Vector Mechanics For Engineers: Statics
Twelfth Edition
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Chapter 2
Statics of Particles
Contents
Application
Sample Problem 2.3
Introduction
Equilibrium of a Particle
Forces on a Particle: Resultant of Two
Forces
Free-Body Diagrams and Problem
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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@ Michael Doolittle/Alamy
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
It states that the sum of the
of application that cannot be changed without
squares of the lengths of the
affecting an analysis.
four sides of a parallelogram
equals the sum of the squares • Free vectors may be freely moved in space
of the lengths of the two
without changing their effect on an analysis.
diagonals.
• 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:
A barge is pulled by two
tugboats. If the resultant of the
forces exerted by the tugboats
is 5000 lb directed along the
axis of the barge, determine the
tension in each of the ropes
when  = 45o.
Discuss with a neighbor how
you would solve this problem.
© 2019 McGraw-Hill Education.
• 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.
• 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.
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
  Fx
Ry  Py  Qy  S y
  Fy
• To find the resultant magnitude and direction,
R  Rx2  Ry2
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  tan 1
Ry
Rx
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.
Four forces act on bolt A as
shown. Determine the resultant
of the force on the bolt.
• Calculate the magnitude and
direction of the resultant.
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Sample Problem 2.3
2
Analysis:
• Resolve each force into rectangular components.
Force
mag
x − comp
F1
150
+129.9
+75.0
F2
80
−27.4
+75.2
F3
110
0
−110.0
F4
100
+96.6
−25.9
Rx  199.1
Ry  14.3
blank
Reflect and Think:
Arranging data in a table not
only helps you keep track of
the calculations, but also
makes things simpler for
using a calculator on similar
computations.
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blank
y − comp
• Determine the components of the resultant by
adding the corresponding force components.
• Calculate the magnitude and direction.
R  199.12  14.32
tan  
14.3 N
199.1 N
R  199.6N
  4.1
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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
two forces:
• equal magnitude.
• same line of action.
• opposite sense.
Particle acted upon by three or more forces:
• graphical solution yields a closed polygon.
• algebraic solution.
R  F  0
F
x
0
F
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y
0
Free-Body Diagrams and Problem Solving
(a) Space diagram
Space Diagram: A sketch showing
the physical conditions of the
problem, usually provided with the
problem statement, or represented
by the actual physical situation.
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(b) Free-body diagram
Free Body Diagram: A sketch
showing only the forces acting on
the selected particle. This must
be created by you.
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.
In a ship-unloading operation, a
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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• Apply trigonometric relations to
determine the unknown force
magnitudes.
Sample Problem 2.4
2
Analysis:
• Apply the conditions for equilibrium
and solve for the unknown force
magnitudes.
Law of Sines:
Modeling:
T
TAB
3500lb
 AC 
sin120 sin 2 sin 58
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:
It is desired to determine the drag
force at a given speed on a prototype
sailboat hull. A model is placed in a
test channel and 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.
R  F  0
F
x
F
y
0
• The two equations means we can
solve for, at most, two unknowns.
Since there are 4 forces involved
(tensions in 3 cables and the drag
force), it is easier to resolve all
forces into components and apply
the equilibrium conditions
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0
Sample Problem 2.6
2
Modeling and Analysis:
• The correct free body diagram is
shown and the unknown angles are:
1.5 ft
7 ft
tan


 0.375
tan  
 1.75
4 ft
4 ft
  20.56
  60.25
• 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
 F
x
y
 0  34.73  0.3512TAC  FD  0
 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
is contained in
the plane OBAC.
• Resolve F
into horizontal and
vertical components.
Fy  F cos y
Fh  F sin y
• Resolve F
into rectangular
components.
Fx  Fh cos
 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 xi  d y j  d z k
d
Fd y
Fd x
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.
The tension in the guy wire is 2500 N.
Determine:
a) components Fx, Fy, Fz of the force
acting on the bolt at A,
• Noting that the components of the
unit vector are the direction
cosines for the vector, calculate the
corresponding direction angles.
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
AB 
40m  80m  30m
2
2
2
 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
© 2019 McGraw-Hill Education.
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
What are the components of the
force in the wire at point B? Can
you find it without doing any
calculations?
Give this some thought and
discuss this with a neighbor.
© 2019 McGraw-Hill Education.
Reflect and Think: It makes sense that,
for a given geometry, only a certain set
of components and angles characterize a
given resultant force. The methods in
this section allow you to translate back
and forth between forces and geometry.
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End of Chapter 2
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