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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
Department of Mechanical Engineeri
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.
Department of Mechanical Engineeri
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
Department of Mechanical Engineeri
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
Department of Mechanical Engineeri
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:
Department of Mechanical Engineeri
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
=== ∑
Department of Mechanical Engineeri
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
Department of Mechanical Engineeri
Example: coplanar parallel forces
If each of the light weighs 150 lb, determine the resultant and the location w.r.t A
Department of Mechanical Engineeri
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
Department of Mechanical Engineeri
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
Department of Mechanical Engineeri
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
Department of Mechanical Engineeri
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
Department of Mechanical Engineeri
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
Department of Mechanical Engineeri
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
Department of Mechanical Engineeri
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.
Department of Mechanical Engineeri
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.