u:". Now click on the matrix icon on the Vector and MatrixToolbar. You can also selectInsert...Matrix or use the shortcut CNTRL+M. In

the popup menu select 3 rows and 1 column. After clicking OK, you should see

the following

u

!

"#

$

&

:=

Now fill in the placeholders with the x, y, and z components of the vector. For

example,

u

3

2

6

!

"#

$

&

:=

Once the vector has been defined you can refer to the components of the vectorsby typing the name of the vector with an index. Indices start at 0 in Mathcad soan index of 0, 1, and 2 correspond to x, y, and z. Indices are entered by typing

"[". Don't confuse an index with a subscript, which is obtained by typing ".". Forexample, to print the y component of utype "u[1=".

u1 2=

To find the magnitude of u, select the absolute value icon (|x|) from the Vectorand Matrix Toolbar. In the placeholder type "u=". You should see the following.

u 7=


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In Statics, we often need to find unit vectors. A unit vector in the direction of u

can be obtained by dividing uby the magnitude ofu.

nu

u:=

n

0.429

0.286

0.857

!

"#

$=

As an example of the above, suppose that you have a force Fwith magnitude 100

lb and with a line of action passing from point A (2, 0, 3) toward B (7, -2, 5). We

can represent F as a Cartesian vector in Mathcad as follows.

rA

2

0

3

!

"#

$:= rB

7

2

5

!

"#

$:=

rAB rA rB:=

F 100rAB

rAB:=

F

87.039

34.816

34.816

!

"#

$

&

=

Dot and cross product operators can also be selected from the Vector and MatrixToolbar.Shortcuts are *for dot product and CNTRL+8for cross product. Hereare a few examples using the vectors we have already defined above.

u F 539.641=

rA rB

6

11

4

!

"#

$

&

=

M rA F:= M

104.447

191.485

69.631

!

"#

$=


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Vector operations can also be carried out symbolically. You will, of course, use

the symbolic equals sign instead of the evaluation equal sign =. Here are a fewexamples.

u

a

b

c

!

"#

$

&

:=

a

v

p

3

4

!

"#

$

&

:=

p

w

x

5

4

!

"#

$:=

x

After typing in the above three vectors you probably noticed that some of the

variables appear in red since they have not been defined. This would obviouslycreate a problem if you were going to evaluate some numerical results, however,it has no effect on symbolic calculations as can be seen from the following.

u v a p 3b+ 4 c+

v

v

p

p( )2 25+

1

2

!#

$&

3

p( )2 25+

1

2

!#

$&

4

p( )2 25+

1

2

!#

$&

'

+++++++++

(

)

,,,,,,,,,

*

u v

4b 3 c

c p 4 a

3 a b p

!

"#

$

&

w u v( ) 0 solve x,5 c p 32 a 4b p+( )

4b 3 c( )


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1.6 Differentiation and Integration

Mechanics problems often require integration and/or differentiation. In Mathcad,you can perform these operations either numerically or symbolically. Before we

get started you will want to open the Calculus Toolbar. You can open this by

pressing the icon in the math toolbar or by selecting View...Toolbars from the

main menu. The icons we will be using are those for the first and nth derivativeand the definite and indefinite integral. The definite integral has a and b asintegration limits. You may also want to open the Symbolic Toolbar.

Let's get started with a simple example of symbolic differentiation. Start by

selecting the icon for the first derivative. Here's what you should see.

d

d

Note that there are two placeholders. Into the placeholder on the right hand sidetype the expression that you would like to differentiate (for this example, type"(a*sec(b*t))"). Then click on the placeholder in the denominator and enter the

variable that you would like to differentiate with respect to. You should see thefollowing.

ta sec b t( )( )d

d

Now click anywhere on this expression and click on the symbolic evaluation icon

() in the Symbolic Toolbar. After clicking away you will see the result of the

symbolic differentiation.

ta sec b t( )( )d

da sec b t( ) tan b t( ) b

Higher order derivatives follow the same procedure except that there is anadditional placeholder to fill in for the order of differentiation. See if you canreproduce the following result.

3x

a ln b x+( )d

d

3

2a

b x+( )3

You can also use derivatives in defining functions. As an example, suppose aparticle moves in a straight line and its position s is known as a function of time.

From your elementary physics course you probably know that the first and

second derivatives of the position give the velocity and acceleration of the

particle.


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s t( ) 10 t 20 t2 2 t3+:=

v t( )

t

s t( )d

d

:=

a t( )2

ts t( )

d

d

2

:=

Now you can evaluate the velocity and acceleration at any time.

v 1( ) 24= a 1( ) 28=

v 7( ) 24= a 7( ) 44=

You can also plot the results.

t 0 0.1, 10..:=

0 2 4 6 8 10400

200

0

200

400

s t( )

v t( )

a t( )

t

While the above is very convenient, especially when you want to numerically

evaluate or plot the results after differentiation, it fails to provide the symbolicresults. If you would like to have a record of these you can consider somethinglike the following.

Note the assignmentequals sign ":=".

Note the evaluationequals sign "=".


25/142


s t( ) 10 t 20 t2 2 t3+:= position

v t( )ts t( )d

d:=

velocity,

ts t( )d

d10 40 t 6 t2+

a t( )2

ts t( )

d

d

2

:= acceleration,

2

ts t( )

d

d

2

40 12 t+

or,tv t( )

d

d40 12 t+

The procedure for performing integrations is very similar to that fordifferentiation. For example, to perform a symbolic integration: (a) click on theicon for either a definite or indefinite integral (or use the shortcut key Shift+7(or

&) for definite and Cntrl+i for indefinite), (b) fill in the placeholders, (c) click

anywhere on the expression and (d) click the Symbolic Evaluation icon (or usethe short cut Cntrl+.). See if you can reproduce the following integrals.

xsin b x( )-./

dcos b x( )

b

c

d

xsin b x( )-./

dcos d b( )

b

cos c b( )

b+

xln x( )-./

d x ln x( ) x c

d

xln x( )-./

d d ln d( ) d c ln c( ) c+

If a definite integral contains no unknown parameters either in the integrand or

the integration limits, the above procedure will provide numerical answers. Hereare a few examples.

0

3

xx 3 x3+( )

-./

d261

4

2

5

xln x( )-./

d 5 ln 5( ) 3 2 ln 2( )

Note that Mathcad will try to return an exact result when the SymbolicEvaluation procedure is used. This results in fractions or functions as in the

above examples. This is very useful in some situations, however, one often wants

to know the numerical answer without having to evaluate a result such as theabove with a calculator. Thus, Mathcad also allows you to obtain results fornumerical integration as floating point numbers. This can be accomplished by

following the same procedure outlined above except that for step (d) you press


26/142


the equals sign "=" on your keyboard instead of clicking the Symbolic Evaluation

icon. To illustrate, we will repeat the same two integrals above.

0

3

xx 3 x3+( )-.

/d 65.25=

2

5

xln x( )-./

d 3.661=

1.7 Solving Equations

Solving a single equation symbolically can be accomplished in a manner verysimilar to other symbolic operations considered earlier. As an example, try typing

the following equation on to your worksheet, being sure to type Cntrl = for theequals sign (you should see a bold equals sign =).

a x2 b x+ c+ 0

Click anywhere on the equation and then click solveon the Symbolic Toolbar .Type the variable you wish to solve for (in this example x) in the placeholder and

the click away. You should see the following.

a x2 b x+ c+ 0 solve x,

1

2 a( )b b2 4 a c( )

1

2

!#

$&

+

'

(

)

*

1

2 a( )

b b2 4 a c( )

1

2

!#

$&

'

(

)

*

'

++++

(

)

,,,,

*

Note that Mathcad has found both solutions to the (hopefully) familiar quadraticequation. If the equals sign is omitted, Mathcad will assume that the expression is

set equal to zero, i.e. Mathcad will find the roots of the expression. Here is an

alternative way to obtain the above result.

a x2 b x+ c+ solve x,

1

2 a( )b b2 4 a c( )

1

2

!#

$&

+

'

(

)

*

1

2 a( ) b b2

4 a c( )

1

2

!#

$&

'

(

)

*

'

++++

(

)

,,,,

*

If the variable being solved for is the only unknown in the equation, Mathcad

will return a number as the result. Here are a couple of examples.


27/142


2 x2 4 x+ 12 solve x,

1 7+

1 7

!

#

$

&

5 sin( ) cos ( ) 1 solve ,

atan5

12

!#

$&

!

"#

$

&

You can also solve equations using Given...Find. For the symbolic case, one

starts with the basic Given...Findformat shown below.

Given

a x2

b x+ c+ 0

Find(x)

Now click on "Find(x)" and then click on the Symbolic Evaluation icon ().After clicking away you should see the following result.

Given

a x2 b x+ c+ 0

Find x( )1

2 a( )b b2 4 a c( )

12!# $&

+'(

)*

1

2 a( )b b2 4 a c( )

12!# $&

'(

)*

'

(

)

*

For a numerical solution you would use the same procedure but type "Find(x)=".

Here's an example.

g x( ) 2 x2 1+ 10 sin x( ):=

x 0:=

Given

g x( ) 0

Find x( ) 0.102=


28/142


x 2:=

Given

g x( ) 0

Find x( ) 2.008=

Given...Find can also be used to solve simultaneous equations eithersymbolically or numerically. The approach is essentially the same as that

described above for a single equation except, of course, that more than one

equation will appear between the GivenandFind statements. Also, for numerical

solutions, an initial guess should be provided for all unknowns. Following areseveral examples. An easy way to tell at a glance whether the solution is

symbolic or numerical is to see whether the symbolic evaluation symbol ()appears afterFind.

Given

P sin ( ) Bx+ Ax 0

Ay P cos ( )+ w a 0

P a cos ( ) P b sin ( ) Bx c+1

2w a2 0

Find Ax Ay, Bx,( )

1

2

2 P sin( ) c 2 P a cos ( ) 2 P b sin ( )+ w a2+( )c

P cos ( ) w a+

1

2

2 P a cos ( ) 2 P b sin( )+ w a2+( )c

'

+++

+(

)

,,,

,*

Given

A subscript can be obtained by typing "."before the subscript. For example, by typing

"A.x".


29/142


x2

y2+ 12 x y 4

Find x y,( ) 5 15 1+

1 51 5

5 1+5 1

1 51 5

!#

$&

Note that each column in the last result represents a solution. Thus, in the last

example, Mathcad has found four solutions, the first being x = 15 and y =

15+ .

x 0:= y 0:= z 0:=

Given

x2

y+ 12 x y 4 x y z

Find x y, z,( )

3.284

1.218

2.065

!

"#

$=

x 0:= y 5:= z 5:=

Given

x2

y+ 12 x y 4 x y z

Find x y, z,( )

0.337

11.887

11.55

!

"#

$

&

=

This is our initial guess for a numerical

solution

Now let's try another guess for the same set

of equations.


30/142


Finding Numerical Solutions with root

Numerical solutions to single equations can be obtained with root. This is

particularly useful in those situations where solve fails to find a solution. We willillustrate by finding a numerical solution to the equation 10sin(x) = 2x^2 + 1.

Before using the root function you should first provide a value in the

neighborhood of the solution you are seeking. This is especially important if, as

in the present case, there is more than one solution. Well, you may be wonderinghow we can determine the neighborhood of a solution if we do not yet know thesolution. Actually, this is very easy to do. First we define a function g(x) whose

roots will be the solution to the equation of interest. Next, we plot this function inorder to estimate the location of points where g(x) = 0.

g x( ) 2 x2 1+ 10 sin x( ):=

x 1 0.9, 3..:=

1 0 1 2 310

5

0

5

10

15

20

g x( )

0

x


31/142


From the graph above we see that g(x) = 0 in two places, about x = 0 and x = 2.

These results provide our initial guesses for the root command. Here's how itworks.

x 0:= root g x( ) x,( ) 0.102=

x 2:= root g x( ) x,( ) 2.008=

Or, equivalently,

x 0:= x1 root g x( ) x,( ):= x1 0.102=

x 2:= x2 root g x( ) x,( ):= x2 2.008=


32/142


33/142

KINEMATICS OFPARTICLES

Kinematics involves the study of the motion of bodies irrespective of the forcesthat may produce that motion. Mathcad can be very useful in solving particle

kinematics problems. Problem 2.1 is a rectilinear motion problem illustrating

symbolic integration. The formulation of this problem results in an equation thatcannot be solved exactly except with some rather sophisticated mathematics.

When this occurs it is generally easiest to obtain either a graphical or numericalsolution. This problem illustrates both approaches with the numerical result being

obtained with GivenFind. Problem 2.2 is a rectangular coordinates problemthat illustrates GivenFindas well as symbolic differentiation. Problem 2.3 is a

relatively straightforward problem where Mathcad is used to generate a plot thatmight be useful in a parametric study. The path of a particle is depicted using a

parametric plot and a polar plot in problem 2.4. In problem 2.5, the r-components of the velocity are determined using symbolic differentiation. Theproblem also illustrates how computer algebra can simplify what might normally

be a rather tedious algebra problem. Symbolic differentiation is further illustrated

in problems 2.6 and 2.7. Problem 2.7 is particularly interesting in that it requiresdifferentiation with respect to time of a function whose explicit time dependenceis unknown. This happens rather frequently in Dynamics so it is useful to know

how to accomplish this with Mathcad.

2


34/142

34 CH. 2 KINEMATICS OF PARTICLES

2.1 Sample Problem 2/4 (Rectilinear Motion)

A freighter is moving at a speed of 8 knots whenits engines are suddenly stopped. From this time

forward, the deceleration of the ship is

proportional to the square of its speed, so that2kva = . The sample problem in your text

shows that it is rather easy to determine the

constant kby measuring the speed of the boat at

some specified time. Show how k could befound by (a) measuring the speed after some

specified distance and (b) measuring the timerequired to travel some specified distance. In

both cases let the initial speed be v0.

Problem Formulation

(a) Since time is not involved, the easiest approach is to integrate the equation

vdv= ads.

dskvadsvdv 2== 00 =sv

v

dskv

dv

00

%&

$"#

!=

v

vks 0ln

With this result it is easy to find k given v at some specified s. To illustrate,

assume that v0= 8 knots and that the speed of the boat is determined to be 3.9knots after it has traveled one nautical mile.

%&

$"#

!=

9.3

8ln)1(k k= 0.718 mi-1

(b) Here we follow the general approach in the sample problem. Integrating a=

dv/dt yields

00 =tv

v

dtkv

dv

0

2

0

0

0

vv

vvkt

=

0

0

1 ktv

vv

+=

To obtain the distancesas a function of time we integrate v= ds/dt

0 00 +===t ts

dtktv

vvdtsds

0 0 0

0

01

( )01ln1

ktvk

s +=


35/142

KINEMATICS OF PARTICLES 35

This equation turns out to be very difficult to solve for k. A good mathematicianor someone familiar with symbolic algebra software might be able to find the

general solution for kin terms of the so-called LambertW function (LambertW(x)is the solution of the equation yey= x). Even if this solution were found it would

be of little use in most practical situations. For example, you would have to spendsome time familiarizing yourself with the function. Once this is done you would

still have to use a program like Maple or a mathematical handbook to evaluate

the function.

For these reasons it is probably easiest to find keither graphically or numerically.

Obtaining a numerical solution with Mathcad is so easy that there is little reasonnot to use this approach. It is generally advisable though to use a graphical

approach even when a numerical solution is being obtained. This is the best way

to identify whether there are multiple solutions to the problem and also serves as

a useful check on the numerical results. Thus, both approaches are illustrated inthe worksheet below.

The usual way to generate a graphical solution is to rearrange the equation so asto give a function that is zero at points that are solutions to the original equation.

Rearranging the equation above in this manner yields,

( ) 01ln 0 =+= ktvksf

Given values ofs, t, and v0,fcan be plotted versus k. The value of kat whichf=

0 provides the solution to the original equation.

Mathcad Worksheet

Although the integrations are simple in this problem, we'll go ahead and evaluate

them symbolically for purposes of illustration.

s_a1

kv0

v

x1

x

-../

d:=

v

s_a1

kln v( ) ln v0( )( )

s_b

0

t

xv0

1 k v0 x+

-../

d:=

t

s_bln 1 t k v0+( )

k

To illustrate the graphical solution, take v0 = 8 knots and assume that the boat isfound to move 1.1 nautical miles after 10 minutes.


36/142


v0 8:= s 1.1:= t10

60:=

f k( ) k s ln 1 k v0 t+( ):=

k 0 0.01, 0.5..:=

0 0.1 0.2 0.3 0.4 0.50.02

0

0.02

0.04

f k( )

0

k

The above graph shows that k is about 0.34 mi-1. Now let's try a symbolicsolution.

f k( ) 0 solve k , f k( )

No solution was found, so we'll try Given...Find.

Given

f k( ) 0

Find k( ) 0 .33923053342470867736( )

Note that we have used the symbolic Given...Find. When this is done, Mathcadfirst looks for a symbolic result. If this fails, it will then automatically try a

numerical approach. This is indeed what happened in the present case asevidenced by the floating-point answer.


37/142


2.2 Problem 2/87 (Rectangular Coordinates)

A long-range rifle is fired atAwith the projectilehitting the mountain at B. (a) If the muzzle

velocity is u 400 m/s, determine the two

angles of elevation which will permit theprojectile to hit the mountain target B and plot

the two trajectories. (b) Determine the smallest

muzzle velocity that will allow the projectile tostrike at B and the angle at which it must be

fired. Repeat the plot for part (a) and include thetrajectory of the projectile for this minimum

initial velocity.

Problem Formulation

Place a coordinate system atAwithxpositive to the right andypositive up. Theinitial components of the velocity are,

( ) cos0 uvx = sin0 uvy =

The acceleration is constant with components 0=xa and gay = . Integrating

these two accelerations twice and applying initial conditions yields (see page 44of your text if you need additional details),

x= ucost y= usint -1/2gt2

Plotting y in terms of x for different times t will yield the trajectory of theprojectile. This type of plot is called aparametric plotsince the items plotted (x

andy) are each known in terms of another parameter (t).

Anytime you have a projectile motion problem and you know the coordinates ofa point on the trajectory (our point B) you should solve for xandy (as we have

done above) and then obtain two equations by substituting the coordinates of thepoints. These two equations can then be solved for two unknowns. Note that in

most cases one of the two unknowns will be the time of flight.

Part (a)Substitutingx= 5,000 m,y= 1,500 m and u= 400 m/s gives

5000 = 400cost 1500 = 400sint -1/2(9.81)t2


38/142


We will let MathCad solve these two equations simultaneously. MathCad

actually finds four solutions, however two of the four can be discarded since theyinvolve negative times. The other two solutions correspond to the two solutions

shown in the illustration accompanying the problem statement. The results are,

1= 26.6and 2= 80.6

Part (b) It should be intuitively obvious why there must be a minimum initial

velocity below which the projectile cannot reachB. How do we go about findingit? We still have the two equations for the coordinates of pointB,

5000 = ucost 1500 = usint -1/2(9.81)t2

however there are now three unknowns (u, , t). Suppose for the moment that thelaunch angle were given and we were asked to calculate the required initial

speed uso that the projectile strikesB. In this case we would have two equationsand two unknowns. From this observation we see that u is a function of fromwhich we get our general solution strategy:

(a) Eliminate t from the above two equations and solve for u as a

function of .(b) Differentiate this function with respect to to find the location of

the minimum.

Solving the first equation for u gives

cos

5000

tu=

Substituting into the second yields 2

2

1tan50001500 gt= . This equation is

now solved for ( ) gt /1500tan50002 = which can be substituted back intouto give

( ) gu

/1500tan50002cos

5000

=

We will let MathCad differentiate this equation and solve for the minimum speed

and the associated launch angle. The result is

umin= 256.8 m/s at = 53.3


39/142


MathCad Worksheet

Part (a)

x t,( ) 400 cos ( ) t:= y t,( ) 400 sin ( ) t9.81 t

2

2:=

Given

5000 400 cos ( ) t

1500 400 sin ( ) t9.81

2t2

Find(, t)

MathCad finds four solutions but two can be excluded because they have time

being negative. The two positive times and corresponding angles are

t B1= 13.9205 secs 1= 0.4557 rads (26.6)

and t B2= 76.4520 secs 2= 1.4066 rads (80.6)

1 0.4557:= 2 1.4066:=

t1 0 .05, 13.9205..:= t2 0 .05, 76.452..:=

Note that we need to set up two different time scales. This is necessary to ensure that the plotsstop at pointB.

Lengthy output is suppressed


40/142


0 1000 2000 3000 4000 50000

2000

4000

6000

8000

Plot for Part (a)

x (m)

y(m) y 1 t1,( )

y 2 t2,( )

x1 t1,( ) x2 t2,( ),

Part (b)

u ( )5000

cos ( ) 25000 tan ( ) 1500

9.81

:=

Given

u ( )

d

d0

Find( ) .63966976615851476362 .93112656063638185562( )

Only one of the two solutions is between 0 and 90.

m 0.9311265:=

um u m( ):= um 256.758=


41/142


xm t( ) um cos m( ) t:= ym t( ) um sin m( ) t1

29.81 t

2:=

We still need the time of flight for this path. This can be found by setting x=5000 m and solving for t.

5000

um cos m( )32.623=

t3 0 0.05, 32.623..:=

0 1000 2000 3000 4000 50000

2000

4000

6000

8000 Plot for Part (b)

x (m)

y(m)

y 1 t1,( )

y 2 t2,( )

ym t3( )

x1 t1,( ) x2 t2,( ), xm t3( ),


42/142


2.3 Problem 2/120 (n-t Coordinates)

A baseball player releases a ball with initialconditions shown in the figure. Plot the radius of

curvature of the path just after release and at the

apex as a function of the release angle . Explainthe trends in both results as approaches 90.

Problem Formulation

Just after release

20cos

vgan ==

cos

20

g

v=

At the apex

At the apex, vy= 0 and v= vx= v0cos. Sincevis horizontal, thenormal direction is vertically downward so that an=g.

( )

2

0 cosvgan == ( )

g

v2

0 cos=

Mathcad Worksheet

v0 100:= g 32.2:=

i( )v0

2

g cos ( ):= a ( )

v0 cos ( )( )2

g:=

0 0.01,

2..:=


43/142


0 15 30 45 60 75 900

200

400

600

800radius of curvature (ft)

theta (degrees)

i ( )

a ( )

180

Note that as approaches 90, the initial goes to infinity while at the apexapproaches zero. When = 90, the ball travels along a straight (vertical) path.As you recall, straight paths have a radius of curvature of infinity. At the apex ofthis vertical path the velocity will be zero giving a radius of curvature of zero.


44/142


2.4 Sample Problem 2/9 (Polar Coordinates)

Rotation of the radially slotted arm is governedby = 0.2t+ 0.02t3, where is in radians and tis in seconds. Simultaneously, the power screwin the arm engages the slider B and controls its

distance from O according to r = 0.2 + 0.04t2,

where r is in meters and t is in seconds.Calculate the magnitudes of the velocity andacceleration of the slider as a function of time t.

(a) Plot v, vrand vfor tbetween 0 and 5 sec. (b)

Plot a, arand a for t between 0 and 5 sec. (c)Plot the path of the slider B and compare with

the result in your book.

Problem Formulation

The first part of this problem solution will be identical to that in the SampleProblem in your text except that everything will be left in terms of t. To

summarize,

204.02.0 tr += tr 08.0=! 08.0=r!!

302.02.0 tt+= 206.02.0 t+=! t12.0=!!

Now all we have to do is substitute these expressions into the definitions for thevelocity and acceleration. As usual, there is no need to make an explicit

substitution when using the computer.

trvr 08.0== ! !rv = 22vvv r+=

2!!! rrar = !!!! rra 2+= 22aaa r+=

The plot for part (c) can be found in two ways. The first is to use the suggestion in your book andwrite

cosrx= sinry=

Now we have the x and y coordinates of the slider in terms of a common parameter t. Thissuggests that we can use a parametric plot. Also, the fact that we are using polar coordinates


45/142


would indicate that we might use MathCadspolar plot. This provides us with the second plotting

method.

MathCad Worksheet

r t( ) 0.2 0.04 t2

+:=

rd t( ) 0.08 t:= rdd 0.08:=

t( ) 0.2 t 0.02 t3

+:=

d t( ) 0.2 0.06 t2

+:= dd t( ) 0.12 t:=

vr t( ) 0.08 t:= v t( ) r t( ) d t( ):=

v t( ) vr t( )2

v t( )2

+:=

ar t( ) rdd r t( ) d t( )2

:= a t( ) r t( ) dd t( ):=

a t( ) ar t( )2

a t( )2

+:=

If, for some reason, you would like to see the actual expressions shown explicitly

as a function of time you can use the symbolic arrow ,

ar t( ) .8e-1 .2 .4e-1 t2

+( ) .2 .6e-1 t2+( )2

or

ar t( ) expand .72e-1 .64e-2 t2

.168e-2 t4

.144e-3 t6

t 0 .01, 5..:=


46/142


0 1 2 3 4 50

0.5

1

1.5

2

2.5

Part (a) Velocity (m/s)

time (s)

vr t( )

v t( )

v t( )

t

0 1 2 3 4 54

2

0

2

4

Part (b) Acceleration (m/s)

time (s)

ar t( )

a t( )

a t( )

t


47/142


x t( ) r t( ) cos t( )( ):= y t( ) r t( ) sin t( )( ):=

1.5 1 0.5 0 0.50.5

0

0.5

1 Part (c) method 1: Parametric Plot

x (m)

y(m)

y t( )

x t( )

0

30

60

90

120

150

180

210

240

270

300

330

1

0.8

0.60.4

0.2

Part (c) method 2: Polar Plot

r t( )

t( )


48/142


2.5 Sample Problem 2/10 (Polar Coordinates)

A tracking radar lies in the vertical plane of the

path of a rocket which is coasting in unpowered

flight above the atmosphere. For the instant

when = 30, the tracking data give r= 25(104)

feet, r! = 4000 ft/s, and ! = 0.8 deg/s. Let thisinstant define the initial conditions at time t = 0

and plot vr and vas a function of time for thenext 150 seconds. You may assume that g

remains constant at 31.4 ft/s2 during this time

interval.

Problem Formulation

Place a Cartesian coordinate system at the radar withxpositive to the right

andypositive up. Since the rocket is coasting in unpowered flight we canuse the equations for projectile motion,

( )tvxx cos00+=

( ) 2002

1sin gttvyy +=

Wherex0= 25(104)sin(30) ft,y0= 25(10

4)cos(30)

ft, v0 is the initial speed (5310 ft/sec, see the

sample problem) andis the angle that v0makeswith the horizontal. From the figure shown to the

right we can find the angle between v0and the r

axis as o1 11.41)4000/3490(tan == . Since

the r axis is 60 from the horizontal, = 60 41.11 = 18.89.

With rand defined as in the sample problem we have, at any time t

22 yxr += ( )yx /tan 1=

Now we find vrand vfrom their definitions.

rvr != !rv =


49/142


Substitution ofxandyinto the above equations and carrying out the derivatives

with respect to time gives vr and v as functions of time. The results are verymessy and will not be given here. Remember, though, that substitutions such as

this can be made automatically when using computer software such as Mathcad.

Mathcad Worksheet

v0 5310:= 18.89

180:= g 31.4:=

x0 25 104 sin 30

180

!#

:= y0 25 104 cos 30

180

!# &

:=

x t( ) x

0

v

0

cos ( ) t+:= y t( ) y0 v0 sin( ) t+1

2

g t2:=

r t( ) x t( )2

y t( )2+:= t( ) atan

x t( )

y t( )

!# &

:=

vr t( )tr t( )

d

d:= v t( ) r t( )

t t( )d

d:=

t 0 0.5, 150..:=


50/142


0 50 100 1502500

3000

3500

4000

4500

5000

vr t( )

v t( )

t


51/142


2.6 Problem 2/183 (Space Curvilinear Motion)

The base structure of the firetruck ladder rotatesabout a vertical axis through Owith a constant

angular velocity =! . At the same time, the

ladder unit OBelevates at a constant rate =! ,and sectionABof the ladder extends from within

section OA at the constant rate =R! . Findgeneral expressions for the components of

acceleration of point B in spherical coordinates

if, at time t = 0, = 0, = 0, and AB = 0.Express your answers in terms of , , , R0and t, where R0 = OA and is constant. Plot the

components of acceleration ofBas a function oftime for the case =10 deg/s, = 7 deg/s, =0.5 m/s, and R0= 9 m. Let t vary between 0 and

the time at which = 90.

Problem Formulation

The components of acceleration in spherical coordinates are,

222 cos!!!! RRRaR =

( ) sin2cos 2 !!! RRdtdRa =

( ) cossin1 22 !! RR

dt

d

Ra +=

The components may be obtained as functions of time by substituting,

tRR += 0 , t= and t=

Differentiation and substitution will be performed in Mathcad. The results are,

( ) ( )( )ttRaR += 2220 cos

( ) )sin(2)cos(2 0 ttRta +=


52/142


( ) ( ) ( )tttRa ++= cossin2 20

Mathcad Worksheet

Symbolic Calculations

t:= t t:= t R R0 t+:= t

Some variables will appear in red on your worksheet since they have not beendefined. This has no effect on the symbolic results.

Even though there are some obvious simplifications in this case, we still write themost general expressions for the spherical components of the acceleration. In this

way we can consider other types of time dependence without modifying the

worksheet.

aR 2t

Rd

d

2

Rtd

d

!

#

$

&

2

Rtd

d

!

#

$

&

2

cos ( )2

:= R

acos ( )

R tR

2

td

d

!

# &d

d 2 R

td

d

!

#

td

d

!

# & sin( ):=

R

a1

R tR

2

td

d

!

#

$

&d

d R

td

d

!

#

$

&

2

sin( ) cos ( )+:=R

The results of the symbolic operations can be seen by using the symbolic

evaluation sign .

aR R0 t+( ) 2

R0 t+( )2

cos t( )2

a 2 cos t( ) 2 R0 2 t+( ) sin t( )

a 2 R0 t+( )2

sin t( ) cos t( )+

Numerical Results

R0 9:= 10

180:= 7

180:= 0.5:=

Now we can copy and paste to create our functions of time for plotting.


53/142


aR t( ) R0 t+( ) 2

R0 t+( )2

cos t( )2

:=

a t( ) 2 cos t( ) 2 R0 2 t+( ) sin t( ):=

a t( ) 2 R0 t+( )2

sin t( ) cos t( )+:=

tf

2:= tf 12.857= time at which = /2.

t 0 0.05, tf..:=

0 5 10 150.8

0.6

0.4

0.2

0

0.2

acceleration (m/s^2)

(sec)

aR t( )

a t( )

a t( )

t


54/142


2.7 Sample Problem 2/16 (Constrained Motion of Connected Particles)

The tractorAis used to hoist the baleBwith thepulley arrangement shown. If A has a forward

velocity vA, determine an expression for the

upward velocity vBof the bale in terms ofx. Put

the result in nondimensional form by

introducing the velocity ratio = vB/vA andnondimensional position=x/h. Plot versusfor 0 2.

Problem Formulation

The lengthLof the cable can be written

( ) ( ) tsconsxhyhtsconslyhL tan2tan2 22 +++=++=

Now, 0=L! will be used to obtain a relation between vA(= x! ) and vB(=y! ).

2220

xh

xxyL

++==

!!!

222

1

xh

xvv AB

+=

The non-dimensional result is now obtained by substituting AB vv = and

hx= .

212

1

+=

Even though these operations are rather easily performed by hand, it isinstructive to have Mathcad do them. In particular, it will be instructive to see

how to evaluate L! even thoughxandyare not known explicitly as functions of

time.

Mathcad Worksheet

L 2 h y t( )( ) h2 x t( )2++:= x

Note that we need to differentiate L with respect to time. Both x and y depend ontime, however, exactly how they depend on time is not known. It turns out that


55/142


this is not a problem. All we need to do is let Mathcad know x and y depend on

time by writing x(t) and y(t).

LdottLd

d:= L

Ldot 2ty t( )

d

d

1

h2

x t( )2

+( )

1

2

x t( )tx t( )

d

d+

Now we make our substitutions. The first substitutes vB= vAfor y! (= )(tydt

d).

Ldot

substitutety t( )d

d vA,

substitutetx t( )

d

dvA,

substitute x t( ) h,

2 vA1

h2

2h

2+( )

1

2

h v+

Now we can copy and paste to solve the equation Ldot = 0 for .

2 vA

1

h2

2h

2+( )12

!#

$&

h vA

+'

+

(

)

,

*

0 solve ,1

2 h2

2h

2+( )12

h

We note finally that the h cancels in the above expression yielding the resultgiven in the problem formulation section above. Now we can produce the

required plot.

( )1

2

1 2

+

:=

0 0.01, 2..:=


56/142


0 0.5 1 1.5 20

0.2

0.4

0.6vB / vA

x/h

( )


57/142

KINETICS OFPARTICLES

The kinetics of particles is concerned with the motion produced by unbalancedforces acting on a particle. This chapter considers three approaches to the

solution of particle kinetics problems: (1) direct application of Newtons second

law, (2) work and energy, and (3) impulse and momentum. Problem 3.1 is arectilinear motion problem where three equations are solved symbolically forthree unknowns. In problem 3.2, polar plotting is used to plot the absolute path of

a particle. This problem also illustrates how Mathcad can be used to solve a

second order differential equation with initial conditions numerically. Problem3.3 uses Mathcad to study the effect of initial spring stretch upon the velocity of aslider. A physical interpretation of the results is also required. Problem 3.4 is a

typical ballistic pendulum problem requiring both work/energy and conservation

of momentum to relate the velocity of a projectile to the maximum swing angleof a pendulum. Problem 3.5 is a relatively straightforward conservation of

angular momentum problem where Mathcad is used to generate a plot that mightbe useful in a parametric study. In problem 3.6, two equations are solvedsymbolically for two unknowns using GivenFind. The maximum value of a

function is then determined.

3


58/142

58 CH. 3 KINETICS OF PARTICLES

3.1 Sample Problem 3/3 (Rectilinear Motion)

The 250-lb concrete block A is released fromrest in the position shown and pulls the 400-lb

log up the 30 ramp. Plot the velocity of theblock as it hits the ground at Bas a function of

the coefficient of kinetic frictionkbetween thelog and the ramp. Let kvary between 0 and 1.Why does the computer not plot results for the

entire range specified?

Problem Formulation

The constant length of the cable is L = 2sC+ sA

(see figure). Differentiating this expressiontwice yields a relation between the acceleration

ofAand C(note that aC= aLOG).

AC aa += 20 (1)

From the free-body diagram for the log

0== yy maF 0)30cos(400 =N

[ ]xx maF = Ck aTN2.32

400)30sin(4002 =+

Substituting N yields,

Ck aT2.32

400)30sin(4002)30cos(400 =+ (2)

From the free-body diagram for blockA

maF= AaT2.32

250250 = (3)

Mathcad will be used to solve the three equations above for aA, aCand Tin termsof k. Since the accelerations are constant, dav AA 2

2 = where d is the verticaldistance through which block A has fallen. Thus, the velocity ofAwhen it strikesthe ground (d= 20 ft) is


59/142

KINETICS OF PARTICLES 59

AAf av 40=

Mathcad Worksheet

Given

0 2 aC aA+

400k cos 30

180

!# &

2 T 400 sin 30

180

!# &

+400

32.2aC

250 T250

32.2a

Find T aA, aC,( )

142.85714285714285714 123.71791482634837811k+

15.934867429633671100 k 13.800000000000000000+

7.9674337148168355502k 6.9000000000000000000

!""

#

%%

&

Note that the accelerations may be either positive or negative depending on the

value of k

. The largest value of k

for which the block will move up can thus

be found by solving the equation aA= 0 for k . This yields k = 13.8/15.935 =

0.866.

aAk( ) 15.9349 k 13.8+:=

vAf k( ) 40 aAk( ):=

If you would like to see the symbolic result use the symbolic equals sign

vAf k( ) 637.3960 k 552.0+( )

1

2

k 0 0.001, 1..:=


60/142


0 0.2 0.4 0.6 0.8 10

10

20

vAfk( )

k

Note that results are not plotted beyond the limiting value for k

(0.866) that

was determined above. From a numerical point of view this occurs becauseMathcad will not plot imaginary answers. Whenever imaginary or complex

values result there is usually some physical explanation. In this problem, the

physical explanation is that the log will not slide up the incline if the coefficient

of friction is too large.


61/142


3.2 Problem 3/98 (Curvilinear Motion)

The particle Pis released at time t= 0 from theposition r = r0 inside the smooth tube with no

velocity relative to the tube, which is driven at

the constant angular velocity 0 about thevertical axis. Determine the radial velocity vr,the radial position r, and the transverse velocity

v as functions of time t. Plot the absolute pathof the particle during the time that it is inside the

tube for r0= 0.1 m, l= 1 m, and 0= 1 rad/s.

Problem Formulation

From the free-body diagram to the right,

( )20 !!! rrmmaF rr ===

20

2 rrr == !!!

Any book on differential equations will have the solution to this

equation in terms of the hyperbolic sine and cosine,

)cosh()sinh( 00 tBtAr +=

The constantsAandBare found from the initial conditions. These conditions arethat r = r0and 0=r! at t= 0. The second condition comes from the fact that the

particle has no velocity (initially) relative to the tube. Before evaluating thiscondition we must first differentiate r with respect to time.

)sinh()cosh( 0000 tBtAr +=!

BBArtr =+=== )0cosh()0sinh()0( 0

000 )0sinh()0cosh(0)0( ABAtr =+===!

From the above we haveB= r0andA= 0. Thus,

)cosh( 00 trr=

From this we can obtain the radial and transverse velocities,


62/142


63/142


0

30

6090

120

150

180

210

240270

300

330

0.8

0.6

0.4

0.2

0

path of the particle

r( )

Numerical Solution

t0

0

0:=

Given

2t r t( )

d

d

2

02

r t( )

r 0( ) r0r' 0( ) 0

r Odesolve t t0,( ):=

Since we have our results versus time it is necessary to use a parametric plot.

t( ) 0 t:=

Specifies the initial conditions. Note that the prime

indicates differentiation.

Numerically solves the differential equation out to

time t = t0

Time at which the particle leaves the tube.


64/142


0

30

60

90

120

150

180

210

240

270

300

330

0.8

0.6

0.4

0.2

0

path of the particle

r t( )

t( )

3.3 Sample Problem 3/17 (Potential Energy)

The 10-kg slider A moves with negligible

friction up the inclined guide. The attached

spring has a stiffness of 60 N/m and is stretched

m at position A, where the slider is releasedfrom rest. The 250-N force is constant and the

pulley offers negligible resistance to the motion

of the cord. Plot the velocity of the slider as it

passes Cas a function of the initial spring stretch

. Let vary between 0.4 and 0.8 m andexplain the results when exceeds a value ofabout 0.65 m.

Problem Formulation

The change in the elastic potential energy is

( ) ( )( )222122 2.12

1

2

1 +== kxxkVe

The other results in the sample problem are unchanged,

150)6.0(25021 ==U J ( ) 2202 )10(2

1

2

1vvvmT ==


65/142


9.58)30sin2.1)(81.9(10 === hmgVg J

( ) ( ) ( )( )2

2221 2.1602

19.58102

1150 +++== vU

This equation can be solved for veither by hand or by using Mathcad.

144095810

1=v

Mathcad Worksheet

First we solve the work/energy equation symbolically for v.

150 5 v2 58.9+ 30 1.2 +( )

2

2+ solve v,

110

958 1440( )

1

2

1

10958 1440( )

1

2

'

++++

(

)

,,,,

*

v ( )1

10958 1440:=

v ( ) 0 solve,479

720

479

7200.665=

0.4 0.399, 0.8..:=

0.4 0.2 0 0.2 0.4 0.6

1

2

3

4

v ( )


66/142


Note that no results are plotted beyond 0.65 m. Why? One reason is toobserve from the above equation that vbecomes imaginary when > 958/1440 =0.665 m. No results are plotted because Mathcad does not plot imaginary

numbers. But this is a numerical reason instead of a physical explanation.Usually, imaginary answers signify a situation that is physically impossible forsome reason. One way of understanding this is as follows. If the spring is initiallycompressed it will, at least for some part of the motion, be pushing up and thus

be aiding the 250 N force in overcoming the weight of the slider. If the spring is

initially stretched, it will always be pulling back on the slider. Thus the 250 Nforce will have to overcome not only the weight but also the spring force. It

stands to reason then that there will be some value for the initial spring stretchbeyond which the 250 N force will not be able to pull the slider all the way to C.This value is found from the limiting case where v = 0. Thus, the block never

reaches Cif > 0.665 m.

3.4 Problem 3/218 (Linear Impulse/Momentum)

The ballistic pendulum is a simple device tomeasure the projectile velocity v by observing

the maximum angle to which the box of sandwith embedded particle swings. As an aid for alaboratory technician, make a plot of the

velocity v in terms of the maximum angle .Assume that the weight of the box is 50-lb whilethe weight of the projectile is 2-oz.

Problem Formulation

(1) Impulse/Momentum

During impact, 0=G and 21 GG =

bvv %&

$"#

! +=%

&

$"#

!+%

&

$"#

!

2.32

5016/2)0(

2.32

50

2.32

16/2

bvv 401=

where vis the velocity of the projectile while vbis the velocity of the box of sandimmediately after impact.


67/142


(2) Work/Energy

Now we use the work/energy equation with our initialposition being the position where the pendulum is still

vertical (= 0) and the final position is that where thependulum has rotated through the maximum angle .

( ) hmgvmVTU bg +=+== 2221 02

10

where m is the combined mass of the box and the

projectile.

( )cos1)6)(2.32(22 == hgvb

cos17882401 == bvv

Mathcad Worksheet

v ( ) 7882 1 cos ( ):=

0 0.01,

2..:=

0 15 30 45 60 75 900

2000

4000

6000

8000projectile velocity (ft/s)

(degrees)

v ( )

180


68/142


3.5 Problem 3/250 (Angular Impulse/Momentum)

The assembly of two 5-kg spheres is rotatingfreely about the vertical axis at 40 rev/min with

= 90. The force F that maintains the givenposition is increased to raise the base collar and

reduce the angle from 90to an arbitrary angle. Determine the new angular velocity and

plot as a function of for 0 90. Assumethat the mass of the arms and collars is

negligible.

Problem Formulation

Since the summation of moments about thevertical axis is zero we have conservation ofangular momentum about that axis. The spheres

are rotating in a circular path about the vertical

axis. The angular momentum of a particlemoving in a circular path of radius r with

angular velocity is 2mrH= . Thus, from theconservation of angular momentum we have,

202

0 22 mrmr = 02

20

r

r=

where )45cos(6.01.0 00 +=r and ( )2/cos6.01.0 +=r

Mathcad Worksheet

0 402

60:= 0 4.189=

( )0.1 0.6 cos

4

!#

$&

+!#

$&

2

0.1 0.6 cos2

!#

$&

+!#

$&

2

0:=

0 0.05,

2..:=


69/142


0 15 30 45 60 75 90

2

2.5

3

3.5

4

4.5angular velocity (rad/s)

(degrees)

( )

180

This diagram begins at = 90where = 0= 40(2)/60 = 4.19 rad/s.


70/142


3.6 Problem 3/365 (Curvilinear Motion)

The 26-in. drum rotates about a horizontal axiswith a constant angular velocity = 7.5 rad/sec.The small blockAhas no motion relative to the

drum surface as it passes the bottom position =0. Determine the coefficient of static friction sthat would result in block slippage at an angular

position ; plot your expression for 0 180.Determine the minimum required coefficient

value minthat would allow the block to remainfixed relative to the drum throughout a fullrevolution. For a friction coefficient slightly less

than min, at what angular position wouldslippage occur?

Problem Formulation

From the free body and mass acceleration diagrams wehave,

[ ]nn maF = 2cos = mrmgN

[ ]tt maF = 0sin = mgF

For impending slip we have NF s= . Substituting Finto the above and solving gives,

cos8925.1

sin

cos

sin2 +

=+

=rg

gs

The last two questions can be answered only after plottingsas a function of .

Mathcad Worksheet

Given

N m g cos ( ) m r 2

sN m g sin( ) 0


71/142


Finds N,( )g

sin( )

g cos ( ) r2

+( )

m g cos ( ) m r 2

+

'

+

+(

)

,

,*

g 32.2:= r13

12:= 7.5:=

s ( ) gsin( )

g cos ( ) r2

+( ):=

0 0.02, ..:=

0 30 60 90 120 150 1800

0.2

0.4

0.6

0.8coefficient of static friction

(degrees)

s ( )

180

If the block is not to slip at any angle , the coefficient of friction must be greaterthan or equal to any value shown on the plot above. Thus, the minimum required

coefficient value min that would allow the block to remain fixed relative to the

drum throughout a full revolution is equal to the maximum value in the plotabove. The location where this maximum occurs can be found by solving the

equation 0/ = dd s for . This can then be substituted into s to yield the

required value formin. Heres how we do this with Mathcad.


72/142


xs x( )

d

d0 solve x,

2.1275232852017454951

2.1275232852017454951

!

# &

s 2.1275( ) 0.622=

From the above we see thatmin= 0.622. Ifsis slightly less than this value, theblock will slip when = 2.128 rads (121.9).


73/142

KINETICS OF SYSTEMSOF PARTICLES

This chapter concerns the extension of principles covered in chapters two andthree to the study of the motion of general systems of particles. The chapter first

considers the three approaches introduced in chapter 3 (direct application ofNewtons second law, work/energy, and impulse/momentum) and then moves toother applications such as steady mass flow and variable mass. Problem 4.1

considers an application of the conservation of momentum to a system comprised

of a small car and an attached rotating sphere. Mathcad is used to plot the

velocity of the car as a function of the angular position of the sphere. Theabsolute position of the sphere is also plotted. Problem 4.2 uses the concept ofsteady mass flow to study the effects of geometry upon the design of a sprinkler

system. One of the main purposes of this problem is to illustrate how a problem

can be greatly simplified using non-dimensional analysis. In particular, anequation containing seven parameters is reduced to a non-dimensional equationwith only three parameters. Problem 4.3 is a variable mass problem in which

Mathcad is used to integrate the kinematic equation adxvdv= . Symbolic solveandsimplifyare also illustrated.

4


74/142

74 CH. 4 KINETICS OF SYSTEMS OF PARTICLES

4.1 Problem 4/26 (Conservation of Momentum)

The small car, which has a mass of 20 kg, rollsfreely on the horizontal track and carries the 5-

kg sphere mounted on the light rotating rod with

r = 0.4 m. A geared motor drive maintains a

constant angular speed ! = 4 rad/s of the rod. If

the car has a velocity v = 0.6 m/s when = 0,plot vas a function of for one revolution of therod. Also plot the absolute position of the sphere

for two revolutions of the rod. Neglect the mass

of the wheels and any friction.

Problem Formulation

Since 0= xF we have conservation of momentum in the x

direction. The diagram to the right shows the system at = 0

and at an arbitrary angle . From the relative velocityequation, the velocity of the sphere is the vector sum of thevelocity of the car (v) and the velocity of the sphere relative

to the car ( !r ).

( ) 15)6.0(5)6.0(200

=+==xG Ns

( ) ( ) sin825sin520 =+= vrvvGx!

Setting ( ) ( ) xx

GG ==0 and solving yields,

sin32.06.0 +=v

Now let time t= 0 be the time when = 0 and place anx-ycoordinate system atthe center of the car as shown in the diagram so that x(t) is the position of the

center of the car. Since v= dx/dtand = 4twe have,

( )( ) ( )( )0 0 +=+==t t

ttdttvdtx

0 0

4cos108.06.04sin32.06.0

Thexandycomponents of the sphere can now be determined as,


75/142


76/142


xs t( ) 0.08 0.6 t+ 0.32 cos 4 t( )+:= ys t( ) 0.4 sin 4 t( ):=

t 0 0.01, ..:=

0 0.5 1 1.5 2 2.50.5

0

0.5position of the sphere

ys t( )

xs t( )


77/142

KINETICS OF SYSTEMS OF PARTICLES 77

4.2 Problem 4/62 (Steady Mass Flow)

The sprinkler is made to rotate at the constantangular velocity and distributes water at thevolume rate Q. Each of the four nozzles has anexit areaA. Write an expression for the torqueM

on the shaft of the sprinkler necessary tomaintain the given motion. Here we would liketo study the effects of the geometry of the

sprinkler upon this torque. To this end, it is

helpful to introduce the non-dimensional

parameters M = M/4Aru2, = r/u, and =b/rwhere u= Q/4A is the velocity of the water

relative to the nozzle and is the density of the

water. Plot the non-dimensional torque Mversus for = 0.5, 1, and 2. Let 0 be thenon-dimensional velocity at which thesprinkler will operate with no applied torque.

Plot 0 versus . For both plots let rangebetween 0 and 1.

Problem Formulation

The figure to the right shows the three components of theabsolute velocity of the water at the exit. u(= Q/4A) is the

velocity of the water relative to the nozzle. The mass flow

rate m=Q. Taking clockwise as positive, the applicationof equation 4/19 of your text yields,

( )0220 +== urbrQMM

( )( )22 brurQM +=

Now we want to introduce the non-dimensional parameters defined in theproblem statement. For many undergraduate students, non-dimensional analysis

is a very confusing topic. It is important to realize that the difficulty is really thatof determining which non-dimensional parameters are appropriate for a particular

problem. If these parameters have already been defined, as in this problem, all

you have to do is substitute. In this case we merely substituteM= 4Aru2M, = u/r, and b= rinto the equation above. When this is done many terms willcancel yielding,


78/142


( )211 +=M

SettingM= 0 we can solve for 0,

201

1

+=

Mathcad W

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