Engineering Mechanics:
Statics
Chapter 2: Force Systems
Force Systems
Part A: Two Dimensional Force Systems
Force

An action of one body on another
Vector quantity

External and Internal forces

Mechanics of Rigid bodies: Principle of Transmissibility

• Specify magnitude, direction, line of action
• No need to specify point of application

Concurrent forces
• Lines of action intersect at a point
Vector Components

A vector can be resolved into several vector components

Vector sum of the components must equal the original vector

Do not confused vector components with perpendicular
projections
Rectangular Components
2D force systems

• Most common 2D resolution of a force vector
• Express in terms of unit vectors ˆi , ˆj
y
j
F
Fy
F  Fx  Fy  Fxˆi  Fy ˆj

i
Fx  F cos  ,
Fy  F sin
x
Fx
F  F  Fx 2  Fy 2
  tan1
Fy
Fx
Scalar components – can
be positive and negative
2D Force Systems

Rectangular components are convenient for finding
the sum or resultant R of two (or more) forces which
are concurrent
R  F1  F2  (F1xˆi  F1yˆj )  (F2 xˆi  F2 yˆj )
= (F1x  F2 x )ˆi  (F1y  F2 y )ˆj
Actual problems do not come with reference axes. Choose the most convenient one!
Example 2.1

The link is subjected to two forces F1
and F2. Determine the magnitude
and direction of the resultant force.
Solution
FR 
 236.8N2   582.8N2
 629 N
  tan1  582.8N 
 236.8N 
 67.9
Example 2/1 (p. 29)
Determine the x and y scalar components of each of the three forces
Rectangular components
Unit vectors

•n
y
j
= Unit vector in direction of V
n
V
Vy
n 
y
x
i
Vx
V
V

Vxˆi  Vyˆj
V

V
Vx ˆ
i  y ˆj
V
V
 cos xˆi  cos yˆj
x
Vx
V
 cos x  direction cosine
cosx2  cosy 2  1
Problem 2/4
The line of action of the 34-kN force runs through the points A and
B as shown in the figure.
(a) Determine the x and y scalar component of F.
(b) Write F in vector form.
Moment



In addition to tendency to move a body
in the direction of its application, a force
tends to rotate a body about an axis.
The axis is any line which neither intersects
nor is parallel to the line of action
This rotational tendency is known as the
moment M of the force
 Proportional to force F and the
perpendicular distance from the axis
to the line of action of the force d
 The magnitude of M is
M = Fd
Moment


The moment is a vector M perpendicular
to the plane of the body.
Sense of M is determined by the righthand rule




Direction of the thumb = arrowhead
Fingers curled in the direction of the
rotational tendency
In a given plane (2D),we may speak of
moment about a point which means
moment with respect to an axis normal to
the plane and passing through the point.
+, - signs are used for moment directions –
must be consistent throughout the
problem!
Moment


A vector approach for moment
calculations is proper for 3D problems.
Moment of F about point A maybe
represented by the cross-product
M=rxF
where r = a position vector from point A to
any point on the line of action of F
M = Fr sin a = Fd
Example 2/5 (p. 40)
Calculate the magnitude of the moment
about the base point O of the 600-N
force by using both scalar and vector
approaches.
Problem 2/43
(a) Calculate the moment of the 90-N force
about point O for the condition  = 15º.
(b) Determine the value of  for which the
moment about O is (b.1) zero (b.2) a
maximum
Couple

Moment produced by two equal, opposite,
and noncollinear forces = couple
M = F(a+d) – Fa = Fd


Moment of a couple has the same value
for all moment center
Vector approach
M = rA x F + rB x (-F) = (rA - rB) x F = r x F

Couple M is a free vector
Couple

Equivalent couples
 Change of values F and d
 Force in different directions but parallel plane
 Product Fd remains the same
Force-Couple Systems


Replacement of a force by a force and a couple
Force F is replaced by a parallel force F and a
counterclockwise couple Fd
Example Replace the force by an equivalent system at point O
Also, reverse the problem by the replacement of
a force and a couple by a single force
Problem 2/67
The wrench is subjected to the 200-N force and
the force P as shown. If the equivalent of the two
forces is a for R at O and a couple expressed as
the vector M = 20 kN.m, determine the vector
expressions for P and R
Resultants


The simplest force combination which can
replace the original forces without
changing the external effect on the rigid
body
Resultant = a force-couple system
R  F1  F2  F3 
 F
Rx  Fx , Ry  Fy , R  (Fx )2  (Fy )2
-1 Ry
 = tan
Rx
Resultants



Choose a reference point (point O) and
move all forces to that point
Add all forces at O to form the resultant
force R and add all moment to form the
resultant couple MO
Find the line of action of R by requiring R to
have a moment of MO
R  F
MO  M  (Fd)
Rd = MO
Problem 2/79
Replace the three forces acting on the bent pipe
by a single equivalent force R. Specify the
distance x from point O to the point on the x-axis
through which the line of action of R passes.
Force Systems
Part B: Three Dimensional Force Systems
Three-Dimensional Force System

Rectangular components in 3D
• Express in terms of unit vectors
ˆi, ˆj, kˆ
F  Fxˆi  Fyˆj  Fz kˆ
Fx  F cosx ,
Fy  F cosy , Fz  F cos  z
F  Fx 2  Fy 2  Fz 2
• cosx, cosy , cosz are the direction cosines
• cosx = l, cosy = m, cos z= n
F  F(liˆ  mjˆ  nkˆ)
Three-Dimensional Force System

Rectangular components in 3D
• If the coordinates of points A and B on the line
of action are known,
(x2  x1)ˆi  (y2  y1)ˆj  (z2  z1)kˆ
AB
F  FnF  F
F
AB
(x2  x1)2  (y2  y1)2  (z2  z1)2
• If two angles  and f which orient the line of
action of the force are known,
Fxy  F cos f ,
Fz  F sinf
Fx  F cos f cos  ,
Fy  F cos f sin
Problem 2/98

The cable exerts a tension of 2 kN on the fixed bracket at A.
Write the vector expression for the tension T.
Three-Dimensional Force System

Dot product
P  Q  PQ cos a


Orthogonal projection of Fcosa of F in the direction of Q
Orthogonal projection of Qcosa of Q in the direction of F

We can express Fx = Fcosx of the force F as Fx = F  i

If the projection of F in the n-direction is F  n
Example

Find the projection of T along the line OA
Moment and Couple

Moment of force F about the axis through point O is
MO = r x F




r runs from O to any point on the line of action of F
Point O and force F establish a plane A
The vector Mo is normal to the plane in the direction
established by the right-hand rule
Evaluating the cross product
ˆi
MO  rx
ˆj
ry
kˆ
rz
Fx
Fy
Fz
Moment and Couple

Moment about an arbitrary axis
M  (r  F  n)n
known as triple scalar product (see appendix C/7)

The triple scalar product may be represented by the
determinant
rx
ry
rz
M  M  Fx
Fy
Fz
l
m
n
where l, m, n are the direction cosines of the unit vector n
Sample Problem 2/10
A tension T of magniture 10 kN is applied to the
cable attached to the top A of the rigid mast
and secured to the ground at B. Determine the
moment Mz of T about the z-axis passing through
the base O.
Resultants

A force system can be reduced to a resultant force and a
resultant couple
R  F1  F2  F3
 F
M  M1  M2  M3 
 (r  F)
Wrench Resultants

Any general force systems can be represented by a wrench
Problem 2/143


Replace the two forces and single couple by an equivalent
force-couple system at point A
Determine the wrench resultant and the coordinate in the xy
plane through which the resultant force of the wrench acts
Resultants

Special cases
• Concurrent forces – no moments about point of
concurrency
• Coplanar forces – 2D
• Parallel forces (not in the same plane) – magnitude of
resultant = algebraic sum of the forces
• Wrench resultant – resultant couple M is parallel to the
resultant force R
• Example of positive wrench = screw driver
Problem 2/142


Replace the resultant of the force system acting on the pipe
assembly by a single force R at A and a couple M
Determine the wrench resultant and the coordinate in the xy
plane through which the resultant force of the wrench acts