ENGR 0135
Chapter 5 –2
Equivalent force-couple system
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Engineering
Topics
 Couples
– Definition of Couples
– Characteristics of Couples
– Equivalent force-couple system
 Resultant
of non-concurrent force
system
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Equivalent force-couple
systems

A force F can be replaced by a parallel force and a couple  an equivalent
force-couple system acting on a point O
Original system of a single force F acting on A.
The vector position of A relative to O is r.
Original system in the moment plane
Replacing the two opposite and
equal forces F that are separated
by distance d with a couple
new system of a couple
and a force F at O
Adding up a pair of two equal but
opposite forces F at O  no effect
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Example Problem
 Replace
the force that acts on A by a force at
point B and a couple
 Use scalar and vector analysis
 Basically, it asks us to compute MB and
translate the force from A to B
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Another example problem:
 The
magnitude of F is 780 lb. Replace the F
by a force Fo at the O and a couple C.
Two tasks:
1. Compute Mo due to Fo
2. Translate Fo to O.
Note:
In performing the cross product,
both moment arm rOA or rOB are valid.
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Coplanar force system
- non-concurrent force system

The system can be replaced by a resultant force R
acting at a distance dR (from a selected point O)

The method for finding the resultant of concurrent
forces can be used to find the resultant R and its
direction cosines
The resultant R can be written in the Cartesian vector
form
The distance dR (or xR, or yR) can be found utilizing
the Varignon’s theorem
If the sum of forces R is zero, the resultant is a couple
C whose direction is perpendicular to the plane
(provided that the couple is non-zero)
If the sum of moment about a point is zero, then the
line of action of the R passes that point
When both C and R are zeros, then the system is in
equilibrium





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Coplanar force system
- non-concurrent force system
 The
resultant R
n
n
n
i =1
i =1
i =1
R = ∑ Fi = ∑ Fx i + ∑ Fy j = Re
R= R =
magnitude
(∑ F ) + (∑ F )
2
2
x
y
e = cos θ x i + cos θ y j
cos θ x
 The
F
∑
=
x
R
cos θ y
distance dR
F
∑
=
Gives the direction
y
R
dR =
∑M
O
R
∑ M O = F1d1 + F2 d 2 + ...
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Example Problem
 What
is the
magnitude of the
resultant and its
direction?
n
n
n
i =1
i =1
i =1
R = ∑ Fi = ∑ Fx i + ∑ Fy j = Re
R= R =
(∑ F ) + (∑ F )
2
2
x
y
Magnitude:
e = cos θ x i + cos θ y j
cos θ x
F
∑
=
x
R
cos θ y
F
∑
=
y
R
Rx = 300 + 500 cos 60 = 550 N
R y = 200 + 500 sin 60 = 633 N
R = 550 2 + 6332 = 839 N
Direction:
θ x = cos −1
550
= 49.0o
839
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Example Problem
 How
dR =
to get dR and xR?
∑M
xR=
O
R
∑ M O = −300(0.4) − 500 cos 60(0.7)
− 200(0.450) + 600 = 215 Nm
215
= 0.256m
dR =
839
xR =
∑M
Ry
O
=
dR =
215
= 0.340m
633
xR = the distance from O to the intercept of the line of
action of the resultant with the x-axis
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Coplanar force system – parallel forces

The system can be replaced by a
resultant force R acting at a distance
dR (from a selected point )

The resultant R is simply the sum of
the parallel forces

The distance xR can be calculated
using the Varignon’s theorem
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Coplanar force system – parallel forces
 The
resultant R is given by
R = ∑ Fi
direction is vertical
 The distance xR can be
calculated using the
Varignon’s theorem
O
 The
xR
M
∑
=
O
O
R
∑ M O = F1 x1 + F2 x2 + ...
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Example: coplanar parallel forces
 If
each of the light
weighs 150 lb,
determine the resultant
and the location w.r.t A
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Non-coplanar parallel force system

The resultant R is given by

And the position of R is
given by
R = ∑ Fi
xR
M
∑
=−
y
yR
M
∑
=
x
R
R
∑ M x = F1 y1 + F2 y2 + ...
− ∑ M y = F1 x1 + F2 x2 + ...
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Example Problem 5-12:


Determine the R and the
location w.r.t O
Tasks:
– Find the sum of the forces
– Find the sum of the moments
about y and x axis
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General 3D non-concurrent force system
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General 3D non-concurrent force system

General 3D force system can be
replaced by
–
–
A system of non-coplanar, concurrent
forces through the origin O that have the
same magnitude and directions as the
forces of the original system
+
A system of non-coplanar couples
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General 3D force system –
1. The concurrent force system
 The
resultant of the concurrent force
system:
R = R x + R y + R z = Rx i + R y j + Rz k = Re
Rx = ∑ Fx
R y = ∑ Fy
Rz = ∑ Fz
R = Rx2 + R y2 + Rz2
e = cos θ x i + cos θ y j + cos θ z k
Rx
cos θ x =
R
cos θ y =
Ry
R
Rz
cos θ z =
R
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General 3D force system – 2. The noncoplanar
couple system
 The
resultant of the non-coplanar couples
C = ∑ C x + ∑ C y + ∑ C z = ∑ C x i + ∑ C y j + ∑ C z k = Ce
C=
∑C x + ∑C y + ∑C z
2
2
2
e = cos θ x i + cos θ y j + cos θ z k
C = ∑ (rOi × Fi )
cos θ x
C
∑
=
C
x
cos θ y
C
∑
=
C
y
cos θ z
C
∑
=
z
C
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Example

Replace the force system with a
force R through point O and a
couple C

Tasks:
– Find the force vectors FA, FB, FC
– Find the resultant R
– Obtain the total moment Mo (or
C) in vectorial
– Obtain the magnitude of Mo
– Determine the direction cosines
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Example:

Four forces are applied to a
truss as shown. Determine the
magnitude and direction of the
resultant of the four forces and
the perpendicular distance dR
from point A to the line of
action of the resultant.
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Example:

Four parallel forces act on a
concrete slab as shown.
– Determine the resultant of the
forces and locate the intersection
of the line of action of the resultant
with the xy-plane.
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Example:

The homogeneous plate shown has
a mass of 90 kg. The magnitude of
the force T in cable BC is 800 N.
– Replace the weight and cable forces
by an equivalent force-couple system
at hinge A.
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