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The force acting on a body has two effects:
the first one is the tendency to push or pull the body in the direction
of the force, and the second one is to rotate the body about any fixed
axis which does not intersect nor is parallel to the line of the force.
This dual effect can more easily be represented by replacing the
given force by an equal parallel force and a couple to compensate for
the change in the moment of the force.
F
Let us consider for acting at point A in a rigid body. It is possible
to slide force along its line of action, but it is not possible to
directly move it to point B without changing the external effect on the
rigid body.
F
F
In order to do this, two equal and opposite forces and are added to
point B without introducing any net external effects on the body. It is
seen that, the original force at A and and the equal and opposite one at
B constitute the couple M=Fd, which is counterclockwise for this case.
F
F
Therefore, we have replaced the original force at A by the same
force acting at a different point B and a couple, without altering
the external effects of the original force on the body. Since is
a free vector, its location is of no concern. The combination of
the force and couple is referred to as a force-couple system.
M
By reversing this process, we can combine a given couple and a
force which lies in the plane of the couple (normal to the couple
vector) to produce a single, equivalent force. Force can be moved
to a point by constructing a moment equal in magnitude and
opposite in direction . The magnitude and direction of
remains the same, but its new line of action will be distance
away from point B.
M
F
F
Md
1. A force F of magnitude 50 N is exerted on the automobile parking-brake
lever at the position x= 250 mm. Replace the force by an equivalent force–
couple system at the pivot point O.
If two force systems are creating the same external effect
on the rigid body they are exerted on, they are said to be
“ ”.
The resultant of a force system is the simplest
combination that they can be reduced without altering the
external effects they produce on the body.
Coplanar Force Systems
If the resultant of all forces lying in a single plane
such as xy is , this resultant is calculated by the vector sum of these
forces.
nFFFFR
...321
nFFFF
...,,,, 321
R
x
y
yx
yyxx
R
R
FFR
FRFR
1
22
tan
The location of the line of action of the resultant force to an
arbitrary point (such as point O is the origin of the xy coordinate
system) can be determined by using the Varignon’s theorem. The
moment of about point O will be equal the sum of the couple
moments constructed by moving its components to point O.
R
R
FdMMo
FR
R
Md o
2. Determine the x- and y-axis intercepts of the line of action of the resultant of
the three loads applied to the gearset.
3. The device shown is part of an automobile seat-back-release
mechanism. The part is subjected to the 4 N force exerted at A and a
300 Nmm restoring moment exerted by a hidden torsional spring.
Determine the y-intercept of the line of action of the single equivalent
force.
Three Dimensional Force Systems
The same principles can be enlarged to three dimensional force
systems. The resultant of forces acting on a
body can be obtained by moving them to a desired point. In this
way, the given force system will be converted to
1) Three dimensional, concurrent forces comprising the same
magnitudes and directions as the original forces,
2) Three dimensional couples.
nFFFF
...,,,, 321
By calculating the resultants of these forces and couples, a single
resultant force and a single couple can be obtained.
The resultant force,
222
321 ...
zyx
zzyyxx
n
FFFR
FRFRFR
FFFFFR
The resultant couple moment,
The selection of point O is arbitrary, but the magnitude and direction
of will depend on this point; whereas, the magnitude and
direction of are the same no matter which point is selected.
CFrM
M
R
C C
As a special case, if the resultant couple is perpendicular to
the resultant force , these two vectors can further be simplified to
obtain a single resultant force . The force can be slided a
distance d to form a moment , which is equal in magnitude
and opposite in direction , so that they will cancel each other
out. The distance d will be equal to d=SM/R.
R
R
R
M
M
M
4. The special-purpose milling cutter is subjected to the force of
1200 N and a couple of 240 N.m as shown. Replace the given force
system with an equivalent force-couple system at O.
5. The pulley wheels are subjected to the loads shown. Determine the equivalent
force-couple system at point O.
When the resultant couple vector is parallel to the resultant
force , the resultant is called a “wrench”.
The wrench is the simplest form in which the resultant of a general
force system may be expressed.
By definition, a wrench is positive if the couple and force vectors
point in the same direction, and negative if they point in opposite
directions.
R
M
Wrench Resultants
A common example of a wrench is found with the application of a
screw driver.
All force systems can be reduced to a wrench acting at a particular
line of action.
M R
//RR
RR//R
MMMM
nnMMM
2
1
Equivalent force-couple system at point O M is resolved into components M1 along the
direction of R and M2 normal to R.
Positive wrench
R
Md 2
M
R
M R
6. For the position shown, the crankshaft of a small two-cylinder compressor is
subjected to the 400 N and 800 N forces exerted by the connecting rods and the 200
N·m couple. Replace this loading system by a force-couple system at point A. Show
that is not perpendicular to . Then replace this force-couple system by a wrench.
Determine the magnitude M of the moment of the wrench, the magnitude of the force
of the wrench and the coordinates of the point in the x-z plane through which the
line of action of the wrench passes.
R
AM
R
7. The force-couple system acting at O is equivalent to the wrench acting at A.
If and , determine . NkjiR
7001400600 mNM //R 1200
oM
8. a) Reduce the general three-dimensional force system to a force-couple system at O.
b) Replace the force-couple system obtained by a wrench and determine the coordinates of the
point in the yz plane through which the line of the wrench passes.
Dimensions in meters.
9. The threading dye is screwed onto the end of the fixed pipe which is bent
through an angle of 20°. Replace the two forces by an equivalent force at O and a
couple . Find and calculate the magnitude M of the moment which tends to screw
the pipe into the fixed block about its angled axis through O.
Nm.k.i.kiMMM
k.i.kcosisinn
kiM
ji.k.kcos.isin.
ji.k.kcos.isin.M
jjjFR
oOC
OC
o
o
68859403408517
9403402020
8517
150250202015020150
200250202015020150
50200150
S
S
S
S
y
x
500 N
1700 N 3400 N
3
4
tan
30 cm 50 cm
34 cm
50 cm
50 cm z
800 N.m
tan 15
8
10. The pulleys and the gear are subjected to the loads shown. For these forces,
determine the equivalent force-couple system at point A.
15
8
jiR
jiijiFR
jijiF
iF
jijiF
35201040
800150050027202040
800150017
81700
17
151700
500
272020405
43400
5
33400
3
2
1
S
kjiCFrM
kC
kjijikjFr
kijFr
kjiFr
jikjijijikFr
CFrM
A
A
6.11391770960
800
75075040080015005.05.0
1505003.0
6.48910201360
272020405.016.03.02720204017
8
17
1534.05.0
33
22
11
11
SS
SS
C3 = 80 Nm (in plane ABCD)
y > 90o
F1 = 30 N
F2 = 75 N
F3 = 40 N
C1 = 60 Nm
C2 = 100 Nm (in yz plane)
O (0, 0, 0) m A (12, 0, 0) m B (in xz plane) C (12, 8, 0) m E (6, 10, -3) m G (10, 4, 4) m
11. Replace the system comprising two forces, two couples and a positive wrench with an
equivalent force-couple acting at point O. Then, reduce the system further into a wrench and
determine the coordinates of point P, of which the line of action of the wrench cuts through
the yz plane.
A
B
C
O
D
E
G
37°
X Y
Z
45°
60° y
53°
30°
6 m
4 m
1C
1F
2C
3F
3C
2F
k1ji.kcos60jcos6icos45F
cos6cos45cos oyy
222
5152121030
6010
1
Force:
C3 = 80 Nm (in plane ABCD) y > 90o
F1 = 30 N
F2 = 75 N
F3 = 40 N
C1 = 60 Nm
C2 = 100 Nm (in yz plane)
A
B
C
O
D
E
G
37°
X Y
Z
45°
60° y
53°
30°
6 m
4 m
1C
1F
2C
3F
3C
2F
Force:
C3 = 80 Nm (in plane ABCD) y > 90o
F1 = 30 N
F2 = 75 N
F3 = 40 N
C1 = 60 Nm
C2 = 100 Nm (in yz plane)
A
B
C
O
D
E
G
37°
X Y
Z
45°
60° y
53°
30°
6 m
4 m
1C
1F
2C
3F
3C
2F
k.j.i.
k)(j)(iF
884688461535
884
8008181275
2222
O (0, 0, 0) m A (12, 0, 0) m B (in xz plane) C (12, 8, 0) m E (6, 10, -3) m G (10, 4, 4) m
Force:
C3 = 80 Nm (in plane ABCD) y > 90o
F1 = 30 N
F2 = 75 N
F3 = 40 N
C1 = 60 Nm
C2 = 100 Nm (in yz plane)
A
B
C
O
D
E
G
37°
X Y
Z
45°
60° y
53°
30°
6 m
4 m
1C
1F
2C
3F
3C
2F
O (0, 0, 0) m A (12, 0, 0) m B (in xz plane) C (12, 8, 0) m E (6, 10, -3) m G (10, 4, 4) m
k.j.i.
...
k.j.i.
R
Rn
k.j.i.FR
k.jikcosjsincosicoscosF
.
222R
7806300150
52968877951
52968877951
52968877951
643416123040536040536040
04124
3
S
kj.2i.k.jikji6Fr
k.j562.56i.k.j.i.ji12Fr
k.j.k1ji.k4ji10Fr
CFrMo
248443439464341612310
84843043758846884615358
1665166551521214
33
22
11
SSMoment:
C3 = 80 Nm (in plane ABCD) y > 90o
F1 = 30 N
F2 = 75 N
F3 = 40 N
C1 = 60 Nm
C2 = 100 Nm (in yz plane)
A
B
C
O
D
E
G
37°
X Y
Z
45°
60° y
53°
30°
6 m
4 m
1C
1F
2C
3F
3C
2F
O (0, 0, 0) m A (12, 0, 0) m B (in xz plane) C (12, 8, 0) m E (6, 10, -3) m G (10, 4, 4) m
C3 = 80 Nm (in plane ABCD) y > 90o
F1 = 30 N
F2 = 75 N
F3 = 40 N
C1 = 60 Nm
C2 = 100 Nm (in yz plane)
A
B
C
O
D
E
G
37°
X Y
Z
45°
60° y
53°
30°
6 m
4 m
1C
1F
2C
3F
3C
2F
O (0, 0, 0) m A (12, 0, 0) m B (in xz plane) C (12, 8, 0) m E (6, 10, -3) m G (10, 4, 4) m
k3i
2432
k24iC
k24ijki6BDBA
iC
kji.4kcosjcosicos45C
223
2
1
26432
80
3248
100
3030422606060
40
kj.i.M
k3iikji.4
kj.2i.k.j562.56i.k.j.M
o
o
8675684186847
2641003030422
2484434394848430437516651665
Moment:
k86756.841i86.847M
k52.9677.88i95.1
o
j
jR
Equivalent force-couple system at point O
k86756.841i86.847M
k52.9677.88i95.1
o
j
jR
Equivalent force-couple system at point O
Reduction to a wrench in yz plane
k.j.ik.j.i..nMM
Nm.k.j.i.kj.i47.86nMM
R////
Ro//
0210402842780630015035133
351337806300150867568418
Positive wrench
k02.10402.48i2M
k52.9677.88i95.1
//
j
jR
O
x y
z
k02.10402.48i2M//
j
k52.9677.88i95.1
jR
oM
M
Positive wrench
The coordinates of point P, of which the line of action of the wrench cuts through the yz plane:
myyk
mzzj
jizjziyky
jj
j
jj
Rr
27.39198.76295.1
66.47458.92595.1
k98.76258.259i49.8688.7745.152.9695.1
k98.76258.259i49.868k52.968.77i95.1kzjy
k98.76258.259i49.868M
k02.10402.48i2k867841.56i47.868MMM
M
//o
O x y
z
k02.10402.48i2M//
j
k52.9677.88i95.1
jR
Positive wrench
P(0;391.27;474.66) r