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Department of Mechanical Engineering
ENGR 0135
Chapter 5 –2Equivalent force-couple system
Department of Mechanical Engineeri
Topics
Couples– Definition of Couples– Characteristics of Couples– Equivalent force-couple system
Resultant of non-concurrent force system
Department of Mechanical Engineeri
Equivalent force-couple systems
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
A force F can be replaced by a parallel force and a couple an equivalent force-couple system acting on a point O
Adding up a pair of two equal but opposite forces F at O no effect
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
Department of Mechanical Engineeri
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 Fo2. 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 Racting 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
The distance dR
( ) ( )
RF
RF
FFR
RFF
yy
xx
yx
yx
n
iy
n
ix
n
i
∑∑
∑∑
∑∑∑
==
+=
+==
=+=====
θθ
θθ
coscos
coscos
22
111
jie
R
ejiFR i
Gives the direction
...2211 ++=
=
∑
∑
dFdFMRM
d
O
OR
magnitude
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Example ProblemWhat is the
magnitude of the resultant and its direction?
( ) ( )
RF
RF
FFR
RFF
yy
xx
yx
yx
n
iy
n
ix
n
i
∑∑
∑∑
∑∑∑
==
+=
+==
=+=====
θθ
θθ
coscos
coscos
22
111
jie
R
ejiFR i
NR
NRNR
y
x
839633550
63360sin50020055060cos500300
22 =+=
=+==+=
Magnitude:
ox 0.49
839550cos 1 == −θ
Direction:
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Example ProblemHow to get dR and xR?
dR =
md
NmM
RM
d
R
O
OR
256.0839215
215600)450.0(200)7.0(60cos500)4.0(300
==
=+−
−−=
=
∑
∑ xR=
xR = the distance from O to the intercept of the line of action of the resultant with the x-axis
mRM
xy
OR 340.0
633215
=== ∑
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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
Department of Mechanical Engineeri
Coplanar force system – parallel forces
The resultant R is given by
The direction is verticalThe distance xR can be
calculated using the Varignon’s theorem
∑= iFR
...2211 ++=
=
∑
∑
xFxFMRM
x
O
OR
O
O
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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
∑= iFR
∑∑
∑∑
++=−
++=
=−=
...
...
2211
2211
xFxFM
yFyFMRM
yRM
x
y
x
xR
yR
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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:
RR
RR
RR
RRRR
FRFRFR
RRRR
zz
yy
xx
zyx
zyx
zzyyxx
zyx
===
++=
++=
===
=++=++=
∑∑∑
θθθ
θθθ
coscoscos
coscoscos
222
kjie
ekjiRRRR zyx
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General 3D force system – 2. The noncoplanar couple system
The resultant of the non-coplanar couples
CC
C
C
CC
CCCC
CCCC
zz
yy
xx
zyx
zyx
zyxy
∑∑∑∑
∑∑∑∑∑∑∑∑∑
===
×=
++=
++=
=++=++=
θθθ
θθθ
coscoscos
)(
coscoscos
222
iOi
zx
FrC
kjie
ekjiCCCC
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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 dRfrom 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.
