School of Mechanical
Engineering
MEE-1002: Engineering Mechanics
Statics of Particle
Dr Rahul S Sikarwar
Associate Professor
School of Mechanical Engineering,
Machine design section
VIT Vellore
Determine the tension in the Cable?
Frictional force?
Motion of the Snowboarder?
Objectives

Draw Free body diagram

Write and solve equilibrium equations for particles and rigid bodies

To analyse the structures.

To know the geometric properties of the different shapes.

To teach energy and momentum methods.
Outcomes
Student will be able to

Solve the engineering problems in case of equilibrium conditions.

Calculate the reaction forces of various supports of different structures.

Solve the problems involving dry friction.

Determine the centroid, centre of gravity and moment of inertia of various
surfaces and solids.

Solve the problems involving dynamics of particles and rigid bodies
Contents
• Unit I
Basics of Statics
• Unit II
Analysis of Structures
• Unit III
Friction
• Unit IV
Properties of Surfaces and Solids
• Unit V
Virtual Work
• Unit VI
Kinematics
• Unit VII
Energy and Momentum Methods
Text Books
1. Ferdinand P. Beer, E. Russell Johnston (2007), “Vector
Mechanics for Engineers: Statics and Dynamics”, McGraw-Hill
International Edition.
2. J. L. Meriam and L. G. Kraige (2006), “Engineering Mechanics:
Statics and Dynamics (6th Edition)”, Wiley Publishers
References
1. Irving H. Shames, (2003), “Engineering Mechanics – Statics and
Dynamics”, Prentice-Hall of India Private limited.
2. Russell C Hibbeler, (2009), “Engineering Mechanics: Statics and
Dynamics (12th Edition)”, Prentice Hall.
3. Anthony M. Bedford and Wallace Fowler (2007), “Engineering
Mechanics: Statics and Dynamics (5th Edition)”, Prentice Hall.



What is Mechanics?
Mechanics is a physical science which deals with bodies at rest
or motion under the action of forces.
Mechanics is an applied science - it is not an abstract or pure
science.
Mechanics is the foundation of most engineering sciences.
Mechanics
Mechanics of
rigid bodies
Statics
Mechanics of
Deformable bodies
Dynamics
Kinematics
Kinetics
Mechanics of fluids
Rigid Body Mechanics
Statics
Deals with bodies at rest
Displacement
Kinematics
Velocity
Acceleration
Dynamics
Force/Torque
Kinetics
Energy/momentum
Engineering Mechanics
Rigid-body Mechanics
Statics: deals with equilibrium of bodies under action of forces (bodies
may be either at rest or move with a constant velocity).
Engineering Mechanics
Rigid-body Mechanics
Dynamics: deals with motion of bodies (accelerated motion)
Fundamental concepts
• Space - associated with the representation of the position of a point P
given in terms of three coordinates measured from a reference point
or origin.
– Length :: needed to locate position of a point in space, &describe
size of the physical system Distances, Geometric Properties
• Time – definition of an event requires specification of the time and
position at which it occurred.
– measure of succession of events basic quantity in Dynamics
• Mass – used to characterize and compare bodies. Two bodies of the
same mass, for example, response to earth’s gravitational attraction
and resistance to changes in translational motion.
– quantity of matter in a body measure of inertia of a body (its
resistance to change in velocity)
– Mass of a body is the quantitative measure of its inertia
Force represents the action of one body on another.
It changes (or) tends to change the state of rest or union motion
of a body
A force is characterized by its point of application, magnitude,
and direction, i.e., a force is a vector.
(Free vector, Sliding Vector, Fixed Vector)
In Newtonian Mechanics, space, time, and mass are absolute
concepts, independent of each other. Force, however, is not
independent of the other three. The force acting on a body is
related to the mass of the body and the variation of its velocity
with time.
Mechanics: Fundamental Concepts
Newtonian Mechanics
Length, Time, and Mass are absolute concepts independent of each other
Force is a derived concept not independent of the other fundamental concepts.
Force acting on a body is related to the mass of the body and the variation of its
velocity with time.
Force can also occur between bodies that are physically separated (Ex: gravitational,
electrical, and magnetic forces)
Remember:
•
Mass is a property of matter that does not change from one location to another.
•
Weight refers to the gravitational attraction of the earth on a body or quantity of
mass. Its magnitude depends upon the elevation at which the mass is located
•
Weight of a body is the gravitational force acting on it.
Mechanics: Idealizations
To simplify application of the theory
Particle: A body with mass but with dimensions that can be neglected.
Size of earth is insignificant
compared to the size of its
orbit. Earth can be modeled
as a particle when studying
its orbital motion
Mechanics: Idealizations
Rigid Body: A combination of large number of particles in which all
particles remain at a fixed distance (practically) from one another
before and after applying a load.
Material properties of a rigid body are not required to be considered
when analyzing the forces acting on the body.
In most cases, actual deformations occurring in structures, machines,
mechanisms, etc. are relatively small, and rigid body assumption is
suitable for analysis
Mechanics: Idealizations
Concentrated Force: Effect of a loading which is assumed to act at a
point (CG) on a body.
Provided the area over which the load is applied is very small compared
to the overall size of the body.
Ex: Contact Force between a wheel and ground.
Mechanics: Newton’s Three Laws of Motion
First Law: A particle originally at rest, or moving in a straight line with
constant velocity, tends to remain in this state provided the particle is
not subjected to an unbalanced force.
First law contains the principle of the
equilibrium of forces main topic of concern
in Statics
Mechanics: Newton’s Three Laws of Motion
Second Law: A particle of mass “m” acted upon by an unbalanced force “F”
experiences an acceleration “a” that has the same direction as the force and a
magnitude that is directly proportional to the force.
Second Law forms the basis for most of the analysis in Dynamics
Mechanics: Newton’s Three Laws of Motion
Third Law: The mutual forces of action and reaction between two
particles are equal, opposite, and collinear.
Third law is basic to our understanding of Force Forces always occur in
pairs of equal and opposite forces.
Mechanics: Newton’s Law of Gravitation
This states that two particles of mass m1 and m2 are mutually
attracted with equal and opposite forces F and –F of magnitude F given
by the formula.
m1m 2
F G 2
r
Where r = distance between the two particles
G = universal constant called the constant of gravitation
Application: The force F exerted by the earth on the particle located on
its surface, is defined as the weight W of the particle.
GMme
W
2
r
GM
g 2
r
W  mg
Mechanics: Fundamental Laws
Parallelogram Law
If two vector quantities are represented by
two adjacent sides or a parallelogram then
the diagonal of parallelogram will be equal to
the resultant of these two vectors
Parallelogram Law
Principle of Transmissibility
The effect of an external force on a rigid body
remains unchanged if that force is moved
along its line of action. Warning:
This principle is not valid for deformable
bodies.
Principle of Transmissibility
Systems of Units
• International System of Units (SI):
The basic units are length, time, and mass
which are arbitrarily defined as the meter
• An equation must be dimensionally
(m), second (s), and kilogram (kg). Force is
Homogeneous, i.e., Left and Right terms
the derived unit,
must be identical
• Kinetic Units: length, time, mass, and
force.
F  ma
 m


1 N  1 kg 1 2 
 s 
• Three of the kinetic units, referred to as
basic units, may be defined arbitrarily. • U.S.
Customary
Units:
The fourth unit, referred to as a derived
The basic units are length, time, and force
unit, must have a definition compatible
which are arbitrarily defined as the foot (ft),
with Newton’s 2nd Law,
second (s), and pound (lb). Mass is the

derived unit,

F  ma
F
a
1lb
1slug 
1ft s
m
Mechanics: Units Prefixes
Method of Problem Solution
• Problem Statement:
Includes given data, specification of
what is to be determined, and a
figure showing all quantities
involved.
• Free-Body Diagrams:
Create separate diagrams for each of
the bodies involved with a clear
indication of all forces acting on
each body.
• Fundamental Principles:
The six fundamental principles are
applied to express the conditions of
rest or motion of each body. The
rules of algebra are applied to solve
the equations for the unknown
quantities.
• Solution Check:
• Test for errors in reasoning by
verifying that the units of the
computed results are correct
• test for errors in computation by
substituting given data and
computed results into previously
unused
equations based on
the six principles
• always apply experience and
physical intuition to assess
whether
results
seem
“reasonable”.
Numerical Accuracy
• The accuracy of a solution depends on 1) accuracy of the given
data, and 2) accuracy of the computations performed. The
solution cannot be more accurate than the less accurate of these
two.
• The use of hand calculators and computers generally makes the
accuracy of the computations much greater than the accuracy of the
data. Hence, the solution accuracy is usually limited by the data
accuracy.
• As a general rule for engineering problems, the data are seldom
known with an accuracy greater than 0.2%. Therefore, it is usually
appropriate to record parameters beginning with “1” with four digits
and with three digits in all other cases, i.e., 40.2 lb and 15.58 lb.
1 - 25
Statics of Particles
Many engineering problems
can be solved by considering
the
equilibrium
of
a
“particle.”
In the case of this excavator,
which is being loaded onto a
ship, a relation between the
tensions in the various
cables involved can be
obtained by considering the
equilibrium of the hook to
which the cables are
attached.
Statics of Particles
 Particle: Size and shape of the body are neglected, but not the
mass.
 Forces in a plane
 Vectors
 Resultant of several concurrent forces (Polygon law)
 Resolution of a force into components
 Rectangular components of a force: Unit vectors
 Addition of forces by summing X and Y components
 Equilibrium of a Particle
Forces in a plane
A force represents the action of one body on another and is
generally characterized by its point of application, its magnitude,
and its direction.
Characteristics of a force are its
1. Magnitude
2. Point of application
3. Direction
Unit: Newton (N)
1 Newton is defined as the force which gives an acceleration of 1
m/s2 to a mass of 1 kg.
•
•
•
•
Concentrated force or a point force: it is the force acting on a very small
area.
Distributed force: force distributed over a length or an area or a volume
Line of action of force: Direction along a straight line through its point of
application, in which the force tends to move a body to which it is
applied.
Graphical representation of force:
• Experimental evidence shows that the
combined effect of two forces may be
represented by a single resultant force.
• The resultant is equivalent to the diagonal
of a parallelogram which contains the two
forces in adjacent legs.
System of forces
• when several forces of various magnitude and direction act upon a body
they are said to form system of forces.
• Classification of system of forces: Classified as per orientation of line of
action of forces.
System of forces
Coplanar ( In Plane)/ 2-D
Concurrent
Non Coplanar(Space)-3-D
Non- concurrent
Parallel
Concurrent
General
Non- concurrent
Parallel
Note:
Concurrent force system: acts on a particle or rigid body
Parallel and General force system: acts mainly on rigid bodies
General
Vectors
Vector: parameters possessing magnitude and direction
which add according to the parallelogram law. Examples:
displacements, velocities, accelerations.
Scalar:
parameters possessing magnitude but not
direction. Examples: mass, volume, temperature
Vector classifications:
- Fixed or bound vectors have well defined points of
application that cannot be changed without affecting
an analysis.
- Free vectors may be freely moved in space without
changing their effect on an analysis.
- Sliding vectors may be applied anywhere along their
line of action without affecting an analysis.
Equal vectors have the same magnitude and direction.
Negative vector of a given vector has the same magnitude
and the opposite direction.
2 - 31
Addition of Vectors
• Trapezoid rule for vector addition
• Triangle rule for vector addition
• Law of cosines,
C
B
C
B
R 2  P 2  Q 2  2 PQ cos B
  
R  PQ
• Law of sines,
sin A sin B sin C


Q
R
A
• Vector addition is commutative,
   
PQ  Q P
• Vector subtraction


 
P  (Q)  P  Q
2 - 32
Parallelogram law
If two forces represented by vectors AB and AC acting under an angle α are applied to
a body at point A. Their action is equivalent to the action of one force, represented by
vector AD, obtained as the diagonal of the parallelogram constructed on the vectors
AB and AC directed as shown in the figure.
Force AD is called the resultant of AB and AC and the forces are called its components.
Lami’s Theorem
If the three forces acting at a point are in equilibrium, each force is proportional to
the sine of the angle between the other two forces.
F1
θ3
θ2
F2
O
F3
θ1
F3
F1
F2


Sin 1 Sin 2 Sin 3
Addition of Vectors
• Addition of three or more vectors through
repeated application of the triangle rule
• The polygon rule for the addition of three or
more vectors.
• Vector addition is associative,
  
     
P  Q  S  P  Q   S  P  Q  S 
• Multiplication of a vector by a scalar
2 - 38
Resultant of Several Concurrent Forces
• Concurrent forces: set of forces which all
pass
through
the
same
point.
A set of concurrent forces applied to a
particle may be replaced by a single
resultant force which is the vector sum of
the applied forces.
• Vector force components: two or more
force vectors which, together, have the same
effect as a single force vector.
2 - 39
Example 1.
SOLUTION:
The two forces act on a bolt at A.
Determine their resultant.
• Graphical solution - construct a
parallelogram with sides in the same
direction as P and Q and lengths in
proportion. Graphically evaluate the
resultant which is equivalent in direction
and proportional in magnitude to the
diagonal.
• Trigonometric solution - use the triangle
rule for vector addition in conjunction
with the law of cosines and law of sines
to find the resultant.
2 - 44
• Graphical solution - A parallelogram with sides
equal to P and Q is drawn to scale. The magnitude
and direction of the resultant or of the diagonal to
the parallelogram are measured,
R  98 N   35
• Graphical solution - A triangle is drawn with P and Q
head-to-tail and to scale. The magnitude and
direction of the resultant or of the third side of the
triangle are measured,
R  98 N   35
2 - 45
Trigonometric solution - Apply the triangle rule.
From the Law of Cosines,
R 2  P 2  Q 2  2 PQ cos B
 40 N 2  60 N 2  240 N 60 N  cos 155
R  97.73N
From the Law of Sines,
sin A sin B

Q
R
sin A  sin B
Q
R
 sin 155
A  15.04
  20  A
  35.04
2 - 46
60 N
97.73N
Example 2.
SOLUTION:
• Find a graphical solution by applying the
Parallelogram Rule for vector addition. The
parallelogram has sides in the directions of the
two ropes and a diagonal in the direction of the
barge axis and length proportional to 5000 lbf.
A barge is pulled by two tugboats. If
the resultant of the forces exerted by
the tugboats is 5000 lbf directed
along the axis of the barge, determine
• Find a trigonometric solution by applying the
Triangle Rule for vector addition. With the
magnitude and direction of the resultant known
and the directions of the other two sides
parallel to the ropes given, apply the Law of
Sines to find the rope tensions.
a) the tension in each of the ropes for
 = 45o,
• The angle for minimum tension in rope 2 is
b) the value of  for which the tension determined by applying the Triangle Rule and
observing the effect of variations in .
in rope 2 is a minimum.
2 - 47
• Graphical solution - Parallelogram Rule with
known resultant direction and magnitude,
known directions for sides.
T1  3700 lbf
T2  2600 lbf
• Trigonometric solution - Triangle Rule with
Law of Sines
T1
T2
5000 lbf


sin 45 sin 30 sin 105
T1  3660 lbf
2 - 48
T2  2590 lbf
• The angle for minimum tension in rope 2 is
determined by applying the Triangle Rule and
observing the effect of variations in .
• The minimum tension in rope 2 occurs when T1
and T2 are perpendicular.
2 - 49
T2  5000 lbf  sin 30
T2  2500 lbf
T1  5000 lbf  cos 30
T1  4330 lbf
  90  30
  60
Rectangular Components of a Force: Unit Vectors
• May resolve a force vector into perpendicular
components so that the resulting parallelogram is a


rectangle. Fx and Fy
are referred to as rectangular
vector components and
 

F  Fx  Fy


• Define perpendicular unit vectors i and j which are
parallel to the x and y axes.
• Vector components may be expressed as products of
the unit vectors with the scalar magnitudes of the
vector components.



F  Fx i  Fy j

Fx and Fy are referred to as the scalar components of F
2 - 50
Addition of Forces by Summing Components
• Wish to find the resultant of 3 or more concurrent
forces,
   
R  PQ S
• Resolve each force into rectangular components








R x i  R y j  Px i  Py j  Q x i  Q y j  S x i  S y j


  Px  Q x  S x i  Py  Q y  S y j


• The scalar components of the resultant are equal to
the sum of the corresponding scalar components of
the given forces.
Rx  Px  Qx  S x
  Fx
R y  Py  Q y  S y
  Fy
• To find the resultant magnitude and direction,
R
2 - 51
Rx2
 R y2
  tan
1
Ry
Rx
Example 3.
SOLUTION:
• Resolve each force into rectangular
components.
• Determine the components of the
resultant by adding the corresponding
force components.
Four forces act on bolt A as shown.
Determine the resultant of the force on
the bolt.
2 - 52
• Calculate the magnitude and direction of
the resultant.
SOLUTION:
• Resolve each force into rectangular components.
force mag

F1 150

F2
80

F3 110

F4 100
x  comp
y  comp
 129.9
 75.0
 27.4
 75.2
0
 110.0
 96.6
 25.9
R x  199.1 R y  14.3
• Determine the components of the resultant by
adding the corresponding force components.
• Calculate the magnitude and direction.
R  199.12  14.32
14.3 N
tan  
199.1 N
2 - 53
R  199.6N
  4.1
Determine the x and y components of each of the forces and resultant of three forces
shown.
Determine the x and y components of each of the forces and resultant of three
forces shown.
Determine the x and y components of each of the forces and resultant of three forces
shown.
Member BD exerts on member ABC a force P directed along line BD. Knowing that
P must have a 960-N vertical component, determine
(a) the magnitude of the force P, (b) its horizontal component.
Knowing that the tension in cable BC is 145 N, determine the resultant of the three
forces exerted at point B of beam AB.
Knowing that α = 65°, determine the resultant of the three forces shown.
α = 65°,
Equilibrium of a Particle
• When the resultant of all forces acting on a particle is zero, the particle is in
equilibrium.
• Newton’s First Law: If the resultant force on a particle is zero, the particle will remain at
rest or will continue at constant speed in a straight line.
• Particle acted upon by two
forces:
- equal magnitude
- same line of action
- opposite sense
2 - 60
• Particle acted upon by three or more forces:
- graphical solution yields a closed polygon
- algebraic solution


R  F  0
 Fx  0
 Fy  0
Free-Body Diagrams
Free body diagram: The sketch of the isolated body which shows
the external force on the body and the reactions on it by the removed
elements.
General Procedure for construction of free body diagram
1. Draw the sketch of body without supporting surfaces.
2. Show all the external or applied forces on this sketch
3. Show all the reactive forces such as those caused by constraints
or supports
4. Show all relevant dimensions , angles and reference axes on the
sketch
External and Internal Forces
• Forces acting on rigid bodies are divided into two groups:
- External forces
- Internal forces
• External forces are shown in a free-body diagram.
• If unopposed, each external force can impart a motion of
translation or rotation, or both.
Types of support
1. Frictionless Support: Reactions are normal to the surface at the
point of contact
W
c
W
A
RA
2. Hinge support or pin jointed support:
RAx
RAy
3. Roller Support :
4. Fixed Support :
RA
RAx
RAy
MA
Draw FDB for following system
WP
WQ
RA
RC
WP
RA
WQ
RD
RB
RC
RD
Sample Free Body Diagrams
Draw FDB for following system
WP
WQ
RA
RC
WP
RA
WQ
RD
RB
RC
RD
Free-Body Diagrams
Space Diagram: A sketch showing the
physical conditions of the problem.
2 - 68
Free-Body Diagram: A sketch showing only
the forces on the selected particle.
Example 4.
SOLUTION:
• Construct a free-body diagram for the
particle at the junction of the rope and
cable.
• Apply the conditions for equilibrium
by creating a closed polygon from the
forces applied to the particle.
• Apply trigonometric relations to
determine the unknown force
magnitudes.
In a ship-unloading operation, a 3500lb automobile is supported by a cable.
A rope is tied to the cable and pulled to
center the automobile over its
intended position. What is the tension
in the rope?
2 - 69
SOLUTION:
• Construct a free-body diagram for the particle at A.
• Apply the conditions for equilibrium.
• Solve for the unknown force magnitudes.
T
T AB
3500 lb
 AC 
sin 120 sin 2 sin 58
TAB  3570 lb
T AC  144 lb
2 - 70
Example 5.
SOLUTION:
• Choosing the hull as the free body, draw a
free-body diagram.
• Express the condition for equilibrium for
the hull by writing that the sum of all
forces must be zero.
It is desired to determine the drag force at
a given speed on a prototype sailboat hull. • Resolve the vector equilibrium equation
into two component equations. Solve for
A model is placed in a test channel and
the two unknown cable tensions.
three cables are used to align its bow on
the channel centerline. For a given speed,
the tension is 40 lb in cable AB and 60 lb in
cable AE.
Determine the drag force exerted on the
hull and the tension in cable AC.
2 - 71
SOLUTION:
• Choosing the hull as the free body, draw a
free-body diagram.
7 ft
 1.75
4 ft
  60.25
tan  
1.5 ft
 0.375
4 ft
  20.56
tan  
• Express the condition for equilibrium for
the hull by writing that the sum of all
forces must be zero.
 



R  TAB  TAC  TAE  FD  0
2 - 72
• Resolve the vector equilibrium equation into two
component equations. Solve for the two unknown
cable tensions.







T AB   40 lb sin 60.26 i  40 lb cos 60.26 j


 34.73 lb i  19.84 lb j



T AC  T AC sin 20.56 i  T AC cos 20.56 j


 0.3512T AC i  0.9363T AC j


T  60 lb i


FD  FD i

R0

  34.73  0.3512T AC  FD  i



 19.84  0.9363T AC  60 j
2 - 73

R0

  34.73  0.3512T AC  FD  i

 19.84  0.9363T AC  60 j
This equation is satisfied only if each component
of the resultant is equal to zero
 Fx  0 0  34.73  0.3512TAC  FD
 Fy  0 0  19.84  0.9363TAC  60
T AC  42.9 lb
FD  19.66 lb
2 - 74
Example 7.
Determine the tension developed in each cord required for equilibrium of the 20-kg
lamp.
Example 7.
Determine the tension developed in each cord required for equilibrium of the 20-kg
lamp.
SOLUTION
Two cables are tied together at C and are loaded as shown. Determine
the tension
(a) in cable AC, (b) in cable BC.
Ans: (a) 2.20 kN, (b) 2.43 kN
Knowing that α = 50° 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.
Ans: 169.7 kN, (b) 348 kN
A boat is pulling a parasail and rider at a constant speed. Knowing that
the rider weighs 550 N and that the resultant force R exerted by the
parasail on the towing yoke A forms an angle of 65° with the
horizontal, determine
(a) the tension in the tow rope AB, (b) the magnitude of R.
Ans: (a) 405 N, (b) 830 N
A load of weight 400 N is suspended from a spring and two cords that
are attached to blocks of weights 3W and W as shown. Knowing that
the constant of the spring is 800 N/m, determine
(a) the value of W, (b) the unstretched length of the spring.
Ans: (a) 62.8 N, (b)758 mm
Two cables tied together at C are loaded as shown. Knowing that W =
190 N, determine the tension (a) in cable AC, (b) in cable BC.
Determine the range of values of W for which the tension will not
exceed 240 N in either cable.
Ans: (a) 169.6 N, (b) 265 N
Two traffic signals are temporarily suspended from a cable as shown.
Knowing that the signal at B weighs 300 N, determine the weight of
the signal at C.
Ans: 97.7 N
Example 8
The cylinders in Figure have the indicated weights and dimensions. Assuming smooth
contact surfaces, determine the reactions at A, B, C, and D on the cylinders.
Free Body Diagram
Rectangular Components in Space

F
The vector
is
contained in the
plane OBAC.

Resolve F into
horizontal and vertical
components.
Fy  F cos  y
Fh  F sin  y
2 - 88
Resolve Fh into
rectangular components
Fx  Fh cos
 F sin  y cos
Fz  Fh sin 
 F sin  y sin 
Rectangular Components in Space

• With the angles between F and the axes,
Fx  F cos  x Fy  F cos  y Fz  F cos  z




F  Fx i  Fy j  Fz k



 F cos  x i  cos  y j  cos  z k

 F




  cos  x i  cos  y j  cos  z k


•  is a unit vector along the line of action of F
and cos  x , cos  y , and cos  z are the direction
cosines for F

2 - 89

Rectangular Components in Space
Direction of the force is defined by
the location of two points,
M  x1 , y1 , z1  and N  x2 , y2 , z 2 

d  vector joining M and N



 d xi  d y j  d z k
2 - 90
d x  x2  x1 d y  y 2  y1 d z  z 2  z1


F  F

 1


  d x i  d y j  d z k 
d
Fd y
Fd x
Fd z
Fx 
Fy 
Fz 
d
d
d
Example 9.
SOLUTION:
• Based on the relative locations of the points
A and B, determine the unit vector pointing
from A towards B.
• Apply the unit vector to determine the
components of the force acting on A.
The tension in the guy wire is 2500 N.
Determine:
a) components Fx, Fy, Fz of the force
acting on the bolt at A,
b) the angles qx, qy, qz defining the
direction of the force
2 - 91
• Noting that the components of the unit
vector are the direction cosines for the
vector, calculate the corresponding angles.
SOLUTION:
• Determine the unit vector pointing from A towards
B.



AB   40 m  i  80 m  j  30 m k
AB 
 40 m 2  80 m 2  30 m 2
 94.3 m
  40    80    30  
 
i  
 j 
k
94
.
3
94
.
3
94
.
3

 
 




 0.424 i  0.848 j  0.318k

• Determine the components of the force.


F  F



 2500 N  0.424 i  0.848 j  0.318k 



  1060 N i  2120 N  j  795 N k
2 - 92
• Noting that the components of the unit vector are
the direction cosines for the vector, calculate the
corresponding angles.



  cos  x i  cos  y j  cos  z k



 0.424 i  0.848 j  0.318k

 x  115.1
 y  32.0
 z  71.5
2 - 93
Example 10
The end of the coaxial cable AE is attached to the pole AB, which is strengthened by the
guy wires AC and AD. Knowing that the tension in wire AD is 85 lb, determine (a) the
components of the force exerted by this wire on the pole, (b) the angles θx, θy, and θz
that the force forms with the coordinate axes.
Solution
Resolve Fh into
rectangular components
Fx  Fh cos
 F sin  y cos
Fz  Fh sin 
 F sin  y sin 
Example 11
A transmission tower is held by three guy wires anchored by bolts at B, C,
and D. If the tension in wire AB is 2100 N, determine the components of the
force exerted by the wire on the bolt at B.
Example 12
A transmission tower is held by three guy wires anchored by bolts at B, C, and D. If the
tension in wire AD is 1260 N, determine the components of the force exerted by the
wire on the bolt at D.
Example 13
A horizontal circular plate is suspended as shown from three wires which are
attached to a support at D and form 30 degrees angles with the vertical. Knowing
that the tension in wire CD is 120 lb, determine (a) the components of the force
exerted by this wire on the plate, (b) the angles θx , θy, and θz that the force forms
with the coordinate axes.
(a) the components of the force exerted by this wire on the plate,
(b) the angles θx , θy, and θz that the force forms with the
coordinate axes.
Example 14
A rectangular plate is supported by three cables as shown. Knowing that the tension in
cable AD is 195 lb, determine the components of the force exerted on the plate at D.
Solution
H.W Problem: Example 14
H.W Problem: Example 15
The support assembly shown is bolted in place at B, C, and D and supports a
downward force P at A. Knowing that the forces in members AB, AC, and AD are
directed along the respective members and that the force in member AB is 29.2 N,
determine the magnitude of P.
ANS: 55.9 N
1.
2.
A transmission tower is held by three guy wires anchored by bolts at B, C, and
D. If the tension in wire AB is 2100 N, determine the components of the force
exerted by the wire on the bolt at B.
A transmission tower is held by three guy wires anchored by bolts at B, C, and
D. If the tension in wire AD is 1260 N, determine the components of the force
exerted by the wire on the bolt at D.
Two cables BG and BH are attached to the frame ACD as shown.
Knowing that the tension in cable BG is 450 N, determine the
components of the force exerted by cable BG on the frame at B.
Two cables BG and BH are attached to the frame ACD as shown.
Knowing that the tension in cable BH is 600 N, determine the
components of the force exerted by cable BH on the frame at B.
The boom OA carries a load P and is supported by two cables as
shown. Knowing that the tension is 510 N in cable AB and 765 N in
cable AC, determine the magnitude and direction of the resultant of the
forces exerted at A by the two cables.
For the transmission tower shown below, determine the tensions in
cables AB and AD knowing that the tension in cable AC is 1770 N
and that the resultant of the forces exerted by the three cables at A
must be vertical.
For the boom shown below, knowing that α = 0°, the tension in cable
AB is 600 N, and the resultant of the load P and the force exerted at
A by the two cables is directed along OA, determine (a) the tension
in cable AC, (b) the magnitude of the load P.
Thank You