Lecture Notes on Engineering Statics.

Engineering Mechanics
Statics
Supported with MATLAB Codes
Dr. Ahmed Momtaz Hosny
PhD in Aircraft Dynamics and Control, BUAA
Lecturer at KMA
Lecture Notes & Solved Examples with MATLAB Applications

List of Contents
Chapter 1 Introduction to Mechanics
Chapter 2
Chapter 3
Chapter 4
Chapter 5
Chapter 6
Statics of Particles
Statics of Rigid Bodies, Equivalent Systems of Forces
Equilibrium of Rigid Bodies
Distributed Forces, Centroids and Center of Gravity
Analysis of Structures, Simple Truss (Pin Joints & Two-Forces Members)
Chapter 7 Shear and Bending Moment of Beams
Chapter 8 Friction, Pulleys, Tension Belts and Journal Bearings

Chapter 1
Introduction to Mechanics
1.1 WHAT IS MECHANICS?
Mechanics can be defined as that science which describes and predicts the
conditions of rest or motion of bodies under the action of forces. It is divided into
three parts: mechanics of rigid bodies, mechanics of deformable bodies, and
mechanics of fluids. The mechanics of rigid bodies is subdivided into statics and
dynamics, the former dealing with bodies at rest, the latter with bodies in motion.
In this part of the study of mechanics, bodies are assumed to be perfectly rigid.
Actual structures and machines, however, are never absolutely rigid and deform
under the loads to which they are subjected. But these deformations are usually
small and do not appreciably affect the conditions of equilibrium or motion of the
structure under consideration. They are important, though, as far as the resistance
of the structure to failure is concerned and are studied in mechanics of materials,
which is a part of the mechanics of deformable bodies. The third division of
mechanics, the mechanics of fluids, is subdivided into the study of incompressible
fluids and of compressible fluids. An important subdivision of the study of
incompressible fluids is hydraulics, which deals with problems involving water.
Mechanics is a physical science, since it deals with the study of physical
phenomena. However, some associate mechanics with mathematics, while many
consider it as an engineering subject. Both these views are justified in part.
Mechanics is the foundation of most engineering sciences and is an indispensable
prerequisite to their study. However, it does not have the empiricism found in some
engineering sciences, i.e., it does not rely on experience or observation alone; by
its rigor and the emphasis it places on deductive reasoning it resembles
mathematics. But, again, it is not an abstract or even a pure science; mechanics is
an applied science. The purpose of mechanics is to explain and predict physical
phenomena and thus to lay the foundations for engineering applications.
1

1.2 SYSTEMS OF UNITS
With the four fundamental concepts introduced in the preceding section are
associated the so-called kinetic units, i.e., the units of length, time, mass, and force.
International System of Units (SI Units). In this system, which will be in
universal use after the United States has completed its conversion to SI units, the
base units are the units of length, mass, and time, and they are called, respectively,
the meter (m), the kilogram (kg), and the second (s). All three are arbitrarily
defined. The unit of force is a derived unit. It is called the newton (N) and is
defined as the force which gives an acceleration of 1 m/s 2 to a mass of 1 kg.
1 N = (1 kg)(1 m/s2
)
Multiples and submultiples of the fundamental SI units may be obtained through
the use of the prefixes defined in Table 1.1 . The multiples and submultiples of the
units of length, mass, and force most frequently used in engineering are,
respectively, the kilometer (km) and the millimeter (mm); the megagram (Mg) and
the gram (g); and the kilonewton (kN). According to Table 1.1 ,
2

Other derived SI units used to measure the moment of a force, the work of a force,
etc., are shown in Table 1.2 . While these units will be introduced in later chapters
as they are needed.
U.S. Customary Units. Most practicing American engineers still commonly use a
system in which the base units are the units of length, force, and time. These units
are, respectively, the foot (ft), the pound (lb), and the second (s).
The U.S. customary units most frequently used in mechanics are listed in Table
1.3 with their SI equivalents.
3

Chapter 2
Statics of Particles
2.1 INTRODUCTION
In this chapter you will study the effect of forces acting on particles. First you will
learn how to replace two or more forces acting on a given particle by a single force
having the same effect as the original forces. This single equivalent force is the
resultant of the original forces acting on the particle. Later the relations which
exist among the various forces acting on a particle in a state of equilibrium will be
derived and used to determine some of the forces acting on the particle. The use of
the word “particle” does not imply that our study will be limited to that of small
corpuscles. What it means is that the size and shape of the bodies under
consideration will not significantly affect the solution of the problems treated in
this chapter and that all the forces acting on a given body will be assumed to be
applied at the same point. Since such an assumption is verified in many practical
applications, you will be able to solve a number of engineering problems in this
chapter. The first part of the chapter is devoted to the study of forces contained in a
single plane, and the second part to the analysis of forces in three-dimensional
space.
2.2 FORCE ON A PARTICLE. RESULTANT OF TWO FORCES
A force represents the action of one body on another and is generally characterized
by its point of application, its magnitude, and its direction. Forces acting on a
given particle, however, have the same point of application. Each force considered
in this chapter will thus be completely defined by its magnitude and direction. The
magnitude of a force is characterized by a certain number of units. As indicated in
Chap. 1, the SI units used by engineers to measure the magnitude of a force are the
newton (N) and its multiple the kilonewton (kN), equal to 1000 N, while the U.S.
customary units used for the same purpose are the pound (lb) and its multiple the
kilopound (kip), equal to 1000 lb. The direction of a force is defined by the line of
action and the sense of the force. The line of action is the infinite straight line
along which the force acts; it is characterized by the angle it forms with some fixed
5

axis (Fig. 2.1). The force itself is represented by a segment of that line; through the
use of an appropriate scale, the length of this segment may be chosen to represent
the magnitude of the force.
Fig. 2.1
Finally, the sense of the force should be indicated by an arrowhead. It is important
in defining a force to indicate its sense. Two forces having the same magnitude and
the same line of action but different sense, such as the forces shown in Fig. 2.1a
and b, will have directly opposite effects on a particle. Experimental evidence
shows that two forces P and Q acting on a particle A (Fig. 2.2a) can be replaced by
a single force R which has the same effect on the particle (Fig. 2.2c). This force is
called the resultant of the forces P and Q and can be obtained, as shown in Fig.
2.2b, by constructing a parallelogram, using P and Q as two adjacent sides of the
parallelogram. The diagonal that passes through A represents the resultant. This
method for finding the resultant is known as the parallelogram law for the addition
of two forces. This law is based on experimental evidence; it cannot be proved or
derived mathematically.
Fig. 2.2
6

2.3 VECTORS
It appears from the above that forces do not obey the rules of addition defined in
ordinary arithmetic or algebra. For example, two forces acting at a right angle to
each other, one of 4 lb and the other of 3 lb, add up to a force of 5 lb, not to a force
of 7 lb. Forces are not the only quantities which follow the parallelogram law of
addition. As you will see later, displacements, velocities, accelerations, and
momenta are other examples of physical quantities possessing magnitude and
direction that are added according to the parallelogram law. All these quantities
can be represented mathematically by vectors, while those physical quantities
which have magnitude but not direction, such as volume, mass, or energy, are
represented by plain numbers or scalars. Vectors are defined as mathematical
expressions possessing magnitude and direction, which add according to the
parallelogram law. Vectors are represented by arrows in the illustrations and will
be distinguished from scalar quantities in this text through the use of boldface type
(P). In longhand writing, a vector may be denoted by drawing a short arrow above
the letter used to represent it (𝑷𝑷��⃗) or by underlining the letter (P)
Two vectors which have the same magnitude and the same direction are said to be
equal, whether or not they also have the same point of application (Fig. 2.4); equal
vectors may be denoted by the
. The last method
may be preferred since underlining can also be used on a typewriter or computer.
The magnitude of a vector defines the length of the arrow used to represent the
vector. In this text, italic type will be used to denote the magnitude of a vector.
Thus, the magnitude of the vector P will be denoted by P. A vector used to
represent a force acting on a given particle has a well-defined point of application,
namely, the particle itself. Such a vector is said to be a fixed, or bound, vector and
cannot be moved without modifying the conditions of the problem. Other physical
quantities, however, such as couples (see Chap. 3), are represented by vectors
which may be freely moved in space; these vectors are called free vectors. Still
other physical quantities, such as forces acting on a rigid body (see Chap. 3), are
represented by vectors which can be moved, or slid, along their lines of action;
they are known as sliding vectors.
same letter.
7

The negative vector of a given vector P is defined as a vector having the same
magnitude as P and a direction opposite to that of P (Fig. 2.5); the negative of the
vector P is denoted by -P. The vectors P and -P are commonly referred to as equal
and opposite vectors. Clearly, we have
P + (-P) = 0
2.4 ADDITION OF VECTORS
We saw in the preceding section that, by definition, vectors add according to the
parallelogram law. Thus, the sum of two vectors P and Q is obtained by attaching
the two vectors to the same point A and constructing a parallelogram, using P and
Q as two sides of the parallelogram (Fig. 2.6). The diagonal that passes through A
represents the sum of the vectors P and Q, consequently we can write
R = P + Q = Q + P
8

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
3
PROBLEM 2.1
Two forces are applied at point B of beam AB. Determine
graphically the magnitude and direction of their resultant using
(a) the parallelogram law, (b) the triangle rule.
SOLUTION
(a) Parallelogram law:
(b) Triangle rule:
We measure: 3.30 kN, 66.6R α= = ° 3.30 kN=R 66.6° 
9
2.1
Solve the above problem replacing 2 kN force with 5 kN and 3 kN force with 10 kN.

4
PROBLEM 2.2
The cable stays AB and AD help support pole AC. Knowing
that the tension is 120 lb in AB and 40 lb in AD, determine
graphically the magnitude and direction of the resultant of the
forces exerted by the stays at A using (a) the parallelogram
law, (b) the triangle rule.
SOLUTION
We measure: 51.3
59.0
α
β
= °
= °
(b) Triangle rule:
We measure: 139.1 lb,R = 67.0γ = ° 139.1lbR = 67.0° 

10
2.2

5
PROBLEM 2.3
Two structural members B and C are bolted to bracket A. Knowing that both
members are in tension and that P = 10 kN and Q = 15 kN, determine
graphically the magnitude and direction of the resultant force exerted on the
bracket using (a) the parallelogram law, (b) the triangle rule.
SOLUTION
(b) Triangle rule:
We measure: 20.1 kN,R = 21.2α = ° 20.1kN=R 21.2° 
11
2.3

6
PROBLEM 2.4
Two structural members B and C are bolted to bracket A. Knowing that
both members are in tension and that P = 6 kips and Q = 4 kips, determine
graphically the magnitude and direction of the resultant force exerted on
the bracket using (a) the parallelogram law, (b) the triangle rule.
SOLUTION
(b) Triangle rule:
We measure: 8.03 kips, 3.8R α= = ° 8.03 kips=R 3.8° 
12
2.4

7
PROBLEM 2.5
A stake is being pulled out of the ground by means of two ropes as shown.
Knowing that α = 30°, determine by trigonometry (a) the magnitude of the
force P so that the resultant force exerted on the stake is vertical, (b) the
corresponding magnitude of the resultant.
SOLUTION
Using the triangle rule and the law of sines:
(a)
120 N
sin30 sin 25
P
=
° °
101.4 NP = 
(b) 30 25 180
180 25 30
125
β
β
° + + ° = °
= ° − ° − °
= °
120 N
sin30 sin125
=
° °
R
196.6 N=R 
13
2.5

8
PROBLEM 2.6
A trolley that moves along a horizontal beam is acted upon by two
forces as shown. (a) Knowing that α = 25°, determine by trigonometry
the magnitude of the force P so that the resultant force exerted on the
trolley is vertical. (b) What is the corresponding magnitude of the
resultant?
SOLUTION
(a)
1600 N
sin 25° sin 75
P
=
°
3660 NP = 
(b) 25 75 180
180 25 75
80
β
β
° + + ° = °
= ° − ° − °
= °
1600 N
sin 25° sin80
R
=
°
3730 NR = 
14
Solve the above problem knowing that P = 1600 N and alfa angle = 110 Degrees.
2.6

13
PROBLEM 2.11
A steel tank is to be positioned in an excavation. Knowing that
α = 20°, determine by trigonometry (a) the required magnitude
of the force P if the resultant R of the two forces applied at A is
to be vertical, (b) the corresponding magnitude of R.
SOLUTION
(a) 50 60 180
180 50 60
70
β
β
+ ° + ° = °
= ° − ° − °
= °
425 lb
sin 70 sin 60
P
=
° °
392 lbP = 
(b)
425 lb
sin 70 sin50
R
=
° °
346 lbR = 
2.7
15

14
PROBLEM 2.12
A steel tank is to be positioned in an excavation. Knowing that
the magnitude of P is 500 lb, determine by trigonometry (a) the
required angle α if the resultant R of the two forces applied at A
is to be vertical, (b) the corresponding magnitude of R.
SOLUTION
(a) ( 30 ) 60 180
180 ( 30 ) 60
90
sin (90 ) sin60
425 lb 500 lb
α β
β α
β α
α
+ ° + ° + = °
= ° − + ° − °
= ° −
° − °
=
90 47.402α° − = ° 42.6α = ° 
(b)
500 lb
sin (42.598 30 ) sin 60
R
=
° + ° °
551 lbR = 
16
2.8

%%%%%%%%%Problem 2.12 How to calculate angle Alfa for different values of P
%%%%%%%%%Considering Vertical Resultant R and Constant Force of 425 with
%%%%%%%%%angle 30 degree
%% Calculating the regular problem
%% Beta=180 - (Alfa+30)- 60 ie Beta= 90- Alfa ;
%% Substituting in Sines Law sin(Beta)/425 = sin(pi/3)/500;
B =(sin(pi/3)/500)*425;
Beta = asin(B)
Beta = Beta * 57.3 %%%% in Radians
Alfa=90 - Beta
%%%%%%%%% Calculating the resultant Force
R=(500/sin(pi/3))* sin((Alfa+30)/57.3)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Performing a loop calculation for P=[500 550 600 650 700 750 800 850 900]
P=[500 550 600 650 700 750 800 850 900]
for I=0:1:8
I=I+1
B =(sin(pi/3)/P(I))*425;
Beta = asin(B)
Beta = Beta * 57.3 %%%% in Degrees
Beta1(I)=Beta
Alfa(I)=90 - Beta
R(I)=(P(I)/sin(pi/3))* sin((Alfa(I)+30)/57.3)
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Plotting Alfa and Beta for Different P
17

figure(1);
plot(P,Alfa);
xlabel('P in LB'),grid on;;ylabel('Alfa in Degrees'),grid on;;
figure(2);
plot(P,Beta1);
xlabel('P in LB'),grid on;;ylabel('Beta in Degrees'),grid on;;
figure(3);
plot(P,R);
xlabel('P in LB'),grid on;;ylabel('R Resultant'),grid on;;
figure(4);
plot(P,Alfa,'+',P,Beta1,'o');
xlabel('P in LB'),grid on;;ylabel('Alfa Beta in Degrees'),grid on;;
18

17
PROBLEM 2.15
Solve Problem 2.2 by trigonometry.
PROBLEM 2.2 The cable stays AB and AD help support pole
AC. Knowing that the tension is 120 lb in AB and 40 lb in AD,
determine graphically the magnitude and direction of the
resultant of the forces exerted by the stays at A using (a) the
parallelogram law, (b) the triangle rule.
SOLUTION
8
tan
10
38.66
6
tan
10
30.96
α
α
β
β
=
= °
=
= °
Using the triangle rule: 180
38.66 30.96 180
110.38
α β ψ
ψ
ψ
+ + = °
° + ° + = °
= °
Using the law of cosines: 2 22
(120 lb) (40 lb) 2(120 lb)(40 lb)cos110.38
139.08 lb
R
R
= + − °
=
Using the law of sines:
sin sin110.38
40 lb 139.08 lb
γ °
=
15.64
(90 )
(90 38.66 ) 15.64
66.98
γ
φ α γ
φ
φ
= °
= ° − +
= ° − ° + °
= ° 139.1 lb=R 67.0° 
2.9
20

%%%%%%%%%Problem 2.15 How to calculate Resultant R and its angle from the
%%%%%%%%%horizontal plane
%% tan(Alfa)=8/10 & tan(Beta)=6/10 then we can get Alfa & Beta;
Alfa=atan(8/10)*57.3
Beta=atan(6/10)*57.3
Epsai = 180 - Alfa - Beta
%%%%% Calculating the resultant from the Law of Cosines
R = ((120^2) + (40^2) - 2*40*120*cos(Epsai/57.3))^0.5
%%%%% Calculating the angle Gama to obtain angle Fai as shown in the figure
%%%%% sin(Gama)/40=sin(Epsai)/R consequently we can obtain sin(Gama) in
%%%%% form of G
G=(sin(Epsai/57.3)/R)*40
Gama=asin(G)*57.3 %%%%%%% in Degrees
Fai=90-Alfa + Gama
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Performing a loop calculation for P=[500 550 600 650 700 750 800 850 900]
T1=[100 120 140 160 180 200 220 240 260];
T2=[30 40 50 60 70 80 90 100 110];
for I=0:1:8
I=I+1;
%%%%% Calculating the resultant from the Law of Cosines
R(I) = ((T1(I)^2) + (T2(I)^2) - 2*T1(I)*T2(I)*cos(Epsai/57.3))^0.5;
%%%%% Calculating the angle Gama to obtain angle Fai as shown in the figure
%%%%% sin(Gama)/40=sin(Epsai)/R consequently we can obtain sin(Gama) in
%%%%% form of G
G=(sin(Epsai/57.3)/R(I))*40;
Gama=asin(G)*57.3; %%%%%%% in Degrees
Fai(I)=90-Alfa + Gama;
21

end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Plotting Resultant R and the Corresponding Angle Fai for different T1 & T2
figure(1);
plot3(T1,T2,R);
xlabel('T1 in LB'), grid on;ylabel('T2 in LB'), grid on;zlabel('R in LB'),
grid on;
figure(2);
plot(R,Fai);
xlabel('R in LB'), grid on;ylabel('Fai in Degrees'), grid on;
22

18
PROBLEM 2.16
Solve Problem 2.4 by trigonometry.
PROBLEM 2.4 Two structural members B and C are bolted to bracket A.
Knowing that both members are in tension and that P = 6 kips and Q = 4 kips,
determine graphically the magnitude and direction of the resultant force
exerted on the bracket using (a) the parallelogram law, (b) the triangle rule.
SOLUTION
Using the force triangle and the laws of cosines and sines:
We have: 180 (50 25 )
105
γ = ° − ° + °
= °
Then 2 2 2
2
(4 kips) (6 kips) 2(4 kips)(6 kips)cos105
64.423 kips
8.0264 kips
R
R
= + − °
=
=
And
4 kips 8.0264 kips
sin(25 ) sin105
sin(25 ) 0.48137
25 28.775
3.775
α
α
α
α
=
° + °
° + =
° + = °
= °
8.03 kips=R 3.8° 
2.10
23

31
PROBLEM 2.29
The guy wire BD exerts on the telephone pole AC a force P directed along
BD. Knowing that P must have a 720-N component perpendicular to the
pole AC, determine (a) the magnitude of the force P, (b) its component
along line AC.
SOLUTION
(a)
37
12
37
(720 N)
12
2220 N
=
=
=
xP P
2.22 kNP = 
(b)
35
12
35
(720 N)
12
2100 N
y xP P=
=
=
2.10 kN=yP 
24
2.11

%%%%%%%%%Problem 2.29 How to calculate Resultant P by knowing one component
%%%%%%%%%Px = 720 N as shown in figure
Px = 720;
%% Taking into consideration that Px = P*sin(theta)where theta = the angle
%% between Py and P -----> sin(theta)= 2.4/(2.4^2 + 7^2)^0.5
s= 2.4/(2.4^2 + 7^2)^0.5;
P = Px/s
%% Taking into consideration that Py = P*cos(theta)where theta = the angle
%% between Py and P -----> cos(theta) = 7/(2.4^2 + 7^2)^0.5
Py = P*(7/(2.4^2 + 7^2)^0.5)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Performing a loop calculation with Length L (from point D to point C
%% varies from L=[1 2.4 4 6 8 10 12 14 16 18]) therefore calculating the
%% corresponding P & Py
L=[1 2.4 4 6 8 10 12 14 16 20 30 40];
for I=0:1:11
I=I+1;
%% Taking into consideration that Px = P*sin(theta)where theta = the angle
%% between Py and P -----> sin(theta)= L(I)/(L(I)^2 + 7^2)^0.5
s= L(I)/(L(I)^2 + 7^2)^0.5;
P(I) = Px/s;
%% Taking into consideration that Py = P*cos(theta)where theta = the angle
%% between Py and P -----> cos(theta) = 7/(L(I)^2 + 7^2)^0.5
Py(I) = P(I)*(7/(L(I)^2 + 7^2)^0.5);
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Plotting Resultant P & Px & Py with respect to length L
25

Px = [720 720 720 720 720 720 720 720 720 720 720 720];
figure(1);
plot(L,P,'o',L,Px,'*',L,Py,'+');
xlabel('L in m '), grid on;ylabel('P & Px & Py in N'), grid on;
26

33
PROBLEM 2.31
Determine the resultant of the three forces of Problem 2.23.
PROBLEM 2.23 Determine the x and y components of each of
the forces shown.
SOLUTION
Components of the forces were determined in Problem 2.23:
Force x Comp. (N) y Comp. (N)
800 lb +640 +480
424 lb –224 –360
408 lb +192 –360
608xR = + 240yR = −
(608 lb) ( 240 lb)
tan
240
608
21.541
240 N
sin(21.541°)
653.65 N
x y
y
x
R R
R
R
R
α
α
= +
= + −
=
=
= °
=
=
R i j
i j
654 N=R 21.5° 
27
2.12
as shown in the table below.
Write a MATLAB Code to resolve the problem above.

%%%%%%%%%Problem 2.31 How to calculate Resultant R by knowing all the
%%%%%%%%%components of the applied forces as shown in figure
%%%Calculating the regular problem as it is stated in the book
%%Calculating the X & Y Components for each applied force as following:-
%%%First force T1 = 800 N & x = 800 mm & y = 600 mm
T1 = 800
T1x = 800*(800/(800^2 + 600^2)^0.5)
T1y = 800*(600/(800^2 + 600^2)^0.5)
%%%Second force T2 = 408 N & x = 480 mm & y = -900 mm
T2 = 408
T2x = 408*(480/(480^2 + 900^2)^0.5)
T2y = 408*(-900/(480^2 + 900^2)^0.5)
%%%Third force T3 = 424 N & x = -560 mm & y = -900 mm
T3 = 424
T3x = 424*(-560/(560^2 + 900^2)^0.5)
T3y = 424*(-900/(560^2 + 900^2)^0.5)
%%%%Calculating the Resultant R as a summation of X components and Y
%%%%components as following:-
Rx = T1x + T2x + T3x
Ry = T1y + T2y + T3y
R=(Rx^2 + Ry^2)^0.5
%%%%Calculating the angle of the resultant R as following:-
theta = atan(Ry/Rx)*57.3 %%%% in Degrees
%%In case of several acting forces or even enormous number of acting forces
%%A Successive Loop should be constructed as following:-
%%T = [T1 T2 T3 T4 T5 .........]
%%x = [x1 x2 x3 x4 x5 .........]
%%y = [y1 y2 y3 y4 y5 .........]
T = [800 408 424]
x = [800 480 -560]
y = [600 -900 -900]
Rx=0;
Ry=0;
for H=0:1:2
H=H+1
Tx(H)= T(H)*(x(H)/(x(H)^2 + y(H)^2)^0.5)
Ty(H)= T(H)*(y(H)/(x(H)^2 + y(H)^2)^0.5)
Rx = Rx + Tx(H)
Ry = Ry + Ty(H)
end
R=(Rx^2 + Ry^2)^0.5
28

29

34
PROBLEM 2.32
PROBLEM 2.21 Determine the x and y components of each of the
forces shown.
SOLUTION
Force x Comp. (N) y Comp. (N)
80 N +61.3 +51.4
120 N +41.0 +112.8
150 N –122.9 +86.0
20.6xR = − 250.2yR = +
( 20.6 N) (250.2 N)
tan
250.2 N
tan
20.6 N
tan 12.1456
85.293
250.2 N
sin85.293
x y
y
x
R R
R
R
R
α
α
α
α
= +
= − +
=
=
=
= °
=
°
R i j
i j
251 N=R 85.3° 
2.13
30

36
PROBLEM 2.34
PROBLEM 2.24 Determine the x and y components of each of the
forces shown.
SOLUTION
Force x Comp. (lb) y Comp. (lb)
102 lb −48.0 +90.0
106 lb +56.0 +90.0
200 lb −160.0 −120.0
152.0xR = − 60.0yR =
( 152 lb) (60.0 lb)
tan
60.0 lb
tan
152.0 lb
tan 0.39474
21.541
α
α
α
α
= +
= − +
=
=
=
= °
x y
y
x
R R
R
R
R i j
i j
60.0 lb
sin 21.541
R =
°
163.4 lb=R 21.5° 
31
2.14

38
PROBLEM 2.36
Knowing that the tension in rope AC is 365 N, determine the
resultant of the three forces exerted at point C of post BC.
SOLUTION
Determine force components:
Cable force AC:
960
(365 N) 240 N
1460
1100
(365 N) 275 N
1460
= − = −
= − = −
x
y
F
F
500-N Force:
24
(500 N) 480 N
25
7
(500 N) 140 N
25
x
y
F
F
= =
= =
200-N Force:
4
(200 N) 160 N
5
3
(200 N) 120 N
5
x
y
F
F
= =
= − = −
and
2 2
2 2
240 N 480 N 160 N 400 N
275 N 140 N 120 N 255 N
(400 N) ( 255 N)
474.37 N
= Σ = − + + =
= Σ = − + − = −
= +
= + −
=
x x
y y
x y
R F
R F
R R R
Further:
255
tan
400
32.5
α
α
=
= °
474 N=R 32.5° 
2.15
32

%%%%%%%%%Problem 2.36 How to calculate Resultant P by knowing all the
%%%%%%%%%components of the applied forces as shown in figure
%%%Calculating the regular problem as it is stated in the book
%%Calculating the X & Y Components for each applied force as following:-
%%%First force T1 = 500 N & x = 24 m & y = 7 m
T1 = 500
T1x = 500*(24/(24^2 + 7^2)^0.5)
T1y = 500*(7/(24^2 + 7^2)^0.5)
%%%Second force T2 = 200 N & x = 4 m & y = -3 m
T2 = 200
T2x = 200*(4/(4^2 + 3^2)^0.5)
T2y = 200*(-3/(4^2 + 3^2)^0.5)
%%%Third force T3 = 365 N & x = -960 mm & y = -1100 mm
T3 = 365
T3x = 365*(-960/(960^2 + 1100^2)^0.5)
T3y = 365*(-1100/(960^2 + 1100^2)^0.5)
%%%%Calculating the Resultant R as a summation of X components and Y
%%%%components as following:-
Rx = T1x + T2x + T3x
Ry = T1y + T2y + T3y
R=(Rx^2 + Ry^2)^0.5
%%In case of several acting forces or even enormous number of acting forces
%%A Successive Loop should be constructed as following:-
%%T = [T1 T2 T3 T4 T5 .........]
%%x = [x1 x2 x3 x4 x5 .........]
%%y = [y1 y2 y3 y4 y5 .........]
for I=1:1:50
%%P is the highest force of 500 N, a range of P will be proposed
%%from 0 : 500 N in the following loop to discuss the effect on the
%%resultant
P(I)=I*10;
T = [P(I) 200 365];
x = [24 4 -960];
y = [7 -3 -1100];
Rx=0;
Ry=0;
for H=0:1:2
H=H+1;
33

Tx(H)= T(H)*(x(H)/(x(H)^2 + y(H)^2)^0.5);
Ty(H)= T(H)*(y(H)/(x(H)^2 + y(H)^2)^0.5);
Rx = Rx + Tx(H);
Ry = Ry + Ty(H);
end
R(I)=(Rx^2 + Ry^2)^0.5;
theta(I) = atan(Ry/Rx)*57.3; %%%% in Degrees
end
figure(1);
plot(P,theta,'o');
xlabel('Required P (The Highest Force)'), grid on;ylabel('theta in Degrees'),
grid on;
figure(2);
plot(P,R,'+');
xlabel('Required P (The Highest Force)'), grid on;ylabel('Resultant in N'),
grid on;
34

39
PROBLEM 2.37
Knowing that α = 40°, determine the resultant of the three forces
shown.
SOLUTION
60-lb Force: (60 lb)cos20 56.382 lb
(60 lb)sin 20 20.521lb
x
y
F
F
= ° =
= ° =
80-lb Force: (80 lb)cos60 40.000 lb
(80 lb)sin 60 69.282 lb
x
y
F
F
= ° =
= ° =
120-lb Force: (120 lb)cos30 103.923 lb
(120 lb)sin30 60.000 lb
x
y
F
F
= ° =
= − ° = −
and
2 2
200.305 lb
29.803 lb
(200.305 lb) (29.803 lb)
202.510 lb
x x
y y
R F
R F
R
= Σ =
= Σ =
= +
=
Further:
29.803
tan
200.305
α =
1 29.803
tan
200.305
8.46
α −
=
= ° 203 lb=R 8.46° 
2.16
36

40
PROBLEM 2.38
Knowing that α = 75°, determine the resultant of the three forces
shown.
SOLUTION
60-lb Force: (60 lb)cos 20 56.382 lb
(60 lb)sin 20 20.521 lb
x
y
F
F
= ° =
= ° =
80-lb Force: (80 lb)cos 95 6.9725 lb
(80 lb)sin 95 79.696 lb
x
y
F
F
= ° = −
= ° =
120-lb Force: (120 lb)cos 5 119.543 lb
(120 lb)sin 5 10.459 lb
x
y
F
F
= ° =
= ° =
Then 168.953 lb
110.676 lb
x x
y y
R F
R F
= Σ =
= Σ =
and 2 2
(168.953 lb) (110.676 lb)
201.976 lb
R = +
=
110.676
tan
168.953
tan 0.65507
33.228
α
α
α
=
=
= ° 202 lb=R 33.2° 
2.17
37

42
PROBLEM 2.40
For the post of Prob. 2.36, determine (a) the required tension in
rope AC if the resultant of the three forces exerted at point C is
to be horizontal, (b) the corresponding magnitude of the
resultant.
SOLUTION
960 24 4
(500 N) (200 N)
1460 25 5
48
640 N
73
x x AC
x AC
R F T
R T
= Σ = − + +
= − + (1)
1100 7 3
(500 N) (200 N)
1460 25 5
55
20 N
73
y y AC
y AC
R F T
R T
= Σ = − + −
= − + (2)
(a) For R to be horizontal, we must have 0.yR =
Set 0yR = in Eq. (2):
55
20 N 0
73
ACT− + =
26.545 NACT = 26.5 NACT = 
(b) Substituting for ACT into Eq. (1) gives
48
(26.545 N) 640 N
73
622.55 N
623 N
= − +
=
= =
x
x
x
R
R
R R 623 NR = 
2.18
38

45
PROBLEM 2.43
Two cables are tied together at C and are loaded as shown.
Determine the tension (a) in cable AC, (b) in cable BC.
SOLUTION
Free-Body Diagram
1100
tan
960
48.888
400
tan
960
22.620
α
α
β
β
=
= °
=
= °
Force Triangle
Law of sines:
15.696 kN
sin 22.620 sin 48.888 sin108.492
AC BCT T
= =
° ° °
(a)
15.696 kN
(sin 22.620 )
sin108.492
ACT = °
°
6.37 kNACT = 
(b)
15.696 kN
(sin 48.888 )
sin108.492
BCT = °
°
12.47 kNBCT = 
2.19
39

46
PROBLEM 2.44
Two cables are tied together at C and are loaded as shown. Determine
the tension (a) in cable AC, (b) in cable BC.
SOLUTION
3
tan
2.25
53.130
1.4
tan
2.25
31.891
α
α
β
β
=
= °
=
= °
Free-Body Diagram
Law of sines: Force-Triangle
660 N
sin31.891 sin53.130 sin94.979
AC BCT T
= =
° ° °
(a)
660 N
(sin31.891 )
sin94.979
ACT = °
°
350 NACT = 
(b)
660 N
(sin53.130 )
sin94.979
BCT = °
°
530 NBCT = 
2.20
40

47
PROBLEM 2.45
Knowing that 20 ,α = ° determine the tension (a) in cable AC,
(b) in rope BC.
SOLUTION
Free-Body Diagram Force Triangle
Law of sines:
1200 lb
sin 110 sin 5 sin 65
AC BCT T
= =
° ° °
(a)
1200 lb
sin 110
sin 65
ACT = °
°
1244 lbACT = 
(b)
1200 lb
sin 5
sin 65
BCT = °
°
115.4 lbBCT = 
2.21
41
Solve the problem above with replacing alfa angle = 20 Degrees with alfa angle = 80
Degrees, mention your conclusion upon the changing in alfa angle from 20 to 80 Degrees.

48
PROBLEM 2.46
Knowing that 55α = ° and that boom AC exerts on pin C a force directed
along line AC, determine (a) the magnitude of that force, (b) the tension in
cable BC.
SOLUTION
Law of sines:
300 lb
sin 35 sin 50 sin 95
AC BCF T
= =
° ° °
(a)
300 lb
sin 35
sin 95
ACF = °
°
172.7 lbACF = 
(b)
300 lb
sin 50
sin 95
BCT = °
°
231 lbBCT = 
2.22
42

54
PROBLEM 2.52
Two cables are tied together at C and loaded as shown.
Determine the range of values of P for which both cables
remain taut.
SOLUTION
Free Body: C
12 4
0: 0
13 5
x ACTΣ = − + =F P
13
15
ACT P= (1)
5 3
0: 480 N 0
13 5
y AC BCT T PΣ = + + − =F
Substitute for ACT from (1):
5 13 3
480 N 0
13 15 5
BCP T P
  
+ + − =  
  
14
480 N
15
BCT P= − (2)
From (1), 0ACT Ͼ requires 0.P Ͼ
From (2), 0BCT Ͼ requires
14
480 N, 514.29 N
15
P PϽ Ͻ
Allowable range: 0 514 NPϽ Ͻ 
2.23
43

55
PROBLEM 2.53
A sailor is being rescued using a boatswain’s chair that is
suspended from a pulley that can roll freely on the support
cable ACB and is pulled at a constant speed by cable CD.
Knowing that 30α = ° and 10β = ° and that the combined
weight of the boatswain’s chair and the sailor is 900 N,
determine the tension (a) in the support cable ACB, (b) in the
traction cable CD.
SOLUTION
Free-Body Diagram
0: cos 10 cos 30 cos 30 0x ACB ACB CDF T T TΣ = ° − ° − ° =
0.137158CD ACBT T= (1)
0: sin 10 sin 30 sin 30 900 0y ACB ACB CDF T T TΣ = ° + ° + ° − =
0.67365 0.5 900ACB CDT T+ = (2)
(a) Substitute (1) into (2): 0.67365 0.5(0.137158 ) 900ACB ACBT T+ =
1212.56 NACBT = 1213 NACBT = 
(b) From (1): 0.137158(1212.56 N)CDT = 166.3 NCDT = 
44
2.24

56
PROBLEM 2.54
A sailor is being rescued using a boatswain’s chair that is
suspended from a pulley that can roll freely on the support
cable ACB and is pulled at a constant speed by cable CD.
Knowing that 25α = ° and 15β = ° and that the tension in
cable CD is 80 N, determine (a) the combined weight of the
boatswain’s chair and the sailor, (b) in tension in the support
cable ACB.
SOLUTION
Free-Body Diagram
0: cos 15 cos 25 (80 N)cos 25 0x ACB ACBF T TΣ = ° − ° − ° =
1216.15 NACBT =
0: (1216.15 N)sin 15 (1216.15 N)sin 25yFΣ = ° + °
(80 N)sin 25 0
862.54 N
W
W
+ ° − =
=
(a) 863 NW = 
(b) 1216 NACBT = 
2.25
45

57
PROBLEM 2.55
Two forces P and Q are applied as shown to an aircraft
connection. Knowing that the connection is in equilibrium and
that 500P = lb and 650Q = lb, determine the magnitudes of
the forces exerted on the rods A and B.
SOLUTION
Free-Body Diagram
Resolving the forces into x- and y-directions:
0A B= + + + =R P Q F F
Substituting components: (500 lb) [(650 lb)cos50 ]
[(650 lb)sin50 ]
( cos50 ) ( sin50 ) 0B A AF F F
= − + °
− °
+ − ° + ° =
R j i
j
i i j
In the y-direction (one unknown force):
500 lb (650 lb)sin50 sin50 0AF− − ° + ° =
Thus,
500 lb (650 lb)sin50
sin50
AF
+ °
=
°
1302.70 lb= 1303 lbAF = 
In the x-direction: (650 lb)cos50 cos50 0B AF F° + − ° =
Thus, cos50 (650 lb)cos50
(1302.70 lb)cos50 (650 lb)cos50
B AF F= ° − °
= ° − °
419.55 lb= 420 lbBF = 
2.26
46

58
PROBLEM 2.56
Two forces P and Q are applied as shown to an aircraft
connection. Knowing that the connection is in equilibrium
and that the magnitudes of the forces exerted on rods A and B
are 750AF = lb and 400BF = lb, determine the magnitudes of
P and Q.
SOLUTION
Free-Body Diagram
Resolving the forces into x- and y-directions:
0A B= + + + =R P Q F F
Substituting components: cos 50 sin 50
[(750 lb)cos 50 ]
[(750 lb)sin 50 ] (400 lb)
P Q Q= − + ° − °
− °
+ ° +
R j i j
i
j i
In the x-direction (one unknown force):
cos 50 [(750 lb)cos 50 ] 400 lb 0Q ° − ° + =
(750 lb)cos 50 400 lb
cos 50
127.710 lb
Q
° −
=
°
=
In the y-direction: sin 50 (750 lb)sin 50 0P Q− − ° + ° =
sin 50 (750 lb)sin 50
(127.710 lb)sin 50 (750 lb)sin 50
476.70 lb
P Q= − ° + °
= − ° + °
= 477 lb; 127.7 lbP Q= = 
2.27
47

61
PROBLEM 2.59
For the situation described in Figure P2.45, determine (a) the
value of α for which the tension in rope BC is as small as
possible, (b) the corresponding value of the tension.
PROBLEM 2.45 Knowing that 20 ,α = ° determine the tension
(a) in cable AC, (b) in rope BC.
SOLUTION
To be smallest, BCT must be perpendicular to the direction of .ACT
(a) Thus, 5α = ° 5.00α = ° 
(b) (1200 lb)sin 5BCT = ° 104.6 lbBCT = 
2.28
2.21
2.21
48
Write a MATLAB Code to describe the correlation between the Tension in rope BC and the
angle between rope BC and the upward vertical axis.

%%%%%%%%%Problem 2.59 How to calculate value of alfa for which the tension
%%%%%%%%%in rope BC is as small as possible.
%%Measuring alfa from vertical line upward
%%Considering the Law of Sines consequently (Tac/sin(90+90-alfa)) =
%%(Tbc/sin(90+90-5)) = (1200/sin(180 -5-alfa).
for I=1:1:180
alfa(I)=I
Tbc(I) = (1200/sin((5+alfa(I))/57.3))*sin((180-5)/57.3);
Tac(I) = (1200/sin((5+alfa(I))/57.3))*sin((180-alfa(I))/57.3);
end
figure(1);
plot(alfa,Tbc,'o');
xlabel('alfa in Degrees '),grid on;ylabel('Tbc in LB'),grid on;
xlim([0 180])
ylim([0 2000])
figure(2);
plot(alfa,Tac,'+');
xlabel('alfa in Degrees '),grid on;ylabel('Tac in LB'),grid on;
xlim([0 180])
ylim([0 2000])
figure(3);
plot(alfa,Tac,'+',alfa,Tbc,'o');
xlabel('alfa in Degrees '),grid on;ylabel('Tac in LB & Tbc in LB'),grid on;
xlim([0 180])
ylim([0 2000])
49

62
PROBLEM 2.60
For the structure and loading of Problem 2.46, determine (a) the value of α for
which the tension in cable BC is as small as possible, (b) the corresponding
value of the tension.
SOLUTION
BCT must be perpendicular to ACF to be as small as possible.
Free-Body Diagram: C Force Triangle is a right triangle
To be a minimum, BCT must be perpendicular to .ACF
(a) We observe: 90 30α = ° − ° 60.0α = ° 
(b) (300 lb)sin 50BCT = °
or 229.81lbBCT = 230 lbBCT = 
2.29
51

63
PROBLEM 2.61
For the cables of Problem 2.48, it is known that the maximum
allowable tension is 600 N in cable AC and 750 N in cable BC.
Determine (a) the maximum force P that can be applied at C,
(b) the corresponding value of α.
SOLUTION
(a) Law of cosines 2 2 2
(600) (750) 2(600)(750)cos(25 45 )P = + − ° + °
784.02 NP = 784 NP = 
(b) Law of sines
sin sin (25 45 )
600 N 784.02 N
β ° + °
=
46.0β = ° 46.0 25α∴ = ° + ° 71.0α = ° 
2.30
52

64
PROBLEM 2.62
A movable bin and its contents have a combined weight of 2.8 kN.
Determine the shortest chain sling ACB that can be used to lift the
loaded bin if the tension in the chain is not to exceed 5 kN.
SOLUTION
Free-Body Diagram
tan
0.6 m
α =
h
(1)
Isosceles Force Triangle
Law of sines:
1
2
1
2
(2.8 kN)
sin
5 kN
(2.8 kN)
sin
5 kN
16.2602
AC
AC
T
T
α
α
α
=
=
=
= °
From Eq. (1): tan16.2602 0.175000 m
0.6 m
h
h° = ∴ =
Half length of chain 2 2
(0.6 m) (0.175 m)
0.625 m
AC= = +
=
Total length: 2 0.625 m= × 1.250 m 
2.31
53

65
PROBLEM 2.63
Collar A is connected as shown to a 50-lb load and can slide on
a frictionless horizontal rod. Determine the magnitude of the
force P required to maintain the equilibrium of the collar when
(a) 4.5 in.,x = (b) 15 in.x =
SOLUTION
(a) Free Body: Collar A Force Triangle
50 lb
4.5 20.5
P
= 10.98 lbP = 
(b) Free Body: Collar A Force Triangle
50 lb
15 25
P
= 30.0 lbP = 

2.32
54

66
PROBLEM 2.64
Collar A is connected as shown to a 50-lb load and can slide on a
frictionless horizontal rod. Determine the distance x for which the
collar is in equilibrium when P = 48 lb.
SOLUTION
Free Body: Collar A Force Triangle
2 2 2
(50) (48) 196
14.00 lb
N
N
= − =
=
Similar Triangles
48 lb
20 in. 14 lb
x
=
68.6 in.x = 
55
2.33
Write a MATLAB Code to describe the correlation between Distance (X) and the relevant
Tension T in rope BA, explain your final conclusion upon the mentioned correlation.

%%%%%%%%%Problem 2.64 How to calculate value of alfa while the tension in
%%%%%%%%%rope AB is varing from T = 0 LB to 250 LB
%%%%%%%%%
%%Measuring alfa from horizontal line to the right at Point A
%%Considering the Law of Sines consequently (T/sin(90)) = (48/sin(90-alfa))
%%= (N/sin(alfa) consequently Beta = sin(90-alfa)=48/T
for I = 1:1:50
T(I)=I*5
Beta(I) = 48/T(I);
alfa(I) = -asin(Beta(I))*57.3 + 90;
X(I) = 20/tan(alfa(I)/57.3)
N(I) = (T(I))*sin(alfa(I)/57.3)
end
figure(1);
plot(T,X,'o');
xlabel('T in LB '),grid on;ylabel('X in in'),grid on;
xlim([0 250])
ylim([0 100])
figure(2);
plot(T,alfa,'*');
xlabel('T in LB '),grid on;ylabel('alfa in Degrees'),grid on;
xlim([0 250])
ylim([0 100])
figure(3);
plot(T,N,'+');
xlabel('T in LB '),grid on;ylabel('N in LB'),grid on;
xlim([0 250])
ylim([0 200])
56

67
PROBLEM 2.65
Three forces are applied to a bracket as shown. The directions of the two 150-N
forces may vary, but the angle between these forces is always 50°. Determine the
range of values of α for which the magnitude of the resultant of the forces acting at A
is less than 600 N.
SOLUTION
Combine the two 150-N forces into a resultant force Q:
2(150 N)cos25
271.89 N
Q = °
=
Equivalent loading at A:
Using the law of cosines:
2 2 2
(600 N) (500 N) (271.89 N) 2(500 N)(271.89 N)cos(55 )
cos(55 ) 0.132685
α
α
= + + ° +
° + =
Two values for :α 55 82.375
27.4
α
α
° + =
= °
or 55 82.375
55 360 82.375
222.6
α
α
α
° + = − °
° + = ° − °
= °
For 600 lb:R < 27.4 222.6α° < < 
Resolve the problem above by the summation of X, Y Components analytically.
2.34
58

68
PROBLEM 2.66
A 200-kg crate is to be supported by the rope-and-pulley arrangement shown.
Determine the magnitude and direction of the force P that must be exerted on the free
end of the rope to maintain equilibrium. (Hint: The tension in the rope is the same on
each side of a simple pulley. This can be proved by the methods of Ch. 4.)
SOLUTION
Free-Body Diagram: Pulley A
5
0: 2 cos 0
281
cos 0.59655
53.377
xF P P α
α
α
 
Σ = − + = 
 
=
= ± °
For 53.377 :α = + °
16
0: 2 sin53.377 1962 N 0
281
yF P P
 
Σ = + ° − = 
 
724 N=P 53.4° 
For 53.377 :α = − °
16
0: 2 sin( 53.377 ) 1962 N 0
281
yF P P
 
Σ = + − ° − = 
 
1773=P 53.4° 
59
2.35

73
PROBLEM 2.71
Determine (a) the x, y, and z components of the 900-N force, (b) the
angles θx, θy, and θz that the force forms with the coordinate axes.
SOLUTION
cos 65
(900 N)cos 65
380.36 N
h
h
F F
F
= °
= °
=
(a) sin 20
(380.36 N)sin 20°
x hF F= °
=
130.091 N,= −xF 130.1 NxF = − 
sin65
(900 N)sin 65°
815.68 N,
y
y
F F
F
= °
=
= + 816 NyF = + 
cos20
(380.36 N)cos20
357.42 N
= °
= °
= +
z h
z
F F
F 357 NzF = + 
(b)
130.091 N
cos
900 N
x
x
F
F
θ
−
= = 98.3xθ = ° 
815.68 N
cos
900 N
y
y
F
F
θ
+
= = 25.0yθ = ° 
357.42 N
cos
900 N
z
z
F
F
θ
+
= = 66.6zθ = ° 
60
2.36

95
PROBLEM 2.93
Knowing that the tension is 425 lb in cable AB and 510 lb in
cable AC, determine the magnitude and direction of the resultant
of the forces exerted at A by the two cables.
SOLUTION
2 2 2
2 2 2
(40 in.) (45 in.) (60 in.)
(40 in.) (45 in.) (60 in.) 85 in.
(100 in.) (45 in.) (60 in.)
(100 in.) (45 in.) (60 in.) 125 in.
(40 in.) (45 in.) (60 in.)
(425 lb)
85 in.
AB AB AB AB
AB
AB
AC
AC
AB
T T
AB
= − +
= + + =
= − +
= + + =
− +
= = =
i j k
i j k
i j k
T λ



(200 lb) (225 lb) (300 lb)
(100 in.) (45 in.) (60 in.)
(510 lb)
125 in.
(408 lb) (183.6 lb) (244.8 lb)
(608) (408.6 lb) (544.8 lb)
AB
AC AC AC AC
AC
AB AC
AC
T T
AC
 
 
 
= − +
 − +
= = =  
 
= − +
= + = − +
T i j k
i j k
T λ
T i j k
R T T i j k

Then: 912.92 lbR = 913 lbR = 
and
608 lb
cos 0.66599
912.92 lb
xθ = = 48.2xθ = ° 
408.6 lb
cos 0.44757
912.92 lb
yθ = = − 116.6yθ = ° 
544.8 lb
cos 0.59677
912.92 lb
zθ = = 53.4zθ = ° 
2.37
61

%%%%%%%%%Problem 2.93 How to calculate the resultant at point A from
%%different acting forces using the unit vector & vector summation concept,
%%considering the tension in cable AB = 425 LB and in cable AC = 510 LB
%% Calculating each force in vector form by calculating the unit vector for
%% each
AB=[40 -45 60] %%% 40i -45j + 60k
AC=[100 -45 60] %%% 100i - 45j + 60k
lamdaAB=AB./(AB(1)^2 + AB(2)^2 + AB(3)^2)^0.5
Tab=lamdaAB.*425
lamdaAC=AC./(AC(1)^2 + AC(2)^2 + AC(3)^2)^0.5
Tac=lamdaAC.*510
R = Tab + Tac
Rm=(R(1)^2 + R(2)^2 + R(3)^2)^0.5
%%Calculating theta angles with respect to X, Y, Z
thetaX = acos(R(1)/Rm)*57.3
thetaY = acos(R(2)/Rm)*57.3
thetaZ = acos(R(3)/Rm)*57.3
62

112
PROBLEM 2.107
Three cables are connected at A, where the forces P and Q are
applied as shown. Knowing that 0,Q = find the value of P for
which the tension in cable AD is 305 N.
SOLUTION
0: 0A AB AC ADΣ = + + + =F T T T P where P=P i

(960 mm) (240 mm) (380 mm) 1060 mm
(960 mm) (240 mm) (320 mm) 1040 mm
(960 mm) (720 mm) (220 mm) 1220 mm
AB AB
AC AC
AD AD
= − − + =
= − − − =
= − + − =
i j k
i j k
i j k




48 12 19
53 53 53
12 3 4
13 13 13
305 N
[( 960 mm) (720 mm) (220 mm) ]
1220 mm
(240 N) (180 N) (55 N)
AB AB AB AB AB
AC AC AC AC AC
AD AD AD
AB
T T T
AB
AC
T T T
AC
T
 
= = = − − + 
 
 
= = = − − − 
 
= = − + −
= − + −
T λ i j k
T λ i j k
T λ i j k
i j k


Substituting into 0,AΣ =F factoring , , ,i j k and setting each coefficient equal to φ gives:
48 12
: 240 N
53 13
AB ACP T T= + +i (1)
:j
12 3
180 N
53 13
AB ACT T+ = (2)
:k
19 4
55 N
53 13
AB ACT T− = (3)
Solving the system of linear equations using conventional algorithms gives:
446.71 N
341.71 N
AB
AC
T
T
=
= 960 NP = 
2.38
63

113
PROBLEM 2.108
Three cables are connected at A, where the forces P and Q are
applied as shown. Knowing that 1200 N,P = determine the values
of Q for which cable AD is taut.
SOLUTION
We assume that 0ADT = and write 0: (1200 N) 0A AB AC QΣ = + + + =F T T j i
(960 mm) (240 mm) (380 mm) 1060 mm
(960 mm) (240 mm) (320 mm) 1040 mm
AB AB
AC AC
= − − + =
= − − − =
i j k
i j k


48 12 19
53 53 53
12 3 4
13 13 13
AB AB AB AB AB
AC AC AC AC AC
AB
T T T
AB
AC
T T T
AC
 
= = = − − + 
 
 
= = = − − − 
 
T λ i j k
T λ i j k


Substituting into 0,AΣ =F factoring , , ,i j k and setting each coefficient equal to φ gives:
48 12
: 1200 N 0
53 13
AB ACT T− − + =i (1)
12 3
: 0
53 13
AB ACT T Q− − + =j (2)
19 4
: 0
53 13
AB ACT T− =k (3)
Solving the resulting system of linear equations using conventional algorithms gives:
605.71 N
705.71 N
300.00 N
AB
AC
T
T
Q
=
=
= 0 300 NQՅ Ͻ 
Note: This solution assumes that Q is directed upward as shown ( 0),Q Ն if negative values of Q
are considered, cable AD remains taut, but AC becomes slack for 460 N.Q = − 
2.39
64

114
PROBLEM 2.109
A rectangular plate is supported by three cables as shown.
Knowing that the tension in cable AC is 60 N, determine the
weight of the plate.
SOLUTION
We note that the weight of the plate is equal in magnitude to the force P exerted by the support on Point A.
Free Body A:
0: 0AB AC ADF PΣ = + + + =T T T j
We have:
( )
(320 mm) (480 mm) (360 mm) 680 mm
(450 mm) (480 mm) (360 mm) 750 mm
(250 mm) (480 mm) 360 mm 650 mm
AB AB
AC AC
AD AD
= − − + =
= − + =
= − − =
i j k
i j k
i j k



Thus:
( )
8 12 9
17 17 17
0.6 0.64 0.48
5 9.6 7.2
13 13 13
AB AB AB AB AB
AC AC AC AC AC
AD AD AD AD AD
AB
T T T
AB
AC
T T T
AC
AD
T T T
AD
 
= = = − − + 
 
= = = − +
 
= = = − − 
 
T λ i j k
T λ i j k
T λ i j k



Substituting into the Eq. 0FΣ = and factoring , , :i j k
8 5
0.6
17 13
12 9.6
0.64
17 13
9 7.2
0.48 0
17 13
AB AC AD
AB AC AD
AB AC AD
T T T
T T T P
T T T
 
− + + 
 
 
+ − − − + 
 
 
+ + − = 
 
i
j
k
Dimensions in mm
2.40
65

115
PROBLEM 2.109 (Continued)
Setting the coefficient of i, j, k equal to zero:
:i
8 5
0.6 0
17 13
AB AC ADT T T− + + = (1)
:j
12 9.6
0.64 0
7 13
AB AC ADT T T P− − − + = (2)
:k
9 7.2
0.48 0
17 13
AB AC ADT T T+ − = (3)
Making 60 NACT = in (1) and (3):
8 5
36 N 0
17 13
AB ADT T− + + = (1′)
9 7.2
28.8 N 0
17 13
AB ADT T+ − = (3′)
Multiply (1′) by 9, (3′) by 8, and add:
12.6
554.4 N 0 572.0 N
13
AD ADT T− = =
Substitute into (1′) and solve for :ABT
17 5
36 572 544.0 N
8 13
AB ABT T
 
= + × = 
 
Substitute for the tensions in Eq. (2) and solve for P:
12 9.6
(544 N) 0.64(60 N) (572 N)
17 13
844.8 N
P = + +
= Weight of plate 845 NP= = 
2.40
66

%%%%%%%%%Problem 2.109 How to calculate weight of a plate supported by
%%tensions at point A in cables AD & AB & AC knowing that tension in AC is
%%60 N
AB=[-320 -480 360] %%% -320i -450j + 360k
AC=[450 -480 360] %%% 450i - 450j + 360k
AD=[250 -480 -360] %%% 250i -450j + 360k
%%Tab=lamdaAB.*Tab
%%Tac=lamdaAC.*Tac
Tac=lamdaAC.*60
%%Tad=lamdaAD.*Tad
lamdaAD=AD./(AD(1)^2 + AD(2)^2 + AD(3)^2)^0.5
%%%%Consider the weight force = Wj as it is upward
%%%%As long the plate is in equilibrium state therefore the summation of
%%%%forces is equal to zero in all directions X, Y, Z
%%%Calculating the weight of the plate in matrix calculations
%%%[lamdaAB(1) lamdaAD(1) 0] [Tab] [-Tac(1)]
%%%[lamdaAB(2) lamdaAD(2) 1] * [Tad] = [-Tac(2)]
%%%[lamdaAB(3) lamdaAD(3) 0] [ W ] [-Tac(3)]
A = [lamdaAB(1) lamdaAD(1) 0;lamdaAB(2) lamdaAD(2) 1;lamdaAB(3)
lamdaAD(3) 0]
C = [-Tac(1) -Tac(2) -Tac(3)]'
X = inv(A)*C
Tab = X(1)
Tad = X(2)
W = X(3)
-480
-480
-480 -360
67

116
PROBLEM 2.110
A rectangular plate is supported by three cables as shown. Knowing
that the tension in cable AD is 520 N, determine the weight of the plate.
SOLUTION
See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and
(3) below:
8 5
0.6 0
17 13
AB AC ADT T T− + + = (1)
12 9.6
0.64 0
17 13
AB AC ADT T T P− + − + = (2)
9 7.2
0.48 0
17 13
AB AC ADT T T+ − = (3)
Making 520 NADT = in Eqs. (1) and (3):
8
0.6 200 N 0
17
AB ACT T− + + = (1′)
9
0.48 288 N 0
17
AB ACT T+ − = (3′)
Multiply (1′) by 9, (3′) by 8, and add:
9.24 504 N 0 54.5455 NAC ACT T− = =
Substitute into (1′) and solve for :ABT
17
(0.6 54.5455 200) 494.545 N
8
AB ABT T= × + =
Substitute for the tensions in Eq. (2) and solve for P:
12 9.6
(494.545 N) 0.64(54.5455 N) (520 N)
17 13
768.00 N
P = + +
= Weight of plate 768 NP= = 
2.41
68

%%%%%%%%%Problem 2.110 How to calculate weight of a plate supported by
%%tensions at point A in cables AD & AB & AC knowing that tension in AD is
%%520 N
AB=[-320 -480 360] %%% -320i -450j + 360k
AC=[450 -480 360] %%% 450i - 450j + 360k
AD=[250 -480 -360] %%% 250i -450j + 360k
%%Tab=lamdaAB.*Tab
%%Tac=lamdaAC.*Tac
lamdaAD=AD./(AD(1)^2 + AD(2)^2 + AD(3)^2)^0.5
Tad=lamdaAD.*520
%%%%Consider the weight force = -Wj as it is downward
%%%%As long the plate is in equilibrium state therefore the summation of
%%%%forces is equal to zero in all directions X, Y, Z
%%%Calculating the weight of the plate in matrix calculations
%%%[lamdaAB(1) lamdaAC(1) 0] [Tab] [-Tad(1)]
%%%[lamdaAB(2) lamdaAC(2) -1] * [Tac] = [-Tad(2)]
%%%[lamdaAB(3) lamdaAC(3) 0] [ W ] [-Tad(3)]
A = [lamdaAB(1) lamdaAC(1) 0;lamdaAB(2) lamdaAC(2) 1;lamdaAB(3)
lamdaAC(3) 0]
C = [-Tad(1) -Tad(2) -Tad(3)]'
X = inv(A)*C
Tab = X(1)
Tac = X(2)
W = X(3)
69
-480
-480
-480 -360

Chapter 3
Statics of Rigid Bodies
Equivalent Systems of Forces
3.1 INTRODUCTION
In the preceding chapter it was assumed that each of the bodies considered could
be treated as a single particle. Such a view, however, is not always possible, and a
body, in general, should be treated as a combination of a large number of particles.
The size of the body will have to be taken into consideration, as well as the fact
that forces will act on different particles and thus will have different points of
application. Most of the bodies considered in elementary mechanics are assumed to
be rigid, a rigid body being defined as one which does not deform. Actual
structures and machines, however, are never absolutely rigid and deform under the
loads to which they are subjected. But these deformations are usually small and do
not appreciably affect the conditions of equilibrium or motion of the structure
under consideration. They are important, though, as far as the resistance of the
structure to failure is concerned and are considered in the study of mechanics of
materials. In this chapter you will study the effect of forces exerted on a rigid body,
and you will learn how to replace a given system of forces by a simpler equivalent
system. This analysis will rest on the fundamental assumption that the effect of a
given force on a rigid body remains unchanged if that force is moved along its line
of action (principle of transmissibility).
The basic system is called a force-couple system. In the case of concurrent,
coplanar, or parallel forces, the equivalent force-couple system can be further
reduced to a single force, called the resultant of the system, or to a single
couple, called the resultant couple of the system.
3.2 EXTERNAL AND INTERNAL FORCES
Forces acting on rigid bodies can be separated into two groups: (1) external forces
and (2) internal forces.
70

1. The external forces represent the action of other bodies on the rigid body under
consideration. They are entirely responsible for the external behavior of the rigid
body. They will either cause it to move or ensure that it remains at rest. We shall
be concerned only with external forces in this chapter and in Chaps. 4 and 5.
2 . The internal forces are the forces which hold together the particles forming the
rigid body. If the rigid body is structurally composed of several parts, the forces
holding the component parts together are also defined as internal forces. Internal
forces will be considered in Chaps. 6 and 7.
As an example of external forces, let us consider the forces acting on a disabled
truck that three people are pulling forward by means of a rope attached to the front
bumper (Fig. 3.1). The external forces acting on the truck are shown in a free-body
diagram (Fig. 3.2). Let us first consider the weight of the truck. Although it
embodies the effect of the earth’s pull on each of the particles forming the truck,
the weight can be represented by the single force W. The point of application of
this force, i.e., the point at which the force acts, is defined as the center of gravity
of the truck. It will be seen in Chap. 5 how centers of gravity can be determined.
The weight W tends to make the truck move vertically downward. In fact, it would
actually cause the truck to move downward, i.e., to fall, if it were not for the
presence of the ground. The ground opposes the downward motion of the truck by
means of the reactions R 1 and R 2. These forces are exerted by the ground on the
truck and must therefore be included among the external forces acting on the truck.
The people pulling on the rope exert the force F. The point of application of F is
on the front bumper. The force F tends to make the truck move forward in a
straight line and does actually make it move, since no external force opposes this
motion. (Rolling resistance has been neglected here for simplicity.) This forward
motion of the truck, during which each straight line keeps its original orientation
(the floor of the truck remains horizontal, and the walls remain vertical), is known
as a translation. Other forces might cause the truck to move differently. For
example, the force exerted by a jack placed under the front axle would cause the
truck to pivot about its rear axle. Such a motion is a rotation. It can be concluded,
therefore, that each of the external forces acting on a rigid body can, if unopposed,
impart to the rigid body a motion of translation or rotation, or both.
71

3.3 VECTOR PRODUCT OF TWO VECTORS
In order to gain a better understanding of the effect of a force on a rigid body, a
new concept, the concept of a moment of a force about a point, will be introduced
at this time. This concept will be more clearly understood, and applied more
effectively, if we first add to the mathematical tools at our disposal the vector
product of two vectors. The vector product of two vectors P and Q is defined as
the vector V which satisfies the following conditions.
1. The line of action of V is perpendicular to the plane containing P and Q (Fig.
3.3 a).
2. The magnitude of V is the product of the magnitudes of P and Q and of the sine
of the angle u formed by P and Q (the measure of which will always be 180° or
less); we thus have
V = PQ sin θ
3. The direction of V is obtained from the right-hand rule. Close your right hand
and hold it so that your fingers are curled in the same sense as the rotation through
u which brings the vector P in line with the vector Q ; your thumb will then
indicate the direction of the vector V (Fig. 3.3 b ). Note that if P and Q do not have
a common point of application, they should first be redrawn from the same point.
The three vectors P , Q, and V —taken in that order—are said to form a right-
handed rule.
As stated above, the vector V satisfying these three conditions (which define it
uniquely) is referred to as the vector product of P and Q ; it is represented by the
mathematical expression in vector form as:
V = P x Q
72

Fig. 3.3
We can now easily express the vector product V of two given vectors P and Q in
terms of the rectangular components of these vectors. Resolving P and Q into
components, we first write:
V = P x Q = (Pxi + Pyj + Pzk) x (Qxi + Qyj + Qz
V = (P
k)
yQz - PzQy)i + (PzQx - PxQz)j + (PxQy - PyQx
The rectangular components of the vector product V are thus found to be:
)k
Which is more easily memorized by the following determinate:
73

161
PROBLEM 3.1
A 20-lb force is applied to the control rod AB as shown. Knowing that the length of
the rod is 9 in. and that α = 25°, determine the moment of the force about Point B
by resolving the force into horizontal and vertical components.
SOLUTION
Free-Body Diagram of Rod AB:
(9 in.)cos65
3.8036 in.
(9 in.)sin65
8.1568 in.
x
y
= °
=
= °
=
(20 lb cos25 ) ( 20 lb sin 25 )
(18.1262 lb) (8.4524 lb)
x yF F= +
= ° + − °
= −
F i j
i j
i j
/ ( 3.8036 in.) (8.1568 in.)A B BA= = − +r i j

/
( 3.8036 8.1568 ) (18.1262 8.4524 )
32.150 147.852
115.702 lb-in.
B A B= ×
= − + × −
= −
= −
M r F
i j i j
k k
115.7 lb-in.B =M 
3.1
74

%%%%%%%%%Problem 3.1 How to calculate the moment of force about point B as
%%%%%%%%%shown in figure.
%%Calulating the acting force of 20 LB in vector form as following:-
%%F = (Fx)i + (Fy)j
%%F = 20*cos(25/57.3)i - 20*sin(25/57.3)j
F=[ 20*cos(25/57.3) -20*sin(25/57.3)]
%%% F = 20*cos(25/57.3)i - 20*sin(25/57.3)j
%%Calulating the moment arm BA in vector form as shown in figure.
%% BA = (x)i + (y)j
%% BA = -9*cos(65/57.3)i + 9*sin(65/57.3)j
BA=[-9*cos(65/57.3) 9*sin(65/57.3)]
%%% BA = -9*cos(65/57.3)i + 9*sin(65/57.3)j
%%Calculating the moment acting about point B by force of 20 LB as
%%Mb = BA x F by the cross product method as following:-
%% | i j k |
%%determinate of | BA(1) BA(2) 0 |
%% | F(1) F(2) 0 |
%%
Mb = BA(1)*F(2) - F(1)*BA(2) %%%%Mb= (Mb)k
%%Negative sign of the moment magnitude represents a clock wise acting
%%moment
75

166
PROBLEM 3.6
A 300-N force P is applied at Point A of the bell crank shown.
(a) Compute the moment of the force P about O by resolving
it into horizontal and vertical components. (b) Using the
result of part (a), determine the perpendicular distance from
O to the line of action of P.
SOLUTION
(0.2 m)cos40
0.153209 m
(0.2 m)sin 40
0.128558 m
x
y
= °
=
= °
=
/ (0.153209 m) (0.128558 m)A O∴ = +r i j
(a) (300 N)sin30
150 N
(300 N)cos30
259.81 N
x
y
F
F
= °
=
= °
=
(150 N) (259.81 N)= +F i j
/
(0.153209 0.128558 ) m (150 259.81 ) N
(39.805 19.2837 ) N m
(20.521 N m)
O A O= ×
= + × +
= − ⋅
= ⋅
M r F
i j i j
k k
k 20.5 N mO = ⋅M 
(b) OM Fd=
20.521 N m (300 N)( )
0.068403 m
d
d
⋅ =
= 68.4 mmd = 
3.2
76

%%%%%%%%%Problem 3.6 How to calculate the moment of force about point O as
%%%%%%%%%shown in figure.
%%Calulating the acting force of 300 N in vector form as following:-
%%F = (Fx)i + (Fy)j
%%F = 300*sin(30/57.3)i + 300*cos(30/57.3)j
F=[ 300*sin(30/57.3) 300*cos(30/57.3)]
%%% F = 300*sin(30/57.3)i + 300*cos(30/57.3)j
%%Calulating the moment arm OA in vector form as shown in figure.
%% OA = (x)i + (y)j
%% OA = 200*cos(40/57.3)i + 200*sin(40/57.3)j
OA=[200*cos(40/57.3) 200*sin(40/57.3)]
%%% OA = 200*cos(40/57.3)i + 200*sin(40/57.3)j
%%Calculating the moment acting about point O by force of 300 N as
%%Mo = OA x F by evaluating the cross product as following:-
%% | i j k |
%%Mo =determinate of | OA(1) OA(2) 0 |
%% | F(1) F(2) 0 |
%%
Mo = OA(1)*F(2) - F(1)*OA(2) %%%%Mo= (Mo)k in N.mm
Mo = Mo/1000 %%%%Mo in N.m
%%Calculating the perpendicular distance from point O to the line of the
%%acting force of the 300 N as following:-
D = Mo/300 %%%% D in meter
77

171
PROBLEM 3.11
A winch puller AB is used to straighten a fence post. Knowing that the tension in cable BC is 1040 N and
length d is 1.90 m, determine the moment about D of the force exerted by the cable at C by resolving that
force into horizontal and vertical components applied (a) at Point C, (b) at Point E.
SOLUTION
(a) Slope of line:
0.875 m 5
1.90 m 0.2 m 12
EC = =
+
Then
12
( )
13
ABx ABT T=
12
(1040 N)
13
960 N
=
= (a)
and
5
(1040 N)
13
400 N
AByT =
=
Then (0.875 m) (0.2 m)
(960 N)(0.875 m) (400 N)(0.2 m)
D ABx AByM T T= −
= −
760 N m= ⋅ or 760 N mD = ⋅M 
(b) We have ( ) ( )D ABx ABxM T y T x= +
(960 N)(0) (400 N)(1.90 m)
760 N m
= +
= ⋅ (b)
or 760 N mD = ⋅M 
3.3
78

%%%%%%%%%Problem 3.11 How to calculate the moment of force about point D as
%%%%%%%%%shown in figure using unit vector method to resolve the tension in
%%%%%%%%%cable BC into the horizontal and vertical components
%%unit vector of CB = unit vector of CE ---> lamdaCB = lamdaCE therefore
%%lamdaCE can be easily calculated from the geometry
%%CE = ((-d)i + 0j) - ((o.2)i + (0.857)j)
d=1.9 %%%%d = 1.9 meter
CE = [(-d - 0.2) (0 - 0.875)]
lamdaCE =CE./(CE(1)^2 + CE(2)^2)^0.5
%%Calulating the acting tension of 1040 N in vector form as following:-
%%T = (Fx)i + (Fy)j
%%T = 1040*lamdaCE
T = 1040*lamdaCE
%%Calulating the moment arm DC in vector form as shown in figure.
%% DC = (x)i + (y)j
%% DC = (0.2)i + (0.875)j
DC = [(0.2) (0.875)]
%%Calculating the moment acting about point D by force of 1040 N as
%%MD = DC x T by evaluating the cross product as following:-
%% | i j k |
%%MD =determinate of | DC(1) DC(2) 0 |
%% | T(1) T(2) 0 |
%%
MD1 = DC(1)*T(2) - T(1)*DC(2) %%%%MD1 = (MD1)k in N.m
%%Calulating the moment arm DE in vector form as shown in figure.
%% DE = (x)i + (y)j
%% DE = (-1.9)i (0)j
DE = [(-1.9) (0)]
%%Calculating the moment acting about point D by force of 1040 N as
%%MD = DE x T by evaluating the cross product as following:-
%% | i j k |
%%MD =determinate of | DE(1) DE(2) 0 |
%% | T(1) T(2) 0 |
%%
MD2 = DE(1)*T(2) - T(1)*DE(2) %%%%MD2= (MD2)k in N.m
79

172
PROBLEM 3.12
It is known that a force with a moment of 960 N · m about D is required to straighten the fence post CD. If
d = 2.80 m, determine the tension that must be developed in the cable of winch puller AB to create the
required moment about Point D.
SOLUTION
Slope of line:
0.875 m 7
2.80 m 0.2 m 24
EC = =
+
Then
24
25
ABx ABT T=
and
7
25
ABy ABT T=
We have ( ) ( )D ABx AByM T y T x= +
24 7
960 N m (0) (2.80 m)
25 25
1224 N
AB AB
AB
T T
T
⋅ = +
= or 1224 NABT = 
3.4
80

173
PROBLEM 3.13
It is known that a force with a moment of 960 N · m about D is required to straighten the fence post CD. If the
capacity of winch puller AB is 2400 N, determine the minimum value of distance d to create the specified
moment about Point D.
SOLUTION
The minimum value of d can be found based on the equation relating the moment of the force ABT about D:
max( ) ( )D AB yM T d=
where 960 N mDM = ⋅
max max( ) sin (2400 N)sinAB y ABT T θ θ= =
Now
2 2
2 2
0.875 m
sin
( 0.20) (0.875) m
0.875
960 N m 2400 N ( )
( 0.20) (0.875)
d
d
d
θ =
+ +
 
 ⋅ =
 + + 
or 2 2
( 0.20) (0.875) 2.1875d d+ + =
or 2 2 2
( 0.20) (0.875) 4.7852d d+ + =
or 2
3.7852 0.40 0.8056 0d d− − =
Using the quadratic equation, the minimum values of d are 0.51719 m and 0.41151 m.−
Since only the positive value applies here, 0.51719 md =
or 517 mmd = 
3.5
81

175
PROBLEM 3.15
Form the vector products B × C and B′ × C, where B = B′, and use the results
obtained to prove the identity
1 1
sin cos sin ( ) sin ( ).
2 2
α β α β α β= + + −
SOLUTION
Note: (cos sin )
(cos sin )
(cos sin )
B
B
C
β β
β β
α α
= +
′ = −
= +
B i j
B i j
C i j
By definition, | | sin ( )BC α β× = −B C (1)
| | sin ( )BC α β′× = +B C (2)
Now (cos sin ) (cos sin )B Cβ β α α× = + × +B C i j i j
(cos sin sin cos )BC β α β α= − k (3)
and (cos sin ) (cos sin )B Cβ β α α′× = − × +B C i j i j
(cos sin sin cos )BC β α β α= + k (4)
Equating the magnitudes of ×B C from Equations (1) and (3) yields:
sin( ) (cos sin sin cos )BC BCα β β α β α− = − (5)
Similarly, equating the magnitudes of ′×B C from Equations (2) and (4) yields:
sin( ) (cos sin sin cos )BC BCα β β α β α+ = + (6)
Adding Equations (5) and (6) gives:
sin( ) sin( ) 2cos sinα β α β β α− + + =
or
1 1
sin cos sin( ) sin( )
2 2
α β α β α β= + + − 
3.6
82

176
PROBLEM 3.16
The vectors P and Q are two adjacent sides of a parallelogram. Determine the area of the parallelogram when
(a) P = −7i + 3j − 3k and Q = 2i + 2j + 5k, (b) P = 6i − 5j − 2k and Q = −2i + 5j − k.
SOLUTION
(a) We have | |A = ×P Q
where 7 3 3= − + −P i j k
2 2 5= + +Q i j k
Then 7 3 3
2 2 5
[(15 6) ( 6 35) ( 14 6) ]
(21) (29) ( 20)
× = − −
= + + − + + − −
= + −
i j k
P Q
i j k
i j k
2 2 2
(20) (29) ( 20)A = + + − or 41.0A = 
(b) We have | |A = ×P Q
where 6 5 2= − −P i j k
2 5 1= − + −Q i j k
Then 6 5 2
2 5 1
[(5 10) (4 6) (30 10) ]
(15) (10) (20)
× = − −
− −
= + + + + −
= + +
i j k
P Q
i j k
i j k
2 2 2
(15) (10) (20)A = + + or 26.9A = 
3.7
82

177
PROBLEM 3.17
A plane contains the vectors A and B. Determine the unit vector normal to the plane when A and B are equal
to, respectively, (a) i + 2j − 5k and 4i − 7j − 5k, (b) 3i − 3j + 2k and −2i + 6j − 4k.
SOLUTION
(a) We have
| |
×
=
×
A B
λ
A B
where 1 2 5= + −A i j k
4 7 5= − −B i j k
Then 1 2 5
4 7 5
( 10 35) (20 5) ( 7 8)
15(3 1 1 )
× = + −
− −
= − − + + + − −
= − −
i j k
A B
i j k
i j k
and 2 2 2
| | 15 ( 3) ( 1) ( 1) 15 11× = − + − + − =A B
15( 3 1 1 )
15 11
− − −
=
i j k
λ or
1
( 3 )
11
= − − −λ i j k 
(b) We have
| |
×
=
×
A B
λ
A B
where 3 3 2= − +A i j k
2 6 4= − + −B i j k
Then 3 3 2
2 6 4
(12 12) ( 4 12) (18 6)
(8 12 )
× = −
− −
= − + − + + −
= +
i j k
A B
i j k
j k
and 2 2
| | 4 (2) (3) 4 13× = + =A B
4(2 3 )
4 13
+
=
j k
λ or
1
(2 3 )
13
= +λ j k 
3.8
83

179
PROBLEM 3.19
Determine the moment about the origin O of the force F = 4i − 3j + 5k that acts at a Point A. Assume that the
position vector of A is (a) r = 2i + 3j − 4k, (b) r = −8i + 6j − 10k, (c) r = 8i − 6j + 5k.
SOLUTION
O = ×M r F
(a) 2 3 4
4 3 5
O = −
−
i j k
M
(15 12) ( 16 10) ( 6 12)= − + − − + − −i j k 3 26 18O = − −M i j k 
(b) 8 6 10
4 3 5
O = − −
−
i j k
M
(30 30) ( 40 40) (24 24)= − + − + + −i j k 0O =M 
(c) 8 6 5
4 3 5
O = −
−
i j k
M
( 30 15) (20 40) ( 24 24)= − + + − + − +i j k 15 20O = − −M i j 
Note: The answer to Part (b) could have been anticipated since the elements of the last two rows of the
determinant are proportional.
3.9
84

180
PROBLEM 3.20
Determine the moment about the origin O of the force F = 2i + 3j − 4k that acts at a Point A. Assume that the
position vector of A is (a) r = 3i − 6j + 5k, (b) r = i − 4j − 2k, (c) r = 4i + 6j − 8k.
SOLUTION
O = ×M r F
(a) 3 6 5
2 3 4
O = −
−
i j k
M
(24 15) (10 12) (9 12)= − + + + +i j k 9 22 21O = + +M i j k 
(b) 1 4 2
2 3 4
O = − −
−
i j k
M
(16 6) ( 4 4) (3 8)= + + − + + +i j k 22 11O = +M i k 
(c) 4 6 8
2 3 4
O = −
−
i j k
M
( 24 24) ( 16 16) (12 12)= − + + − + + −i j k 0O =M 
Note: The answer to Part (c) could have been anticipated since the elements of the last two rows of the
determinant are proportional.
3.10
85

213
PROBLEM 3.53
A single force P acts at C in a direction perpendicular to the
handle BC of the crank shown. Knowing that Mx = +20 N ·
m and My = −8.75 N · m, and Mz = −30 N · m, determine the
magnitude of P and the values of φ and θ.
SOLUTION
(0.25 m) (0.2 m)sin (0.2 m)cos
sin cos
0.25 0.2sin 0.2cos
0 sin cos
C
O C
P P
P P
θ θ
φ φ
θ θ
φ φ
= + +
= − +
= × =
−
r i j k
P j k
i j k
M r P
Expanding the determinant, we find
(0.2) (sin cos cos sin )xM P θ φ θ φ= +
(0.2) sin( )xM P θ φ= + (1)
(0.25) cosyM P φ= − (2)
(0.25) sinzM P φ= − (3)
Dividing Eq. (3) by Eq. (2) gives: tan z
y
M
M
φ = (4)
30 N m
tan
8.75 N m
φ
− ⋅
=
− ⋅
73.740φ = 73.7φ = ° 
Squaring Eqs. (2) and (3) and adding gives:
2 2 2 2 2 2
(0.25) or 4y z y zM M P P M M+ = = + (5)
2 2
4 (8.75) (30)
125.0 N
P = +
= 125.0 NP = 
Substituting data into Eq. (1):
( 20 N m) 0.2 m(125.0 N)sin( )
( ) 53.130 and ( ) 126.87
20.6 and 53.1
θ φ
θ φ θ φ
θ θ
+ ⋅ = +
+ = ° + = °
= − ° = °
53.1Q = ° 
3.11
86

217
PROBLEM 3.57
The 23-in. vertical rod CD is welded to the midpoint C of the
50-in. rod AB. Determine the moment about AB of the 235-lb
force P.
SOLUTION
(32 in.) (30 in.) (24 in.)AB = − −i j k

2 2 2
(32) ( 30) ( 24) 50 in.AB = + − + − =
0.64 0.60 0.48AB
AB
AB
= = − −i k

λ
We shall apply the force P at Point G:
/ (5 in.) (30 in.)G B = +r i k
(21 in.) (38 in.) (18 in.)DG = − +i j k

2 2 2
(21) ( 38) (18) 47 in.DG = + − + =
21 38 18
(235 lb)
47
DG
P
DG
− +
= =
i j k
P

(105 lb) (190 lb) (90 lb)= − +P i j k
The moment of P about AB is given by Eq. (3.46):
/
0.64 0.60 0.48
( ) 5 in. 0 30 in.
105 lb 190 lb 90 lb
AB AB G B P
− −
= ⋅ × =
−
M rλ
0.64[0 (30 in.)( 190 lb)]
0.60[(30 in.)(105 lb) (5 in.)(90 lb)]
0.48[(5 in.)( 190 lb) 0]
2484 lb in.
AB = − −
− −
− − −
= + ⋅
M
207 lb ftAB = + ⋅M 
3.12
87

%%%%%%%%%Problem 3.57 How to calculate the moment of force
%%%%%%%%%about a definite straight line AB
%%%%%%%%%shown in figure using unit vector method to resolve the
%%%%%%%%%force acting P into horizontal and vertical component and resolve
%%%%%%%%%the rod AB into horizontal and vertical compnents of unity
%%unit vector of AB where AB = 32i - 30j - 24j
AB = [32 -30 -24]
lamdaAB =AB./(AB(1)^2 + AB(2)^2 + AB(3)^2)^0.5
%%unit vector of DG where DG = 16i - 38j + 18j
DG = [21 -38 18]
lamdaDG =DG./(DG(1)^2 + DG(2)^2 + DG(3)^2)^0.5
%%Calulating the acting force P of 235 LB in vector form as following:-
%%P = (Fx)i + (Fy)j + (Fz)K
%%P = 235*lamdaDG
P = 235*lamdaDG
%%Calulating the moment arm CD in vector form as shown in figure.
%% CD = (x)i + (y)j + (z)k
%% CD= 0i + 23j + 0k
CD = 23 %%%% DC = 23j
%%Calculating the moment acting about point C by force of 235 LB as
%%MC = CD x P by evaluating the cross product as following:-
%% | i j k |
%%MC =determinate of | 0 CD 0 |
%% | P(1) P(2) P(3)|
%%
MC = [CD*P(3) -CD*P(1)] %%%%MC = (MCx)i + (MCy)k in LB.in
%%Calculating the moment of force P about AB Rod
Mab=lamdaAB(1)*MC(1) + lamdaAB(3)*MC(2)%%%%%%Scalar value of acting
%%moment by force P about the rod AB in LB.in
88

218
PROBLEM 3.58
The 23-in. vertical rod CD is welded to the midpoint C of the
50-in. rod AB. Determine the moment about AB of the 174-lb
force Q.
SOLUTION
(32 in.) (30 in.) (24 in.)AB = − −i j k

2 2 2
(32) ( 30) ( 24) 50 in.AB = + − + − =
0.64 0.60 0.48AB
AB
AB
= = − −i j k

λ
We shall apply the force Q at Point H:
/ (32 in.) (17 in.)H B = − +r i j
(16 in.) (21 in.) (12 in.)DH = − − −i j k

2 2 2
(16) ( 21) ( 12) 29 in.DH = + − + − =
16 21 12
(174 lb)
29
DH
DH
− − −
= =
i j k
Q

(96 lb) (126 lb) (72 lb)Q = − − −i j k
The moment of Q about AB is given by Eq. (3.46):
/
0.64 0.60 0.48
( ) 32 in. 17 in. 0
96 lb 126 lb 72 lb
AB AB H B
− −
= ⋅ × = −
− − −
M r Qλ
0.64[(17 in.)( 72 lb) 0]
0.60[(0 ( 32 in.)( 72 lb)]
0.48[( 32 in.)( 126 lb) (17 in.)( 96 lb)]
2119.7 lb in.
AB = − −
− − − −
− − − − −
= − ⋅
M
176.6 lb ftAB = ⋅M 
89
3.13

%%%%%%%%%Problem 3.58 How to calculate the moment of force
%%%%%%%%%about a definite straight line AB
%%%%%%%%%shown in figure using unit vector method to resolve the
%%%%%%%%%force acting Q into horizontal and vertical component and resolve
%%%%%%%%%the rod AB into horizontal and vertical compnents of unity
%%unit vector of AB where AB = 32i - 30j - 24j
AB = [32 -30 -24]
lamdaAB =AB./(AB(1)^2 + AB(2)^2 + AB(3)^2)^0.5
%%unit vector of DH where DH = -16i - 21j - 12j
DH = [-16 -21 -12]
lamdaDH =DH./(DH(1)^2 + DH(2)^2 + DH(3)^2)^0.5
%%Calulating the acting force Q of 174 LB in vector form as following:-
%%Q = (Fx)i + (Fy)j + (Fz)K
%%Q = 174*lamdaDH
Q = 174*lamdaDH
%%Calulating the moment arm CD in vector form as shown in figure.
%% CD = (x)i + (y)j + (z)k
%% CD= 0i + 23j + 0k
CD = 23 %%%% DC = 23j
%%Calculating the moment acting about point C by force of 174 LB as
%%MC = CD x Q by evaluating the cross product as following:-
%% | i j k |
%%MC =determinate of | 0 CD 0 |
%% | Q(1) Q(2) Q(3)|
%%
MC = [CD*Q(3) -CD*Q(1)] %%%%MC = (MCx)i + (MCy)k in LB.in
%%Calculating the moment of force Q about AB Rod
Mab=lamdaAB(1)*MC(1) + lamdaAB(3)*MC(2) %%%%%%Scalar value of acting
%%moment by force Q about the rod AB in LB.in
90

271
PROBLEM 3.107
The weights of two children sitting at ends A and B of a seesaw
are 84 lb and 64 lb, respectively. Where should a third child sit
so that the resultant of the weights of the three children will
pass through C if she weighs (a) 60 lb, (b) 52 lb.
SOLUTION
(a) For the resultant weight to act at C, 0 60 lbC CM WΣ = =
Then (84 lb)(6 ft) 60 lb( ) 64 lb(6 ft) 0d− − =
2.00 ft to the right ofd C= 
(b) For the resultant weight to act at C, 0 52 lbC CM WΣ = =
Then (84 lb)(6 ft) 52 lb( ) 64 lb(6 ft) 0d− − =
2.31 ft to the right ofd C= 
3.14
91

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