Vector Mechanics For Engineers: Statics

Twelfth Edition

Chapter 2
Statics of Particles

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 Contents
Application Sample Problem 2.3
Introduction Equilibrium of a Particle
Forces on a Particle: Resultant of Two Free-Body Diagrams and Problem
Forces Solving
Vectors Sample Problem 2.4
Addition of Vectors Sample Problem 2.6
Resultant of Several Concurrent Forces Expressing a Vector in 3-D Space
Sample Problem 2.1 Sample Problem 2.7
Sample Problem 2.2
Rectangular Components of a Force:
Unit Vectors
Addition of Forces by Summing X and
Y Components
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 Application
The tension in the cable supporting this
person can be found using the concepts
in this chapter.

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 Introduction
The objective for the current chapter is to investigate the effects of forces on
particles:
• replacing multiple forces acting on a particle with a single equivalent or
resultant force,
• relations between forces acting on a particle that is in a state of equilibrium.

The focus on particles does not imply a restriction to miniscule bodies. Rather,
the study is restricted to analyses in which the size and shape of the bodies is
not significant to the problem under consideration, so that all forces may be
assumed to be applied at a single point.

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 Resultant of Two Forces
• force: action of one body on another;
characterized by its point of
application, magnitude, line of
action, and sense.

• 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.

• Force is a vector quantity.

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 Vectors
Vectors: parameters possessing magnitude and
direction that add according to the parallelogram
law. Examples: forces, displacements, velocities,
accelerations.
Scalars: 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.

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 Addition of Vectors
• Parallelogram law for vector addition
• Triangle rule for vector addition
• Law of cosines,
R 2  P 2  Q 2  2 PQ cos B
R  PQ

• Law of sines (using figure (a) at left),

sin A sin B sin C
 
Q R P

• Vector addition is commutative,

PQ Q P

• Vector subtraction

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 Resultant of Several Concurrent Forces
• Concurrent forces: set of forces that
all pass through the same point.
A set of concurrent forces applied to
a particle may be replaced by a
single resultant force that is the
vector sum of the applied forces.

• Vector force components: two or

more force vectors that, together,
have the same effect as a single force
vector.

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 Sample Problem 2.1 1

Strategy:
• Graphical solution - construct a
parallelogram with sides in the same
direction as P and Q and lengths in
proportion to these forces.
Graphically evaluate the resultant
that is equivalent in direction and
proportional in magnitude to the
diagonal.
The two forces act on a bolt at
A. Determine their resultant. • Trigonometric solution - use the
triangle rule for vector addition in
conjunction with the law of cosines
and law of sines to find the resultant.

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 Sample Problem 2.1 2

Modeling and Analysis:

• Graphical solution - A parallelogram
with sides equal to P and Q is drawn
to scale. The magnitude and direction
of the resultant (the diagonal of 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 (the third side of the
triangle) are measured,
R  98 N   35

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 Sample Problem 2.1 3

• Trigonometric solution - Apply the triangle rule.

From the Law of Cosines,
R 2  P 2  Q 2  2 PQ cos B
  40 N    60 N   2  40 N  60 N  cos155
2 2

R  97.73N

From the Law of Sines,

sin A sin B Q 60N
 ; sin A  sin B  sin155
Q R R 97.73N
A  15.04
  20  A
  35.04
Reflect and Think: An analytical solution using trigonometry provides for
greater accuracy. However, it is helpful to use a graphical solution as a check.

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 Sample Problem 2.2 1

Strategy:
• Find a graphical solution by applying
the Parallelogram Law 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 lb.
A barge is pulled by two
tugboats. If the resultant of the • Find a trigonometric solution by
forces exerted by the tugboats applying the Triangle Rule for vector
is 5000 lb directed along the addition. With the magnitude and
axis of the barge, determine the direction of the resultant known and
tension in each of the ropes the directions of the other two sides
when  = 45o. parallel to the ropes given, apply the
Law of Sines to find the rope tensions.
Discuss with a neighbor how
you would solve this problem.

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 Sample Problem 2.2 2

Modeling and Analysis:

• Graphical solution - Parallelogram
Law with known resultant direction
and magnitude, and known directions
for sides.
T1  3700lb T2  2600lb

• Trigonometric solution - Triangle Rule

with Law of Sines
T1 T2 5000lb
 
sin45 sin30 sin105

T1  3660lb T2  2590lb

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 What if…? 1

• At what value of a would the tension in

rope 2 be a minimum?
Hint: Use the triangle rule and think
about how changing α changes the
magnitude of T2. After considering this,
discuss your ideas with a neighbor.
• The minimum tension in rope 2 occurs
when T1 and T2 are perpendicular.
T2  5000lbsin30 T2  2500lb

T1  5000lbcos30 T1  4330lb

  90  30   60

Reflect and Think: Part (a) is a straightforward application of resolving a vector into
components. The key to part (b) is recognizing that the minimum value of T2 occurs
when T1 and T2 are perpendicular.

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 Rectangular Components of a Force: Unit Vectors
• It’s possible to 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
that are parallel to the x and y axes.
• Vector components can 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

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 Addition of Forces by Summing X and Y
Components
• To find the resultant of 3 (or more) concurrent
forces,
R  PQS
• Resolve each force into rectangular components,
then add the components in each direction:
Rx i  Ry j  Px i  Py j  Qx i  Qy j  S x i  S y j
  Px  Qx  S x  i   Py  Qy  S y  j

• The scalar components of the resultant vector

are equal to the sum of the corresponding scalar
components of the given forces.
Rx  Px  Qx  S x Ry  Py  Qy  S y
  Fx   Fy

• To find the resultant magnitude and direction,

Ry
R  Rx2  Ry2   tan 1
Rx

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 Sample Problem 2.3 1

Strategy:
• Resolve each force into rectangular
components.
• Determine the components of the
resultant by adding the
corresponding force components in
the x and y directions.
• Calculate the magnitude and
Four forces act on bolt A as direction of the resultant.
shown. Determine the resultant
of the force on the bolt.

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 Sample Problem 2.3 2

Analysis:
• Resolve each force into rectangular components.
Force mag x − comp y − comp
F1 150 +129.9 +75.0
F2 80 −27.4 +75.2
F3 110 0 −110.0
F4 100 +96.6 −25.9
blank blank
Rx  199.1 Ry  14.3

• Determine the components of the resultant by

Reflect and Think: adding the corresponding force components.
Arranging data in a table not
only helps you keep track of • Calculate the magnitude and direction.
the calculations, but also
R  199.12  14.32 R  199.6N
makes things simpler for
14.3 N
using a calculator on similar tan     4.1
199.1 N
computations.
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 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 Particle acted upon by three or more forces:

two forces: • graphical solution yields a closed polygon.
• equal magnitude. • algebraic solution.
• same line of action. R  F  0
• opposite sense. F x 0 F y 0
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 Free-Body Diagrams and Problem Solving

(a) Space diagram (b) Free-body diagram

Space Diagram: A sketch showing

Free Body Diagram: A sketch
the physical conditions of the
showing only the forces acting on
problem, usually provided with the
the selected particle. This must
problem statement, or represented
be created by you.
by the actual physical situation.

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 Sample Problem 2.4 1

Strategy:
• 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
In a ship-unloading operation, a magnitudes.
3500-lb 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?

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 Sample Problem 2.4 2

Analysis:
• Apply the conditions for equilibrium
and solve for the unknown force
magnitudes.

Law of Sines:
TAB T 3500lb
 AC 
sin120 sin 2 sin 58
Modeling:
TAB  3570lb
TAC  144lb

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 Sample Problem 2.4 3

Reflect and Think: This is a common problem of knowing one

force in a three-force equilibrium problem and calculating the
other forces from the given geometry. This basic type of problem
will occur often as part of more complicated situations in this
text.

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 Sample Problem 2.6 1

Strategy:
• Decide what the appropriate “body”
is and draw a free body diagram.
• The condition for equilibrium states
that the sum of forces equals 0, or:
R  F  0
It is desired to determine the drag
force at a given speed on a prototype F x 0 F y 0
sailboat hull. A model is placed in a
test channel and three cables are • The two equations means we can
used to align its bow on the channel solve for, at most, two unknowns.
centerline. For a given speed, the Since there are 4 forces involved
tension is 40 lb in cable AB and 60 (tensions in 3 cables and the drag
lb in cable AE. force), it is easier to resolve all
forces into components and apply
Determine the drag force exerted on the equilibrium conditions
the hull and the tension in cable AC.
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 Sample Problem 2.6 2

Modeling and Analysis:

• The correct free body diagram is
shown and the unknown angles are:

7 ft 1.5 ft
tan    1.75 tan    0.375
4 ft 4 ft
  60.25   20.56

• In vector form, the equilibrium

condition requires that the resultant
force (or the sum of all forces) be zero:

R  TAB  TAC  TAE  FD  0

• Write each force vector above in

component form.

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 Sample Problem 2.6 3

• Resolve the vector equilibrium equation

into two component equations. Solve
for the two unknown cable tensions.

TAB    40 lb  sin 60.26 i   40 lb  cos 60.26 j

   34.73 lb  i  19.84 lb  j
TAC  TAC sin 20.56 i  TAC cos 20.56 j
 0.3512 TAC i  0.9363TAC j
TAE    60 lb  j
FD  FD i

R0
  34.73  0.3512 TAC  FD  i
 19.84  0.9363TAC  60  j

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 Sample Problem 2.6 4

R0
  34.73  0.3512 TAC  FD  i
 19.84  0.9363TAC  60  j

This equation is satisfied only if each

component of the resultant is equal to zero

 F x  0  34.73  0.3512TAC  FD  0

 F y  0 19.84  0.9363TAC  60  0

TAC  42.9 lb
FD  19.66 lb
Reflect and Think: In drawing the free-body diagram, you assumed a sense
for each unknown force. A positive sign in the answer indicates that the
assumed sense is correct. You can draw the complete force polygon (above) to
check the results.
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 Expressing a Vector in 3-D Space 1

If angles with some of the axes are given:

• The vector F • Resolve F • Resolve F

is contained in into horizontal and into rectangular
the plane OBAC. vertical components. components.
Fy  F cos y Fx  Fh cos
Fh  F sin y  F sin y cos
Fz  Fh sin
 F sin y sin
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 Expressing a Vector in 3-D Space 2

If the direction angles are given:

• 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
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 Expressing a Vector in 3-D Space 3

If two points on the line of action are given:

Direction of the force is defined

by the location of two points,
M  x1 , y1 , z1  and N  x2 , y2 , z2 

d  vector joining M and N

 d xi  d y j  d z k
d x  x2  x1 d y  y2  y1 d z  z2  z1
F  F


1
d

d xi  d y j  d z k
Fd x Fd y Fd z
Fx  Fy  Fz 
d d d

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 Sample Problem 2.7 1

Strategy:
• 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.
• Noting that the components of the
The tension in the guy wire is 2500 N. unit vector are the direction
Determine: cosines for the vector, calculate the
a) components Fx, Fy, Fz of the force corresponding direction angles.
acting on the bolt at A,
b) the angles x, y, z defining the
direction of the force (i.e., the
direction angles)
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 Sample Problem 2.7 2

Modeling and Analysis:

• Determine the unit vector pointing from A
towards B.

AB   40m i  80m j  30mk

40m  80m  30m

2 2 2
AB 
 94.3 m

 40   80   30 
  i    j  k
 94.3   94.3   94.3 
 0.424i  0.848 j  0.318k

• Determine the components of the force.

F  F

  2500 N  0.424 i  0.848 j  0.318k 
  1060N i   2120 N  j  795 N k

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 Sample Problem 2.7 3

• 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

 0.424i  0.848 j  0.318k

 x  115.1
 y  32.0
 z  71.5

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 What if…? 2

• Since the force in the guy wire must

be the same throughout its length, the
force at B (and acting toward A) must
be the same magnitude but opposite
in direction to the force at A.

FBA   FAB
 1060N i   2120 N  j   795 N  k

Reflect and Think: It makes sense that,

What are the components of the for a given geometry, only a certain set
force in the wire at point B? Can of components and angles characterize a
you find it without doing any given resultant force. The methods in
calculations? this section allow you to translate back
and forth between forces and geometry.
Give this some thought and
discuss this with a neighbor.

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 End of Chapter 2

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