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Statics Worked Examples With Integrated Text and Graphics
Mark P. Rossow
Southern Illinois University Edwardsville
2009 Mark P. Rossow
ii
The traditional way to learn in a problem-solving course such
as statics is to solve a large number of homework problems.
This procedure is often inefficient, time-consuming and
frustrating because so much time is spent searching for the
solution that little time is left for learning the principles that
will enable a student to solve other related problems. Recent
educational research has suggested that a way to learn that is
superior to simply solving problems is to study worked
examples. However, the research has also shown that the
superiority of studying worked examples is maintained only
if the text and graphics appearing in the examples are
"integrated," by which is meant that the text on the page is
positioned immediately adjacent to the figure to which the
text refers. Additionally, liberal use is made of arrows
relating specific words or labels to appropriate features of the
figure, and the sequence of steps in the solution procedure is
identified by circled numbers.
The present book, which is intended as a supplement to a
course in statics, contains 445 worked examples in which
text and graphics are integrated. The range of topics covered
by the examples is sufficiently broad to encompass most
statics courses. Each topic section, except the introductory
section, begins with a Procedures and Strategies discussion
that outlines techniques for problem solving. Next are given
problem statements but not the solutions for all the
examples for that topic.
Why Integrated Text and Graphics?
The worked examples then follow the problem statements.
The solutions, as has already been mentioned, contain
integrated text and graphics, but in addition provide much
more detail and explanation than are found in examples in
most conventional textbooks.
The purpose of providing problem statements separated from
the solutions is to allow students, after they have studied
some of the examples in a particular topic section and wish
to test their understanding, to attempt to work other problems
without inadvertently seeing part of the solution. This
approach to learning is discussed further in the next section,
"How to Study Worked Examples."
A final note: although considerable effort has been expended
in checking and proofreading the examples presented herein,
in a work of this size undoubtedly some errors remain.
Readers who find errors are encouraged to report them to the
author at markrossow@gmail.com.
iii
You might think of the present book as an unusually detailed
and easy-to-read solution manual designed for students, who
are learning the subject matter, rather than for faculty
members, who already know the subject matter and are only
looking for numerical details of particular solutions. The
book does indeed provide numerical details of solutions, but
more importantly it also provides the rationale behind the
solution steps so that you can understand the principles and
solution techniques in the field of statics.
Here is how to make the best use of the book: When you
begin a topic, for example, "equilibrium of a particle," read
the first several examples. Then, when you think that you
understand the solution procedure, go on to the later
examples and try to work them. Look at the book's solution
only when you encounter a difficulty. In this way, your
mind will be focused on a specific question that you want
answered, and you will be motivated to study the example
solution much more closely than if you were simply to read
the example without first having attempted to work it. Once
you understand the solution thoroughly, summarize to
yourself the general principle that the example illustrates.
The way not to study worked examples is simply to read
them straight through without ever challenging yourself to
work an example on your own or without ever pausing to
think and crystallize in your mind the general principles
A Note to Students: How to Study Worked Examples
that the examples embody. If you read but do not study, you
may fall into the trap that educational psychologists refer to
as the "illusion of understanding": you think that you
understand but in fact you do not. Worked examples have
the considerable advantage over the traditional approach of
working homework problems in that they can save you time
and frustration, but working homework problems has the
advantage that you get immediate feedback on whether or
not you are learning. That is, when you cannot solve a
problem, you know immediately that you may have failed to
learn something important (or you may have made a simple
calculation error). It is obvious that you have to study more
or seek assistance. To use worked examples as a substitute
for working homework problems, you must continually
challenge yourself (by, for example, working some problems
without looking at the solutions in advance) so that you get
feedback on whether or not you are learning.
By the way, to save time and avoid frustration when you are
trying to find your own solution to the worked examples, it is
essential that you use a scientific graphing calculator or,
better, a calculation program on a computer. Having such a
computational tool will allow you to concentrate on learning
principles of statics rather than spending time on mere
numerical manipulation. See the next section, Using
Scientific Graphing Calculators Effectively, for ideas on how
best to do your numerical work with such tools.
iv
The solutions given in the present book contain detailed
numerical calculations that make it clear how the final
answers are obtained. However, when you are working an
example by yourself without looking at the solution, as
explained previously in the Section "How to Study Worked
Examples" having to deal with numerical details is a
distraction that may prevent you from seeing and learning the
underlying principle that the example is meant to illustrate.
To avoid this danger, you are strongly encouraged to use a
scientific graphing calculator (or a calculation program such
as Mathcad, Matlab, Maple, Mathematica, TK Solver, etc.).
But you are not only encouraged to use one of these
calculation tools, but also to use it effectively that is, in a
manner specifically designed to minimize errors, maximize
speed, and provide confidence in the final answer. Because
the author has encountered few students even very able
ones who have taught themselves to operate a calculator in
the manner just described, guidelines are given below for
using calculators and calculation programs effectively, so
that you can concentrate on learning concepts rather than
spending time puzzling over how a particular number
appearing in an example is calculated.
Guideline 1. Enter information in the order in which it
appears in the original problem formulation.
A Second Note: Using Scientific Graphing Calculators Effectively
For example, if you have an equation such as the following
to solve,
14.78 + 5X + 6(3.42) + 2.23(X 4) = 0,
do not rearrange this equation on paper and solve for X the
way you learned in algebra class. Instead, enter the equation
directly in your calculator's solver, and let the calculator do
the algebra.
Guideline 2. Check for errors by proofreading rather than
by repeating the calculation.
If Guideline 1 has been followed, then the equation on paper
can be compared character by character with the equation
appearing in the calculator display or on the computer screen.
Error checking by means of proofreading is preferable to
error checking by repeating the calculation because repeating
the calculation may introduce new errors, but proofreading
never does.
Guideline 3. Data that appear in several equations should be
entered only once.
For example, if the value of a force, 3.491 N, appears more
than once in equations or expressions to be entered into the
v
Using Scientific Graphing Calculators Effectively (Continued)
calculator, then the number should first be stored under the
name of a variable, and that name should be used in all terms
subsequently entered into the calculator.
Guideline 4. Store intermediate results in the calculator
rather than writing them on paper.
Just as in Guideline 3, storing intermediate results under the
name of a variable in the calculator decreases the number of
possibilities for errors.
Guideline 5. Use cut-and-paste rather than re-type.
Again, the number of possibilities for errors is reduced.
Guideline 6. Introduce intermediate variables to simplify
writing multilevel mathematical expressions.
Consider this multilevel mathematical expression:
Entering this expression as a single line of characters would
produce
(81*37.2/16)/(( /2)*(20^4+(9.87/12)^4)).
This is a very error-prone operation, since many parentheses
must be inserted that are not present or are not easily
identified with their position in the original expression. It is
much better to introduce intermediate variables. For
example, define X and Y as follows:
X
Y
That is,
X = 81*37.2/16 (1)
and
Y = ( /2)*(20^4+(9.87/12)^4). (2)
Eqs. 1 and 2 can be entered in the calculator, and then the
original multilevel expression can be entered in the form
X/Y,
which is much less prone to error.
(81)(37.2)16
2 [204 + ( )4]
9.8712
(81)(37.2)16
2 [204 + ( )4]
9.8712
vi
Using Scientific Graphing Calculators Effectively (Continued)
When should intermediate variables be used? A rule of
thumb is to introduce intermediate variables if parentheses
have to be nested three (or more) deep when the expression is
written on one line
Guideline 7. Know thoroughly how to use the solver.
As was pointed out in the discussion of Guideline 1, the
solver is the key to avoiding errors in algebraic manipulation.
Guideline 8. Use built-in functions whenever possible.
For example, definite integrals should always be evaluated
with the calculator's built-in integration function rather than
by finding the antiderivative and substituting the values of
the limits. Vector operations such as dot product, cross
product, magnitude (norm), and finding a unit vector should
all be done using built-in functions.
Guideline 9. Do not use Reverse Polish Notation.
Reverse Polish Notation (RPN) is an ingenious technique for
reducing the number of keystrokes needed to perform certain
arithmetic operations. As an example, consider the
evaluation of the expression 6(3 + 2) using RPN:
Keystrokes Calculator Display
3 3 2 2 + 5 6 6 * 30
Note that the parentheses do not have to be entered, and thus
the user saves two keystrokes, compared to what would have
been required had RPN not been used. RPN is a standard
feature on some types of calculators and has gained wide
acceptance. However, because the complete arithmetic
expression never is displayed on the screen when RPN is
used, the calculator user cannot easily check for errors by
proofreading. Instead the user must check the calculation by
re-entering the data, thus doubling the number of keystrokes
required to obtain a verified answer, and RPN ends up taking
more keystrokes than a conventional left-to-right data-entry
scheme. Furthermore, if the user makes an error in entering
data during the repeat calculation, then yet another repeat
calculation must be performed and the number of keystrokes
is tripled. The more keystrokes, the more opportunity for
making an error.
vii
Using Scientific Graphing Calculators Effectively (Continued)
Guideline 10. Violate any of the other guidelines when
common sense says to do so.
Guidelines 1 to 9 should be followed in the large majority of
cases, but special situations will arise in which the sensible
thing to do is to violate the guidelines. That is why the word
"guideline" has been used rather than "law.
viii
Acknowledgments
The examples and drawings in this text were developed with
the support of Southern Illinois University Edwardsville
Excellence in Undergraduate Education Grants Nos. 05-25
and 09-29 (Mark P. Rossow, Principal Investigator) and
through the hard work and dedication of the following
people:
Jason Anderson
Paul Cayo
Madan Gyanwali
Tyler Hermann
Vishnu Kesaraju
Jennie Moidel
Binod Neupane
Sanjib Neupane
Ramesh Regmi
Laxman Shrestha
Shikhar Shrestha
Sagun Thapa
ix
Contents
1. Preliminaries: Units and Rounding ............................................................................... 1
2. Force and Position Vectors .......................................................................................... 22
2.1 Adding Forces by the Parallelogram Law ......................................................... 23
2.2 Rectangular Components in Two-Dimensional Force Systems ........................ 60
2.3 Rectangular Components in Three-Dimensional Force Systems .................... 113
2.4 Position Vectors. Use in Defining Force Vectors. ........................................ 146
2.5 Applications of Dot Products .......................................................................... 200
3. Equilibrium of a Particle ............................................................................................ 239
3.1 Particles and Two-Dimensional Force Systems ............................................... 240
3.2 Particles and Three-Dimensional Force Systems ............................................. 261
4. Moments and Resultants of Forces Systems .............................................................. 306
4.1 Moments in Two-Dimensional Force Systems ................................................ 307
4.2 Moments in Three-Dimensional Force Systems .............................................. 340
4.3 Moment of a Couple......................................................................................... 401
4.4 Moment of a Force About a Line. .................................................................. 444
4.5 Equivalent Force-Couple Systems ................................................................... 472
4.6 Distributed Loads on Beams ............................................................................ 556
5. Equilibrium of a Rigid Body ...................................................................................... 605
5.1 Constraints and Static Determinacy ................................................................. 606
5.2 Rigid Bodies and Two-Dimensional Force Systems ....................................... 688
5.3 Rigid Bodies and Three-Dimensional Force Systems ..................................... 731
6. Structural Applications .............................................................................................. 790
6.1 Frames and Machines ....................................................................................... 791
6.2 Trusses: Method of Joints and Zero-Force Members ...................................... 903
6.3 Trusses: Method of Sections ............................................................................ 960
6.4 Space Trusses. .............................................................................................. 1000
6.5 Cables: Concentrated Loads ........................................................................... 1035
6.6 Cables: Uniform Loads .................................................................................. 1082
6.7 Cables: Catenaries .......................................................................................... 1133
x
7. Friction ..................................................................................................................... 1191
7.1 Basic Applications ......................................................................................... 1192
7.2 Wedges ........................................................................................................... 1250
7.3 Square-Threaded Screws ................................................................................ 1286
7.4 Flat Belts ........................................................................................................ 1347
7.5 Thrust Bearings and Disks ............................................................................. 1380
7.6 Journal Bearings ............................................................................................. 1402
7.7 Rolling Resistance .......................................................................................... 1450
8. Internal Forces .......................................................................................................... 1481
8.1 Internal Forces in Structural Members ........................................................... 1482
8.2 Shear and Bending-Moment Diagrams: Equation Form ................................ 1536
8.3 Shear and Bending-Moment Diagrams Constructed by Areas ...................... 1597
9. Centroids and Mass Centers ..................................................................................... 1670
9.1 Centroids by Integration ................................................................................. 1671
9.2 Centroids: Method of Composite Parts .......................................................... 1749
9.3 Theorems of Pappus and Guldinus ................................................................ 1845
9.4 Hydrostatic Pressure on Submerged Surfaces. ............................................. 1924
10. Inertia Properties of Plane Areas............................................................................ 1981
10.1 Moments of Inertia by Integration ............................................................... 1982
10.2 Method of Composite Areas ........................................................................ 2027
10.3 Products of Inertia ........................................................................................ 2067
10.4 Moments of Inertia About Inclined Axes; Principal Moments .................... 2115
11. Energy Methods ..................................................................................................... 2165
11.1 Virtual Work ................................................................................................ 2166
11.2 Potential Energy ........................................................................................... 2270
Appendix: Geometric Properties of Lines, Areas, and Solid Shapes ............................ 2338
Index .............................................................................................................................. 2343
1
1. Preliminaries: Units and Rounding
2
1. Preliminaries: Units and Rounding Problem Statement for Example 1
1. Round off the following numbers
to three significant figures.
a) 54.27 m
b) 2 927 124 m
c) 3.6143 10-3 in.
d) 6.875 km
e) 6.885 km
3
1. Preliminaries: Units and Rounding Problem Statement for Example 2
2. Express the following quantities in proper SI units.
a) 2491 N
b) 0.02491 N
c) 2 491 000 N
d) 0.0987 10-5 m
e) 0.0987 108 km
f) 0.000 987 10-2 mm
4
1. Preliminaries: Units and Rounding Problem Statement for Example 3
3. Evaluate the arithmetic expressions and express the
results to three significant figures and in proper SI
form.
a)
b)
c) 934.2 mm2 2 m
d) 99.7 mN2 3.6 mm
e) 8 kN 12.1 mm
f)
17 mm13 N
13 N 17 kg
43 ms
(82.1 ms)2
5
1. Preliminaries: Units and Rounding Problem Statement for Example 4
4. Express the following forces in pounds (lb).
a) 1,230 kip
b) 0.0230 kip
c) 23.0 oz
6
1. Preliminaries: Units and Rounding Problem Statement for Example 5
5. a) A 20-kg box is set upon a scale that displays
weight in SI force units. What weight will the scale
display?
b) A 20-lb box is set upon a scale that displays
weight in U.S. customary units. What weight will
the scale display?
7
1. Preliminaries: Units and Rounding Example 1, page 1 of 5
1. Round off the following numbers
to three significant figures.
a) 54.27 m
b) 2 927 124 m
c) 3.6143 10-3 in.
d) 6.875 km
e) 6.885 km
1
2
a) 54.27 lies between 54.20 and 54.30.
54.27
54.20 (midpoint
of interval)54.30
3 significant figures 3 significant figures
Because 54.27 lies closer to 54.3, round to that value:
54.27 m 54.3 m Ans.
8
1. Preliminaries: Units and Rounding Example 1, page 2 of 5
4
b) 2 927 124 lies between 2 920 000 and 2 930 000.
2 927 124
2 920 000 (midpoint
of interval)2 930 000
3 significant figures 3 significant figures
Because 2 927 124 lies closer to 2 930 000, round to that
value:
2 927 124 m 2 930 000 m
= 2.93 106 m Ans.
Use engineering
notation in which the
significant figures are
followed by a factor of
10 raised to a power.
To make it easy to select the
correct SI prefix, use an
exponent that is a multiple of 3.
5
6
3
9
1. Preliminaries: Units and Rounding Example 1, page 3 of 5
7
8
c) 3.6143 10-3 lies between 3.610 10-3 and 3.620 10-3 .
3.610 10-3 (midpoint of
interval)3.62 10-3
3 significant figures 3 significant figures
Round to the closer value:
3.6143 10-3 in. 3.61 10-3 in. Ans.
3.6143 10-3
10
1. Preliminaries: Units and Rounding Example 1, page 4 of 5
9
10
d) 6.875 lies precisely halfway between 6.870 and 6.880.
6.870 6.875
(midpoint
of interval)
6.880
3 significant figures 3 significant figures
The rule is "If the number lies at the midpoint of the interval,
then round off so that the last digit is even":
6.875 km 6.88 km Ans.
Even digit (8)
The rationale for the rule is that over a long chain of
calculations, the rule will lead to rounding down approximately
the same number of times as rounding up, and, on balance, these
rounding errors will cancel.
12
11
11
1. Preliminaries: Units and Rounding Example 1, page 5 of 5
14
e) 6.885 lies precisely halfway between 6.880 and 6.890.
6.880 6.885
(midpoint
of interval)
6.890
3 significant figures 3 significant figures
Round off so that the last digit is even:
6.885 km 6.88 km Ans.
Even digit (8)
In this case (6.885), we rounded down to get an even digit; in
the previous case (6.875), we rounded up to get an even digit.16
15
6.8806.870 6.875 6.8906.885
Even digit (8)Odd digit (7) Odd digit (9)
Round up Round down
13
12
1. Preliminaries: Units and Rounding Example 2, page 1 of 4
1
2. Express the following quantities in proper SI units.
a) 2491 N
b) 0.02491 N
c) 2 491 000 N
d) 0.0987 10-5 m
e) 0.0987 108 km
f) 0.000 987 10-2 mm
a) 2491 N = 2.491 103 N
= 2.491 kN Ans.
Choose the exponent of 10 to be a
multiple of 3, because SI prefixes
are defined for 103 , 106 , 109 , ...,
and 10-3 , 10-6 , 10-9 , ... .
Try to keep the value of the
coefficient between 0.1 and 1000.
Here it is possible. In other cases,
such as 2222 mm4 , it is not.
Prefix "k" = "kilo" = 103 .
2
3
4
13
1. Preliminaries: Units and Rounding Example 2, page 2 of 4
5 b) 0.024 91 N = 24.91 10-3 N
= 24.91 mN Ans.
Exponent of 10 is a multiple of 3.
Prefix "m" = "milli" = 10-3 .
7
8
Coefficient lies
between 0.1 and 1000.
6
12 Prefix "M" = "mega" = 106.
c) 2 491 000 N = 2.491 106 N
= 2.491 MN Ans.
9
10 Coefficient lies
between 0.1 and 1000.
11 Exponent of 10 is a multiple of 3.
14
1. Preliminaries: Units and Rounding Example 2, page 3 of 4
d) 0.0987 10-5 m = 0.987 10-6 m
= 0.987 m Ans.
Prefix " " = "micro" = 10-6 14
17 Coefficient lies
between 0.1 and 1000.
19
18
Prefix "G" = "giga" = 109 .
Exponent of 10 is a multiple of 3.
e) 0.0987 108 km = 0.0987 108 (103 ) m
= 0.0987 1011 m
= 9.87 109 m
= 9.87 Gm Ans.
15
Convert to base unit (from km to m).16
13
15
1. Preliminaries: Units and Rounding Example 2, page 4 of 4
23
22
Prefix "n" = "nano" = 10-9 .
Exponent of 10 is a multiple of 3.
f) 0.000 987 10-2 mm = 0.000 987 10-2 (10-3 ) m
= 9.87 10-6 (10-3 ) m
= 9.87 10-9 m
= 9.87 nm Ans.
20
Convert to base unit (from mm to m).21
16
1. Preliminaries: Units and Rounding Example 3, page 1 of 4
a)
=
= (13/17)(103 ) N/m
= 0.765 (103 ) N/m
= 0.765 kN/m Ans.
1
3. Evaluate the arithmetic expressions and express the
results to three significant figures and in proper SI
form.
a)
b)
c) 934.2 mm2 2 m
d) 99.7 mN2 3.6 mm
e) 8 kN 12.1 mm
f)
17 mm13 N
13 N 17 kg
43 ms
(82.1 ms)2
13 N 17 mm
= 17(10-3 ) m
13 N
13(103 ) N
17 m
2 Always express the denominator in base units
(In statics, the base units are m, s, and kg).
17
1. Preliminaries: Units and Rounding Example 3, page 2 of 4
b) = (13/17) N/kg
= 0.765 N/kg Ans.
3 13 N 17 kg
4 "kg" is a base unit and so may appear in the
denominator.
The exponent applies to the
prefix "milli" as well as to
the base unit "meters".
6
5 c) 934.2 mm2 2 m = 934.2 (10-3 m)2 2 m
= (934.2 2)(10-6 m2 ) m
= (1868.4)(10-6 m3 )
= (1.87 103 )(10-6 m3 )
= 1.87 10-3 m3 Ans.
In this example, it is not possible to make
the coefficient lie between 0.1 and 1000
by choosing an SI prefix.
7
18
1. Preliminaries: Units and Rounding Example 3, page 3 of 4
The exponent applies to the
prefix "milli" as well as to
the base unit "newtons".
9
8 d) 99.7 mN2 3.6 mm = 99.7(10-3 N)2 3.6(10-3 m)
= (99.7 3.6)(10-6 N2 )(10-3 m)
= (358.92)(10-6 N2 )(10-3 m)
= (359)(N2 10-9 m)
= 359 N2 nm Ans.
The raised dot indicates
the product of two units.10
Prefix "n" = "nano" = 10-9
e) 8 kN 12.1 mm = 8(103 N) 12.1(10-3 m)
= (8 12.1)(103-3 Nm)
= 96.8 Nm Ans.
12
11
19
1. Preliminaries: Units and Rounding Example 3, page 4 of 4
13 f)
= 6.38 10-3 (106 ) m/s
= 6.38 km/s Ans.
14
43 ms
ms)2 =
10-3 s)2 43 ms
2
43 ms
(10-6 s2 =
Raised dot means
"product of meters and
seconds."
No raised dot so "ms"
means "milli-seconds."
15
20
1. Preliminaries: Units and Rounding Example 4, page 1 of 1
4. Express the following forces in pounds (lb).
a) 1,230 kip
b) 0.0230 kip
c) 23.0 oz
1 a) 1,230 kip = 1,230 1,000 lb
= 1.23 106 lb Ans.
b) 0.0230 kip = 0.0230 1,000 lb
= 23.0 lb Ans.
c) 23.0 oz = 23.0 oz 1 lb/16 oz)
= 1.4375 lb
= 1.44 lb Ans.
"kip" = "kilo-pound" = 1,000 lb2
3
4
21
1. Preliminaries: Units and Rounding Example 5, page 1 of 1
5. a) A 20-kg box is set upon a scale that displays
weight in SI force units. What weight will the scale
display?
b) A 20-lb box is set upon a scale that displays
weight in U.S. customary units. What weight will
the scale display?
1 a) Weight = mass acceleration due to gravity
= (20 kg) (9.81 m/s2 )
= 196.2 kgm/s2
= 196.2 N Ans.
b) Weight = 20 lb Ans.
("lb" is a force unit; no factor corresponding
to the acceleration of gravity is needed.)
2
3
1 kgm/s2 1 N
22
2. Force and Position Vectors
23
2.1 Adding Forces by the Parallelogram Law
24
2.1 Adding Forces by the Parallelogram Law: Procedures and Strategies, page 1 of 2
Procedures and Strategies for Solving Problems Involving
Addition of Forces by the Parallelogram Law
To add two force vectors,
1. make a sketch showing the vectors placed tail-to-tail;
2. construct a parallelogram, using the vectors as two of the
sides;
3. draw the diagonal that goes from the tail-tail vertex to the
opposite vertex (This is the resultant the vector sum of
the two forces); and
4. use the sine and cosine laws and geometrical relations
between angles to calculate the magnitude and direction of
the resultant.
A
A
A
A
B
B
B
B
A
BC
Law of sines
sin a
A
sin b
B
sin c
C= =
Law of cosines
C2 = A2 + B2 - 2AB cos c
+
A + B
b
a
c
25
2.1 Adding Forces by the Parallelogram Law: Procedures and Strategies, page 2 of 2
F
u
v
u
v
F
u
v
F
To resolve a given force into components in two given
directions,
1. make a sketch showing the tail of the force vector at the
intersection of two lines in the given directions;
2. construct a parallelogram with the force vector as a
diagonal and the sides parallel to the given directions;
and
3. use the sine and cosine laws and geometrical relations
between angles to calculate the lengths of the sides of the
parallelogram. The sides are the components of the
force vector.
Fu
Fv
26
2.1 Adding Forces by the Parallelogram Law Problem Statement for Example 1
1. Determine the magnitude and direction
of the resultant of the forces shown.
30
150 N
200 N
27
2.1 Adding Forces by the Parallelogram Law Problem Statement for Example 2
y
x
10
20
2. Determine the magnitude and direction
of the resultant of the forces shown.
3 kN
2 kN
28
2.1 Adding Forces by the Parallelogram Law Problem Statement for Example 3
110
100 N
80 N40
3. Determine the magnitude and direction of the resultant force.
29
2.1 Adding Forces by the Parallelogram Law Problem Statement for Example 4
x
40 lb
60 lb
30
y
4. The resultant of the two forces acting on the screw eye is known to
be vertical. Determine the angle and the magnitude of the resultant.
30
2.1 Adding Forces by the Parallelogram Law Problem Statement for Example 5
5. Determine the magnitude F and the angle , if the
resultant of the two forces acting on the block is to be a
horizontal 80-N force directed to the right.
50 NF
25
31
2.1 Adding Forces by the Parallelogram Law Problem Statement for Example 6
A
B
30 N25 N
6. To support the 2-kg flower pot shown, the resultant of the two
wires must point upwards and be equal in magnitude to the
weight of the flower pot. Determine the angles and , if the
forces in the wires are known to be 25 N and 30 N.
C
32
2.1 Adding Forces by the Parallelogram Law Problem Statement for Example 7
25
40
v
u
120 lb
7. Resolve the 120-lb force into components
acting in the u and v directions.
33
2.1 Adding Forces by the Parallelogram Law Problem Statement for Example 8
8. Resolve the 4-kN
horizontal force into
components along truss
members AB and AC.
35
5
12
A
B C
4 kN
34
2.1 Adding Forces by the Parallelogram Law Problem Statement for Example 9
9. Find two forces, one acting along rod AB and one
along rod CB, which when added, are equivalent to the
200-N vertical force.
200 N
A
B
C
30 40
35
2.1 Adding Forces by the Parallelogram Law Example 1, page 1 of 4
1. Determine the magnitude and direction
of the resultant of the forces shown.
30
150 N
200 N
30
200 N
150 N
30
Construct a parallelogram by drawing two
lines. Each line starts at the tip of one vector
and is parallel to the other vector.Tip
Parallel
Tip
1
36
2.1 Adding Forces by the Parallelogram Law Example 1, page 2 of 4
30
200 N
150 N
30
Since opposite sides of a parallelogram are
equal in length, the length of each line
represents the magnitude of the vector opposite.
200 N
150 N
2
30
R
The resultant R is drawn from the tails of the
vectors to the opposite vertex of the parallelogram.
Tails
150 N
200 N
150 N
200 N
3
37
2.1 Adding Forces by the Parallelogram Law Example 1, page 3 of 4
200 N
150 N
30
R150 N
200 N
150 N
200 N
Tails Head
Heads
4
30
200 N
R150 N
To calculate the magnitude and direction of R, consider
the triangle formed by one half of the parallelogram.5
200 N
150 N
This is not the resultant because it is not
drawn from the intersection of the tails.
38
2.1 Adding Forces by the Parallelogram Law Example 1, page 4 of 4
30
200 N
R 150 N
Use trigonometry to calculate the magnitude and direction
of the resultant.
R2 = (200 N)2 + (150 N)2 2(200 N)(150 N) cos 30
The result is
R = 102.66 N Ans.
=
Solving gives
= 46.9 Ans.
b
a
c
C
A
B
Law of cosines
C2 = A2 + B2 2AB cos c
Law of sines
= =
6
sin a A B
sin b C
sin c
sin 150 N
sin 30 R
= 102.66 N
Trigonometric formulas for a
general triangle are given below.
39
2.1 Adding Forces by the Parallelogram Law Example 2, page 1 of 3
2. Determine the magnitude and direction
of the resultant of the forces shown.
3 kN
2 kN
Construct a parallelogram by drawing
two lines parallel to the forces.
1
y
x
3 kN
2 kN
y
x
3 kN
2 kN
10
20
10
20
40
2.1 Adding Forces by the Parallelogram Law Example 2, page 2 of 3
Draw the resultant R from the tails of the vectors to the
opposite vertex of the parallelogram.
3 kN
2 kN
y
x
3 kN
2 kN
2
Tails
R
2 kN2 kN
R
3 kN
3 kN
y
x
To calculate the magnitude and direction of R, consider
the triangle formed by one half of the parallelogram.
3
R
3 kN
2 kN
10
20
20
10
10
20 20
10
41
2.1 Adding Forces by the Parallelogram Law Example 2, page 3 of 3
R
3 kN
2 kN
Use trigonometry to calculate the magnitude
and direction of the resultant.
4
total angle= 10 + 90 + 20 = 120
Law of cosines
R2 = (3 kN)2 + (2 kN)2 2(3 kN)(2 kN) cos 120
R = 4.359 kN Ans.
sin 120 R
Law of sines
=
Solving gives
= 36.6
sin 3 kN = 4.359 kN
5
6
R = 4.36 kN
x
y
36.6 + 20 = 56.6 Ans.
= 36.6
7 Angle measured with
respect to the vertical axis
20
10
2020
42
2.1 Adding Forces by the Parallelogram Law Example 3, page 1 of 3
110
100 N
80 N
Construct a parallelogram1
100 N
80 N
40
110
y
40
3. Determine the magnitude and direction of the resultant force.
43
2.1 Adding Forces by the Parallelogram Law Example 3, page 2 of 3
Draw the resultant R from the tails of the vectors to
the opposite vertex of the parallelogram.
100 N
80 N
40
110
y
2
100 N
80 N
R
40
110
80 N
100 N
y
R
30
To calculate R, consider the triangle formed
by the lower half of the parallelogram.
3
Calculate angle
180 110 40 = 30
Parallel lines make 30 angle
with vertical direction
4
5
Calculate angle
30 + 40 = 70
6
40
Law of cosines
R2 = (80 N)2 + (100 N)2 2(80 N)(100 N) cos 70
R = 104.54 N Ans.
7
44
2.1 Adding Forces by the Parallelogram Law Example 3, page 3 of 3
y
100 N
R = 104.54 N
70
Calculate the angle that the
resultant makes with the vertical.
Law of sines
=
Solving gives
= 64.0
R sin 70
100 N sin
104.54 N
8
R = 104.54 N
y
Angle measured from the vertical9
40 + 64.0 = 104.0 Ans.
40
64
40
45
2.1 Adding Forces by the Parallelogram Law Example 4, page 1 of 4
x
40 lb
60 lb
30
y
4. The resultant of the two forces acting on the screw eye is known to
be vertical. Determine the angle and the magnitude of the resultant.
46
2.1 Adding Forces by the Parallelogram Law Example 4, page 2 of 4
40 lb
30
x
y
30
40 lb
x
y
30
40 lb
y
x
To determine what needs to be calculated, make some
sketches of several possible parallelograms.
1
R
R
R
Each parallelogram is based on two facts that are given:
1) One side of the parallelogram is known (40 lb at 30), and
2) The resultant R lies on the y axis.
2
47
2.1 Adding Forces by the Parallelogram Law Example 4, page 3 of 4
30
40 lb
x
y
How do we determine the actual parallelogram? We have
to use the additional fact that one of the forces is 60 lb.
3
The point of the intersection of the arc and the
vertical axis must be the vertex of the
parallelogram since it lies on the vertical axis and
also lies a "distance" of 60 lb from the tip of the
40-lb vector.
4
Now the parallelogram is completely defined.
60 lb
40 lb
30
x
y
5
40 lb
60 lb
Radius of circular
arc = 60 lb
48
2.1 Adding Forces by the Parallelogram Law Example 4, page 4 of 4
To calculate the resultant R and and the
angle (see below), analyze the triangle
formed by the left half of the parallelogram.
Angle = 90 30 = 60
Corresponding angles are equal
y30
x
40 lb
60 lb
6
8
7
30
Parallel
60 lb
40 lb
60
+ 30
R
sin 60 60 lb
Law of sines
=
Solving gives
= 35.26
sin 40 lb
Law of sines
=
Solving gives
R = 69.0 lb Ans.
sin 60 60 lb
sin( + 30)
R
9
11 54.74
The sum of the angles of the triangle is 180:
+ ( + 30) + 60 = 180
Solving gives
= 54.74 Ans.
10
35.26
R
49
2.1 Adding Forces by the Parallelogram Law Example 5, page 1 of 2
5. Determine the magnitude F and the angle , if the
resultant of the two forces acting on the block is to be a
horizontal 80-N force directed to the right.
50 NF
Draw the parts of parallelogram that are known:
Two sides are of length 50 N and make
an angle of 25 with the horizontal axis.
The diagonal of the parallelogram (the
resultant) is 80 N long and horizontal.
50 N
50 N
80 N
1
2
3
25
25
25
50
2.1 Adding Forces by the Parallelogram Law Example 5, page 2 of 2
Complete the parallelogram.
25
50 N
50 N
80 N
4
25
25
F
F
Analyze the triangle forming the lower half of the parallelogram.
6
25
F
80 N
50 N Calculate F from the law of cosines.
F2 = (50 N)2 + (80 N)2 2(50 N)(80 N) cos 25
The result is
F = 40.61 N Ans. sin
50 N
Calculate from the law of sines.
=
Solving gives
= 31.4 Ans.
sin 25
F 40.61 N
7
5
51
2.1 Adding Forces by the Parallelogram Law Example 6, page 1 of 2
A
B
30 N25 N
6. To support the 2-kg flower pot shown, the resultant of the two
wires must point upwards and be equal in magnitude to the
weight of the flower pot. Determine the angles and , if the
forces in the wires are known to be 25 N and 30 N.
B
Weight of flower pot
mg = (2 kg)(9.81 m/s2 )
= 19.62 N
1
19.62 N 19.62 N
B
R = 19.62 N
Resultant, R, of forces in wires
balances the weight.2
C
52
2.1 Adding Forces by the Parallelogram Law Example 6, page 2 of 2
B
25 N
30 N
30 N
25 N
The resultant R = 19.62 N must be the diagonal of a
parallelogram with sides 25 N and 30 N long.
3 Analyze the triangle forming the left-hand
half of the parallelogram.
4
25 N
30 N
19.62 N 19.62 N
Law of cosines to calculate
(25 N)2 = (30 N)2 + (19.62 N)2 2(30 N)(19.62 N) cos
Solving gives
= 55.90 Ans.
Law of sines to calculate
=
Solving gives
= 83.6 Ans.
30 N sin
5
sin
25 N 55.90
6
53
2.1 Adding Forces by the Parallelogram Law Example 7, page 1 of 2
25
40
v
u
120 lb
7. Resolve the 120-lb force into components
acting in the u and v directions.
Construct a parallelogram with
the 120-lb force as a diagonal.
Draw another line from the
tip but parallel to u.
Draw a line from the tip of the
force vector parallel to v.120 lb
v
40
u
40
25
1
2
3 R u
R v
Label the components R u and R v.4
54
2.1 Adding Forces by the Parallelogram Law Example 7, page 2 of 2
120 lb40
25R v
R u
Analyze the triangle forming the left-hand half of the
parallelogram.
180 40 25 = 115
5
Calculate R u from the law of sines.
=
Solving gives
R u = 78.9 lb Ans.
120 lb
sin 40 sin 25
R u
6
Calculate R v from the law of sines.
=
Solving gives
R v = 169.2 lb Ans.
sin 40 sin 115
R v 120 lb
7
55
2.1 Adding Forces by the Parallelogram Law Example 8, page 1 of 2
8. Resolve the 4-kN
horizontal force into
components along truss
members AB and AC.
35
5
12
A
B C
4 kN
Extend line AC.
35
Construct a parallelogram with the
4-kN force as the diagonal.
12
5
4 kN
2
B C
Draw another line from
the tip but parallel to AC.
4
5
12
ADraw a line from the tip of the
force vector parallel to AB.
3
1
56
2.1 Adding Forces by the Parallelogram Law Example 8, page 2 of 2
35
12
5
4 kN
B C5
12
A
Label the components R B and R C.5
R B
R C
4 kN
12
5
R C
R B
Analyze the triangle forming the upper half of the
parallelogram (The drawing has been enlarged for clarity).6
35
12 5
Geometry
= tan-1 = 67.38
= 180 35 67.38 = 77.62
7
77.62
Law of sines to calculate R C
=
Solving gives
R C = 3.78 kN Ans.
sin sin R C 4 kN
67.38
Law of sines to calculate R B
=
Solving gives
R B = 2.35 kN Ans.
R B sin 35
4 kN
sin 77.62
8
9
35
57
2.1 Adding Forces by the Parallelogram Law Example 9, page 1 of 3
9. Find two forces, one acting along rod AB and one
along rod CB, which when added, are equivalent to the
200-N vertical force.
200 N
A
B
C
30 40
58
2.1 Adding Forces by the Parallelogram Law Example 9, page 2 of 3
30
40 30
40
30
40 30
200 N
A
B
C
30
1 Construct a parallelogram with
sides parallel to AB and BC and
with the 200-N force as a
diagonal.Draw a line from the tail of the
force vector parallel to AB.
Draw another line from the
tail but parallel to BC.
Extend AB.Extend BC.
4
5
32
200 N
B
R C
R A
Label the sides of the
parallelogram R A and R C.
6
R CR A
59
2.1 Adding Forces by the Parallelogram Law Example 9, page 3 of 3
200 N
R C
30
40
180 50 (30 + 40) = 60
200 N
Law of sines to calculate R A
=
Solving gives
R A = 163.0 N Ans.
R A
sin 50
90 40 = 50
Analyze the triangle
formed by the
left-hand side of the
parallelogram.
Angles are equal
7
11
sin (30 + 40)
Law of sines to calculate R C
=
Solving gives
R C = 184.3 N Ans.
R C
sin 60 sin (30 + 40)
200 N
12
89
10
R A
40
60
2.2 Rectangular Components in Two-Dimensional Force Systems
61
2.2 Rectangular Components in Two-Dimensional Force Systems Procedures and Strategies, page 1 of 1
F
x
y
Fy = F sin
Fx = F cos
x
y
5
12
13
F
Fx = 12F
13
Fy = 5F
13
Procedures and Strategies for Solving Problems Involving
Rectangular Components in Two-Dimensional Force
Systems
1. Two situations commonly arise in which the rectangular
components are to be computed.
a) The force is defined by its magnitude F and the angle
that the force makes with the positive x axis. In this case,
use the equations
Fx = F cos Fy = F sin
b) The force is defined by a magnitude F and a slope
triangle. The components can be computed by multiplying
the magnitude F by the ratios of the slope triangle the ratio
of the horizontal side to the hypotenuse gives the horizontal
component of the force. The ratio of the vertical side gives
the vertical component of the force.
2. To add forces simply
a) add all x components to obtain Rx, the x component of the
resultant R, and
b) add all y components to obtain Ry, the y component of the
resultant R.
The magnitude and direction of the resultant can be found
from a right triangle with R, Rx, and Ry as its sides.
R
x
y
Ry
Rx
R2 = Rx2 + Ry
2
= tan-1Rx
Ry
62
2.2 Rectangular Components in Two-Dimensional Force Systems Problem Statement for Example 1
y
x
1. Express the 5-kN force in terms of x and y components.
F = 5 kN
30
63
2.2 Rectangular Components in Two-Dimensional Force Systems Problem Statement for Example 2
2. Resolve the 20-lb force into x and y components.
x
y
F = 20 lb
3
4
64
2.2 Rectangular Components in Two-Dimensional Force Systems Problem Statement for Example 3
3. Express the 260-N force in terms of components
parallel and perpendicular to the inclined plane.
5
12
F = 260 N
65
2.2 Rectangular Components in Two-Dimensional Force Systems Problem Statement for Example 4
4. Determine the components of the 160-N force
perpendicular and parallel to the axis of the nail.
F = 160 N
20
15
66
2.2 Rectangular Components in Two-Dimensional Force Systems Problem Statement for Example 5
5. The connecting rod AB exerts a 2-kN force on the
crankshaft at B. Resolve this force into components
acting perpendicular to BC and along BC.
A
B
C
F = 2 kN
20
30
67
2.2 Rectangular Components in Two-Dimensional Force Systems Problem Statement for Example 6
6. Guy wire AB exerts a horizontal component of force of 0.5 kN
on the utility pole. Determine the total force from the wire acting
on the point of attachment, A. Assume that the force is directed
along the wire from A to B.
10 m
5 m
A
B
68
2.2 Rectangular Components in Two-Dimensional Force Systems Problem Statement for Example 7
7. If the vertical component of the force F
applied to the ring is 10 lb, determine the
magnitude F and also the horizontal component.
F
30
69
2.2 Rectangular Components in Two-Dimensional Force Systems Problem Statement for Example 8
8. The weight W is supported by the boom AB and
cable AC. Knowing that the horizontal and vertical
components of the cable force at A are 5 kN and 3 kN
as shown, determine the distance d.
d
C
B
10 m
F cable
5 kN
W
3 kN
A
70
2.2 Rectangular Components in Two-Dimensional Force Systems Problem Statement for Example 9
9. Determine the magnitude and direction of
the resultant force acting on the hook.
5
12
20 lb
104 lb
x
y
35
71
2.2 Rectangular Components in Two-Dimensional Force Systems Problem Statement for Example 10
10. Determine the magnitude and direction
of the resultant force acting on the beam.
40
8 kN15 kN
11 kN
3
4
72
2.2 Rectangular Components in Two-Dimensional Force Systems Problem Statement for Example 11
11. Determine the magnitude and direction
of the resultant force acting on the particle.
x
y
(5 m, 3 m)
25 N
80 N
50 N(6 m, 2 m)
( m, 6 m)
73
2.2 Rectangular Components in Two-Dimensional Force Systems Problem Statement for Example 12
12. Three forces support the weight W shown. Determine
the value of F, given that the resultant of the three forces
is vertical. Also determine the value of W.
40
W
30
15
120 NF
y
20 N
x
74
2.2 Rectangular Components in Two-Dimensional Force Systems Problem Statement for Example 13
13. The resultant, R, of the forces A and B acting on the
bracket is known to be a force of magnitude 300 lb
making an angle of 40 with the horizontal direction as
shown. Determine the magnitude of A and B.
x
y
7040
B
A
300 lb (resultant, R)
75
2.2 Rectangular Components in Two-Dimensional Force Systems Problem Statement for Example 14
14. To support the 100-N block as shown, the resultant of
the 50-N force and the force F must be a 100-N force
directed horizontally to the right. Determine F and .
60
100 N
F
50 N
Pulley
76
2.2 Rectangular Components in Two-Dimensional Force Systems Example 1, page 1 of 3
1
y
x
Construct a parallelogram with the
5-kN force as the diagonal and with
sides in the x and y directions.
1. Express the 5-kN force in terms of x and y components.
F = 5 kN
Because the x and y axes are
perpendicular, the parallelogram
is a special case a rectangle.
2
Analyze the triangle
forming the lower half of
the rectangle.
3
F x
F x
F y
F y
5 kN
y component, F y
x component, F x
F = 5 kN
30
30
30
77
2.2 Rectangular Components in Two-Dimensional Force Systems Example 1, page 2 of 3
4 Calculate F x from the definition of the cosine:
cos =
so
A = C cos
In words,
side adjacent (to angle) = hypotenuse
times cosine of angle.
(Memorize this you will use this relation many
times in a course in statics; you don't want to have to
think it out each time)
Applying this equation to the force triangle gives:
F x = (5 kN) cos 30
= 4.33 kN (1)
C
A
A
BC B
C
Similarly calculate F y from the definition of the
sine:
sin =
so
B = C sin
In words ,
side opposite (to angle) = hypotenuse
times sine of angle.
(Memorize this.)
Applying this equation to the force triangle gives:
F y = (5 kN) sin 30
= 2.50 kN (2)
5
78
2.2 Rectangular Components in Two-Dimensional Force Systems Example 1, page 3 of 3
6
y
x
Thus we have resolved the 5-kN force
into x and y components.
In terms of base vectors, the force is
F = 4.33i + 2.50j kN Ans.
7
The minus sign indicates that the x component
points in the negative x direction.8
F x = 4.33 kN
F y = 2.50 kNF = 5 kN
y
xi
j
30
79
2.2 Rectangular Components in Two-Dimensional Force Systems Example 2, page 1 of 4
1 Construct a parallelogram (rectangle)
with the 20-lb force as the diagonal.
2. Resolve the 20-lb force into x and y components.
Analyze the triangle forming the
lower half of the rectangle.
2
F x
F y
F y
F x
F = 20 lb
x
y
4
3
F = 20 lb
3
4
20 lb
F y
F x
Equal angles
3
4
80
2.2 Rectangular Components in Two-Dimensional Force Systems Example 2, page 2 of 4
4 From the "slope triangle," we see
cos (3)
sin
Note that we do not have to calculate ;
we already have what we need, sin and cos .
Side adjacent = hypotenuse cos
or
F x = (20 lb) cos (1)
Similarly, side opposite = hypotenuse sin
or
F y = (20 lb) sin (2)
3
3
4
F y
F x
54
35
32 + 42 = 5
81
2.2 Rectangular Components in Two-Dimensional Force Systems Example 2, page 3 of 4
5 Using Eq. 3 in Eq. 1 gives
F x = (20 lb) ( )
= 16 lb (5)
In general, then, get the horizontal force
component by multiplying the force by
the horizontal side of the slope triangle
divided by the hypotenuse (Memorize
this result; it is used frequently).
6
3
4
F x
54
5
F = 20 lb
Slope triangle
82
2.2 Rectangular Components in Two-Dimensional Force Systems Example 2, page 4 of 4
7 Similarly, using Eq. 4 in Eq. 2 gives
F y = (20 lb) ( )
= 12 lb (6)
In general, get the vertical force
component by multiplying the force by
the vertical side of the slope triangle
divided by the hypotenuse (Memorize
this result).
8
3
4
53
5
F = 20 lb
F y
x
y
16 lb
12 lb
F = 20 lb
j
i
9 Eqs. 5 and 6 now give the components in
terms of base vectors as
F = {16i + 12j} lb Ans.
83
2.2 Rectangular Components in Two-Dimensional Force Systems Example 3, page 1 of 2
1 Introduce an inclined x and y coordinate system.
3. Express the 260-N force in terms of components
parallel and perpendicular to the inclined plane.
5
12
F = 260 N
5
12
F = 260 N
x
y
F y
F y
F x
F x
12
5
Construct a parallelogram (rectangle) with the
260-N force as a diagonal.
2
84
2.2 Rectangular Components in Two-Dimensional Force Systems Example 3, page 2 of 2
x
y
Analyze the triangle forming the lower half
of the rectangle.3
F y
F x
52 + 122 = 13
5 12
13
F = 260 N
F y = (260 N)( ) = 240 N
F x = (260 N)( ) = 100 N4 135
1213
5
j
i
Representation in terms of components:
F = { 100i 240j} N Ans.
6
F x points in the negative x direction, and 7
F y points in the negative y direction.
85
2.2 Rectangular Components in Two-Dimensional Force Systems Example 4, page 1 of 3
1 Introduce an inclined x and y coordinate system.
4. Determine the components of the 160-N force
perpendicular and parallel to the axis of the nail.
F = 160 N
x
y
20
15
20
F = 160 N
15
86
2.2 Rectangular Components in Two-Dimensional Force Systems Example 4, page 2 of 3
2 Geometry
x
y
20
F = 160 N
Equal angles3
15Total angle
= 20 + 15
= 35
4
F = 160 N
y
x
F x
F y
35
5 Draw a parallelogram (rectangle)
with the 160-N force as a diagonal.
15
87
2.2 Rectangular Components in Two-Dimensional Force Systems Example 4, page 3 of 3
6 Analyze the triangle forming the
bottom half of the rectangle.
F = 160 N
y
x
35
7 In terms of base vectors,
F = { 131.1i 91.8j} N Ans.
F y = (160 N) sin 35
= 91.8 N
F x = (160 N) cos 35
= 131.1 N
15j
i
F y
88
2.2 Rectangular Components in Two-Dimensional Force Systems Example 5, page 1 of 3
5. The connecting rod AB exerts a 2-kN force on the
crankshaft at B. Resolve this force into components
acting perpendicular to BC and along BC.
A
B
C
F = 2 kN
20
30
89
2.2 Rectangular Components in Two-Dimensional Force Systems Example 5, page 2 of 3
2 kN
Introduce an inclined x and y
coordinate system.
1Calculate angles2
x
y
30
Equal3
4 Equal
Calculate the sum: 20 + 30 = 505
B
C
2 kN
x
y
C
B
2 kN
(2 kN) sin 50 = 1.532 kN(2 kN) cos 50 = 1.286 kN
Calculate components6
20
30
50
20
30
C
B
A
x
y
20
90
2.2 Rectangular Components in Two-Dimensional Force Systems Example 5, page 3 of 3
In terms of base vectors,
F = 1.532i 1.286j kN Ans.
7
j
y
xi30
91
2.2 Rectangular Components in Two-Dimensional Force Systems Example 6, page 1 of 3
6. Guy wire AB exerts a horizontal component of force of 0.5 kN
on the utility pole. Determine the total force from the wire acting
on the point of attachment, A. Assume that the force is directed
along the wire from A to B.
10 m
5 m
A
B
92
2.2 Rectangular Components in Two-Dimensional Force Systems Example 6, page 2 of 3
10 m
5 m
A
B
Express the guy-wire force F in
terms of rectangular components.
1
F
F y
F x = 0.5 kN
x
y
The horizontal component of
force is known to be 0.5 kN.
2
93
2.2 Rectangular Components in Two-Dimensional Force Systems Example 6, page 3 of 3
10 m
5 m
A
B
Relate F x to F through geometry.3
= tan-1 = 63.4345 m
10 m
AF x = 0.5 kN
F F y
0.5 kN = F cos
63.43
Solving gives
F = 1.118 kN Ans.
5
94
2.2 Rectangular Components in Two-Dimensional Force Systems Example 7, page 1 of 1
1 Express F in terms of rectangular components.
7. If the vertical component of the force F
applied to the ring is 10 lb, determine the
magnitude F and also the horizontal component.
x
y
F
30
30
F
F y
F x
2 Relate F to F y.
30F y = 10 lb
F x
F
10 lb = F sin 30
Therefore,
F = 20 lb Ans.
Relate F x to F.
F x = F cos 30
= (20 lb) cos 30
= 17.32 lb Ans.
3
95
2.2 Rectangular Components in Two-Dimensional Force Systems Example 8, page 1 of 2
8. The weight W is supported by the boom AB and
cable AC. Knowing that the horizontal and vertical
components of the cable force at A are 5 kN and 3 kN
as shown, determine the distance d.
d
C
B
10 m
F cable
5 kN
W
3 kN
A
1 Calculate the angle between F cable
and its horizontal component.
F cable
3 kN
5 kN
d
10 m
C
B
A
W
3 kN5 kN
= tan-1
= 30.96
3 kN
5 kN
F cable
96
2.2 Rectangular Components in Two-Dimensional Force Systems Example 8, page 2 of 2
C
A
d
10 m
5 kN
3 kN
The same result could also have been
obtained by using similar triangles.3
C
A
d
10 m
Use to calculate d.2
d = (10 m) tan 30.96
= 6.0 m Ans.
=
Therefore,
d = 6.0 m (same as before)
3 kN5 kN
d10 m
= 30.96
97
2.2 Rectangular Components in Two-Dimensional Force Systems Example 9, page 1 of 2
1 Express the forces in x and y components.
9. Determine the magnitude and direction of
the resultant force acting on the hook.
2 Calculate the x and y components of the resultant R by summing the
components of the given forces algebraically.
R x = F x: R
x = 16.38 lb + 96 lb = 112.38 lb
5
12
20 lb
104 lb
x
y
35
104 lb
12
20 lb
5
35
13
52 + 122 = 13
(20 lb) sin 35 = 11.47 lb
(20 lb) cos 35 = 16.38 lb
135
(104 lb)( ) = 40 lb
(104 lb)( ) = 96 lb1312
R y = F y: R
y = 11.47 lb 40 lb = 28.53 lb = 28.53 lb
(arrow indicates negative y direction)
+
+
98
2.2 Rectangular Components in Two-Dimensional Force Systems Example 9, page 2 of 2
3 Calculate the magnitude and direction of the resultant R.
x
y
R = (112.38 lb)2 + (28.53 lb)2
= 115.9 lb
= tan-1
= 14.2
112.38 lb28.53 lb
112.38 lb
28.53 lbR
14.2
115.9 lb
Ans.
99
2.2 Rectangular Components in Two-Dimensional Force Systems Example 10, page 1 of 2
1 Resolve the forces into x and y components.10. Determine the magnitude and direction
of the resultant force acting on the beam.
2 Calculate the x and y components of the resultant by summing the
components of the given forces algebraically.
R x = F x: R
x = 6.128 kN 12 kN = 5.872 kN = 5.872 kN
x
y
3
R y = F y: R
y = 5.142 kN 9 kN + 11 kN = 3.142 kN = 3.142 kN +
+
40
8 kN15 kN
11 kN
3
4 15 kN
4
11 kN
40
8 kN
35
(15 kN)( ) = 9 kN
(15 kN)( ) = 12 kN
5
54
(8 kN) sin 40 = 5.142 kN
(8 kN) cos 40 = 6.128 kN
Force points left
Force points down
100
2.2 Rectangular Components in Two-Dimensional Force Systems Example 10, page 2 of 2
3 Calculate the magnitude and direction of the resultant.
x
y
R = (5.872 kN)2 + (3.142 kN)2
= 6.66 kN
= tan-1
= 28.2
5.872 kN
3.142 kN
5.872 kN
3.142 kNR
28.2
6.66 kN
Ans.
101
2.2 Rectangular Components in Two-Dimensional Force Systems Example 11, page 1 of 5
11. Determine the magnitude and direction
of the resultant force acting on the particle.
x
y
(5 m, 3 m)
25 N
80 N
50 N(6 m, 2 m)
( m, 6 m)
102
2.2 Rectangular Components in Two-Dimensional Force Systems Example 11, page 2 of 5
1 We want to compute the x and y components of each
force. To do that, we first must compute some angles.
= tan-1
= 26.57
6 m3 m
80 N
y
x
3 m
6 m
Angle for
80-N force.
80 N
yComponents
of 80-N force.
x
= 26.57
(80 N) cos 26.57 = 71.55 N
(80 N) sin 26.57 = 35.78 N
( m, 6 m)
103
2.2 Rectangular Components in Two-Dimensional Force Systems Example 11, page 3 of 5
2 Angle and components for 50-N force.
= tan-1
= 30.96
5 m3 m
Angle for
50-N force
Components
of 50-N force.
= 30.96
(50 N) sin 30.96 = 25.72 N
(50 N) cos 30.96 = 42.88 N
50 N
x
y
5 m
3 m x
y
50 N
104
2.2 Rectangular Components in Two-Dimensional Force Systems Example 11, page 4 of 5
3 Angle and components for 25-N force.
= tan-1
= 18.43
6 m2 m
y
x
6 m
Angle for
25-N force.
y
Components
of 25-N force.
x
(25 N) cos 18.43 = 23.72 N
(25 N) sin 18.43 = 7.90 N
25 N
(6 m, 2 m)
25 N2 m
= 18.43
105
2.2 Rectangular Components in Two-Dimensional Force Systems Example 11, page 5 of 5
4 Sum the components algebraically.
5
R x = F x: R
x = 35.78 N 42.88 N + 23.72 = 16.62 N
+
+
y
xR y = F
y: R
y = 71.55 N 25.72 N 7.90 N = 37.93 N
35.78 N
42.88 N
23.72 N
71.55 N
25.72 N
7.90 N
Calculate the magnitude and direction of the resultant R.
R = (37.93 N)2 + (16.62)2
= 41.4 N
= tan-1
= 66.3
37.93 N
16.62 N
R
16.62 N
37.93 N
66.3
41.4 N
Ans.
106
2.2 Rectangular Components in Two-Dimensional Force Systems Example 12, page 1 of 3
1 Express the forces in x and y components (For clarity, the
components of the unknown force, F, are shown separately).
12. Three forces support the weight W shown. Determine
the value of F, given that the resultant of the three forces
is vertical. Also determine the value of W.
40
(20 N) sin 15 = 5.176 N
(20 N) cos 15 = 19.32 N
W
30
15
120 NF
20 N
x
120 N
y
W
F
y
20 N
x
y
x
(120 N) sin 30 = 60 N
(120 N) cos 30 = 103.92 N
F sin 40
F cos 40
W
30
15
40
107
2.2 Rectangular Components in Two-Dimensional Force Systems Example 12, page 2 of 3
3 Use the fact that the resultant is known to be
vertical, so R x = 0.
2 Sum the components algebraically.
R x = F x: R
x = 103.92 N 19.32 N F sin 40 (1)
R y = F y: R
y = 60 N + 5.176 N + F cos 40 (2) +
+
Eq. 1 becomes
R x = 103.92 N 19.32 N F sin 40
0
Solving gives
F = 131.61 N Ans.
x
y
W
R = R y
R x = 0
Substitute this value of F into Eq. 2 and compute R y:
R y = 60 N + 5.176 N + F cos 40 = 166.0 N
131.61 N
4
108
2.2 Rectangular Components in Two-Dimensional Force Systems Example 12, page 3 of 3
5 The resultant upward force must balance the weight W, so
W = R y = 166.0 N Ans.
W
R = R y = 166.0 N
109
2.2 Rectangular Components in Two-Dimensional Force Systems Example 13, page 1 of 2
1 Express the forces in x and y components.
70
B
y
40
300 lb
Ax
B sin 70
B cos 70
(300 lb) sin 40 = 192.84 lb R y (y component of 300-lb resultant)
(300 lb) cos 40 = 229.81 lb R x (x component of 300-lb resultant)
300 lb (resultant, R)
A
B
4070
y
x
13. The resultant, R, of the forces A and B acting on the
bracket is known to be a force of magnitude 300 lb
making an angle of 40 with the horizontal direction as
shown. Determine the magnitude of A and B.
110
2.2 Rectangular Components in Two-Dimensional Force Systems Example 13, page 2 of 2
4 Solving Eqs. 1 and 2 simultaneously gives:
A = 300 lb Ans.
B = 205 lb Ans.
2 R x is the algebraic sum of x components of A and B:
R x = F x: R
x = A B cos 70 (1)
229.81 lb
Similarly for R y:
R y = F y: R
y = B sin 70 (2)
192.84 lb
+
+
3
111
2.2 Rectangular Components in Two-Dimensional Force Systems Example 14, page 1 of 2
1 Express the forces in x and y components.
14. To support the 100-N block as shown, the resultant of
the 50-N force and the force F must be a 100-N force
directed horizontally to the right. Determine F and .
60
100 N
F
50 N
Pulley
F
60
50 N
x
y
F sin F cos
(50 N) cos 60 = 25 N
(50 N) sin 60 = 43.30 N
112
2.2 Rectangular Components in Two-Dimensional Force Systems Example 14, page 2 of 2
2 Sum the components algebraically.
R x = F x: R
x = F cos 25 N (1)
R y = F
y: R
y = F sin 43.30 N (2)
Because the resultant is known to be horizontal,
R y = 0, and the magnitude R is thus equal to the
horizontal component R x alone, that is, R = R x.
We also know, however, that the magnitude of
the resultant is 100 N, so R = R x = 100 N. Thus
Eqs. 1 and 2 become
100 N = F cos + 25 N (3)
0 = F sin 43.30 N (4)
The best way to solve these equations is to use a
calculator that can solve two simultaneous
nonlinear equations. Alternatively, solve Eq. 4
for F:
F = (5)
+
+
sin
43.30 N
Ans.
86.6 N
30
cos
sin
43.30 N
sin tan 1
tan
43.30 N
43.30 N
sin
3 And then substitute for F in Eq. 3:
100 N = F cos + 25 N
Substituting
=
gives
100 N = + 25 N
and solving gives
= 30.0
Using the result in Eq. 5 gives
F = = 86.6 N
113
2.3 Rectangular Components in Three-Dimensional Force Systems
114
2.3 Rectangular Components in Three-Dimensional Force Systems Procedures and Strategies, page 1 of 1
y
x
z
y
z x
F
Fx
y
x
z
FF cos
(F sin ) cos
F sin
(F sin ) sin
Procedures and Strategies for Solving Problems Involving
Rectangular Components in Three-Dimensional Force Systems
1. If the magnitude F of a force and its direction angles, x y, and z, are
known, then compute the components of the force from the equations
Fx = F cos x Fy = F cos y Fz = F cos z
If only two angles are known, then find the third angle from the equation
cos2 x + cos2 y + cos
2 z = 1
2. If the magnitude F is known and the direction of the force is defined
through its projection on a horizontal plane, then compute the horizontal
components by projecting the projection onto the horizontal axes.
3. If the rectangular components Fx, Fy, and Fz are known, then compute
the magnitude F of the force from the equation
F = Fx2 + Fy
2 + Fz2
and the direction angles from
cos x = Fx/F cos y = Fy/F cos z = Fz/F
4. To compute the resultant of several force, express each force in
rectangular component and add the components:
Rx = Fx Ry = Fy Rz = Fz
115
2.3 Rectangular Components in Three-Dimensional Force Systems Problem Statement. for Example 1
F = 200 lb
O
z
y
x
60
45
120
1. Express the force F in terms of x, y, and z components.
116
2.3 Rectangular Components in Three-Dimensional Force Systems Problem Statement for Example 2
2. Express F in terms of x, y, and z components.
x
y
z
A
B
O
F = 50 N
40
35
117
2.3 Rectangular Components in Three-Dimensional Force Systems Problem Statement. for Example 3
3. Express F in terms of x, y, and z components.
x
y
z
A
B
O
F = 8 kN
70
25
118
2.3 Rectangular Components in Three-Dimensional Force Systems Problem Statement. for Example 4
4. Determine the x, y, and z components of the 26-N force shown.
Also determine the coordinate direction angles of the force.
135
12
20
F = 26 N
O
B
A
z
y
x
119
2.3 Rectangular Components in Three-Dimensional Force Systems Problem Statement. for Example 5
30
20
80x
y
z
O
F 1 = 100 N
F 2 = 60 N
F 3 = 40 N
5. Determine the magnitude and coordinate direction angles
of the resultant of the three forces acting on the mast.
120
2.3 Rectangular Components in Three-Dimensional Force Systems Problem Statement. for Example 6
6. Determine the magnitude and coordinate direction angles
of the resultant of the forces acting on the eye-bolt.
x
y
z
O
F 1 = 650 N
F 2 = 800 N
F 3 = 300 N
30
70 50
12
513
121
2.3 Rectangular Components in Three-Dimensional Force Systems Problem Statement. for Example 7
7. A 300-lb vertical force is required to pull the pipe out of the ground. Determine
the magnitude and direction angles of the force F 2 which, when applied together
with the 150-lb force F 1 shown, will produce a 300-lb vertical resultant.
F 2
F 1 = 150 lb
O
z
y
x
60
60
45
122
2.3 Rectangular Components in Three-Dimensional Force Systems Problem Statement. for Example 8
8. Two forces, F 1 and F 2 act on the bracket as shown. If the
resultant of F 1 and F 2 lies in the xy plane, determine the
magnitude of F 2. Also determine the magnitude of the resultant.
50
60
x
y
z
O
F 2
F 1 = 60 N
123
2.3 Rectangular Components in Three-Dimensional Force Systems Example 1, page 1 of 2
View of plane formed by the x axis and F.1
O x
F = 200 lb
The component points in the negative direction, so
F x = 100 lb Ans.
5
d
4
3
Ox
F = 200 lb
Calculate the x component.2
= 180 120 = 60
d = (200 lb) cos 60 = 100 lb
120
120
F = 200 lb
O
z
y
x
60
45
120
1. Express the force F in terms of x, y, and z components.
124
2.3 Rectangular Components in Three-Dimensional Force Systems Example 1, page 2 of 2
F = 200 lb
O
View of plane formed by the y axis and F6
7
F y
View of plane formed by the z axis and F.8
F = F xi + F yj + F
zk
= { 100i + 141.4j + 100k} lb Ans.
y component
F y = (200 lb) cos 45 = 141.4 lb Ans.
z component
F z = (200 lb) cos 60 = 100 lb Ans.
10
y
z
45
60
F = 200 lb
O
9
Fz
125
2.3 Rectangular Components in Three-Dimensional Force Systems Example 2, page 1 of 2
2. Express F in terms of x, y, and z components.
x
y
z
A
B
O
F = 50 N
View of plane formed by OA, F, and the y axis.
F y = (50 N) cos 40 = 38.3 N Ans.
F OA = (50 N) sin 40 = 32.14 N
O
y
A
F = 50 N
F y
1
2
3
40
40
F OA
35
126
2.3 Rectangular Components in Three-Dimensional Force Systems Example 2, page 2 of 2
View of xz plane from above
Oyx
z
BF x
F BA32.14 N
F BA = (32.14 N) sin 35 = 18.4 N
F x = (32.14 N) cos 35 = 26.3 N Ans.
4
6
5
F z = 18.4 N7
negative direction
F = F xi + F yj + F
zk
= {26.3i + 38.3j 18.4k} N Ans.
8
A
35
127
2.3 Rectangular Components in Three-Dimensional Force Systems Example 3, page 1 of 2
3. Express F in terms of x, y, and z components.
x
y
z
A
B
O
F = 8 kN
70
F = 8 kN
O
y
AF OA
F y
View of the plane formed by OA, F, and the y axis.
F OA = (8 kN) sin 70 = 7.518 kN
(8 kN) cos 70 = 2.74 kN
F y = 2.74 kN Ans.
Negative direction
1
2
3
4
70
25
128
2.3 Rectangular Components in Three-Dimensional Force Systems Example 3, page 2 of 2
7.518 kN
O
yB
F x
View of the xz plane from above.
F x = (7.518 kN) cos 25 = 6.81 kN
F z = (7.518 kN) sin 25
= 3.18 kN Ans.
z
x
F z
5
6
7
F = F xi + F yj + F
zk
= {6.81i 2.74j + 3.18k} kN Ans.
8
25
129
2.3 Rectangular Components in Three-Dimensional Force Systems Example 4, page 1 of 4
4. Determine the x, y, and z components of the 26-N force shown.
Also determine the coordinate direction angles of the force.
View of the plane formed by OA, F, and the y axis.
F OA = (26 N)( )
= 24 N
1
2
3OA
F = 26 N
y
F OA
F y
12 13
F y = (26 N)( )
= 10 N Ans.
5 13
512
13
135
12
20
F = 26 N
O
B
A
z
y
x
130
2.3 Rectangular Components in Three-Dimensional Force Systems Example 4, page 2 of 4
xy, O
z
A
24 N
F z = (24 N) cos 20 = 22.6 N Ans.
F x
(24 N) sin 20 = 8.21 N
F x = 8.21 N Ans.
Negative direction
4
5
6
View of the xz plane as seen from above
OA
zB
F = 26 N
y
x
F x = 8.21 N
Determine the x coordinate direction angle, .
View of the plane formed by the x axis and the force F.
7
8
O
26 N
8.21 Nx
The direction angle is measured
from the the positive part of the
axis. Here it is , not .
= 180
= 180 cos-1
= 180 71.59
= 108.4 Ans.
9
10
8.21 N 26 N
20
F z
131
2.3 Rectangular Components in Three-Dimensional Force Systems Example 4, page 3 of 4
OA
zB
F = 26 N
y
x
F y = 10 N
Determine the y coordinate direction angle, .
View of the plane formed by the y axis and the force F.
O
26 N
= cos-1
= 67.4 Ans.
10 N 26 N
11
12
10 N
y
13
132
2.3 Rectangular Components in Three-Dimensional Force Systems Example 4, page 4 of 4
OA
zB
F = 26 N
y
x
F z = 22.6 N
Determine the z coordinate direction angle, . View of the plane formed by the z axis and the force F.
O
26 N = cos-1
= 29.6 Ans.
22.6 N 26 N
14 15
16
22.6 N
z
Observation: the calculations for , , and can be
summarized by the general formulas
cos =
cos =
cos =
The algebraic signs of F x, F y, and F
z must be included
when using these formulas.
17
F x
F
F y
F
F z
F
133
2.3 Rectangular Components in Three-Dimensional Force Systems Example 5, page 1 of 3
5. Determine the magnitude and coordinate direction angles
of the resultant of the three forces acting on the mast.
30
20
80x
y
z
O
F 1 = 100 N
F 2 = 60 N
F 3 = 40 N
x
y
F 1 = 100 N
z20
O
Express F 1 in rectangular components.
(100 N) sin 80 = 98.48 N
F1x = (98.48 N) sin 20 = 33.68 N
F1z = (98.48 N) cos 20 = 92.54 N
F1y
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