HibBeLer Worked Examples

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 [email protected].

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


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


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


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


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


4. Express the following forces in pounds (lb).

a) 1,230 kip

b) 0.0230 kip

c) 23.0 oz

6


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


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


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


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


Round to the closer value:

3.6143 10-3 in. 3.61 10-3 in. Ans.

3.6143 10-3

10


9

10

d) 6.875 lies precisely halfway between 6.870 and 6.880.

6.870 6.875

(midpoint

of interval)

6.880


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


14

e) 6.885 lies precisely halfway between 6.880 and 6.890.

6.880 6.885

(midpoint

of interval)

6.890


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

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


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


11 Exponent of 10 is a multiple of 3.

14


d) 0.0987 10-5 m = 0.987 10-6 m

= 0.987 m Ans.

Prefix " " = "micro" = 10-6 14

17 Coefficient lies


19

18

Prefix "G" = "giga" = 109 .


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


23

22

Prefix "n" = "nano" = 10-9 .


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


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


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


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


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


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


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


y

x

10

20



3 kN

2 kN

28


110

100 N

80 N40

3. Determine the magnitude and direction of the resultant force.

29


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


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


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


25

40

v

u

120 lb

7. Resolve the 120-lb force into components

acting in the u and v directions.

33


8. Resolve the 4-kN

horizontal force into

components along truss

members AB and AC.

35

5

12

A

B C

4 kN

34


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



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


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


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


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




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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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. Resolve the 20-lb force into x and y components.

x

y

F = 20 lb

3

4

64


3. Express the 260-N force in terms of components

parallel and perpendicular to the inclined plane.

5

12

F = 260 N

65


4. Determine the components of the 160-N force

perpendicular and parallel to the axis of the nail.

F = 160 N

20

15

66


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


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


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


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


9. Determine the magnitude and direction of

the resultant force acting on the hook.

5

12

20 lb

104 lb

x

y

35

71



of the resultant force acting on the beam.

40

8 kN15 kN

11 kN

3

4

72



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


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


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


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


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


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


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


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


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


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


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


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


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 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


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


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 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


In terms of base vectors,

F = 1.532i 1.286j kN Ans.

7

j

y

xi30

91


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


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


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


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


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


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


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


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


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


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



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


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 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


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


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


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


3 Use the fact that the resultant is known to be

vertical, so R x = 0.


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


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



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


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



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



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


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



x

y

z

A

B

O

F = 8 kN

70

25

118


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


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



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


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


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


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



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


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



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


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


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


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


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


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



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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