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ME 201Engineering Mechanics: Statics
Chapter 4 – Part B4.5 Moment of a Force about a Specified Axis
4.6 Moment of a Couple
Moment of a Force about a Specified Axis
A moment and its axis are always perpendicular to the plane defined by the force and its moment arm
Sometimes necessary to find a component of this moment along a specified axis
This is sometimes referred to as projection The dot product can be used to find this
component
Moment of a Force about a Specified Axis
RememberMO = r x Fr is vector from O to line of action of F
Anddefining a unit vector UAand perform a dot product operation
MA = UA ∙ (r x F)MA = MA Ua
zyx
zyx
FFFrrrkji
rxF
zyx
zyx
AzAyAx
A
FFFrrr
UUUM
Example Problem
Given:F = 700 N
FindMAB
z
x
y.2m
F.3m .4mA
B
.6mC D
Example Problem Solution
Given:F = 700 N
Find:MAB
Solution:
CDADA FrM
z
x
y.2m
F.3m .4mA
B
.6mC D
ABABAB M UM
ABAABM UM
mAD jr 2.
N
CD
kji
kjiF
300200600)3.(2.)6.(
3.2.6.700222
NA
kiM
1206030020060002.0
mAB jir 2.4.
NmM AB
7.5304472.8944.120060
mAB jiU 4472.8944.
NmAB
jiM
0.240.4804472.8944.)7.53(
Moment of a Couple
Couple – 2 parallel forces that have the same magnitude, opposite directions, and are separated by a distance
The moment produced by a couple is equivalent to the sum of the moments of both couple forces computed about any arbitrary point in space
Moment of a CoupleClass Exercise
∑ F =
MA =
MB =
MC =
A BC
F
dF
Compute the Moments for the problem below:
x x x
Moment of a Couple: Scalar
M = F d
Where F is magnitude of one force d is the perpendicular distance
(or moment arm) between the forces
direction by right-hand rule+
-
F
dF
Equivalent Couples
Couples are equivalent if: They produce the same moment Both Magnitude & Direction
This means the forces must either lie in the same plane or parallel planes
Example Problem
Given:F = 400 N (couple)
Find equiv couple at DE
.2 mF
F
D E
.1m
Example Problem Solution
Given:F = 400 N (couple)
Find:equiv couple at DE
Solution:
cwNmorNmmNFdM 80802.0400
.2 mF
F
D E
.1m
Nm
NmdMF 800
1.080
F2 -F2
(cw rotation)
Example Problem
Given:F = 150 lb (couple)
Find M
.3 m
F
F
A
B
.1m
45 3
Example Problem SolutionGiven:
F = 150 lb (couple)Find:
MSolution:
.3 m
F
F
A
B
.1m
45 3
Breaking into components may be simpler than finding perpendicular distance.
.3 mFx
A
B
.1mFx
Fy
Fy
Note: There are now 2 couples
Example Problem Solution
Given:F = 150 lb (couple)
.3 m
F
F
A
B
.1m
45 3
.3 mFx
A
B
.1mFx
Fy
Fy
ccwftlb
FFM yx
3903.901.120
3.150531.150
54
3.1.
Moment of a Couple: Vector
M = r x F
Where r is position vector between the
lines of action of the forces F is one forcesThink of taking the moment of one of
the forces about a point lying on the line of action of the other force
Result is referred to as a free vector – it can act at any point
z
x
yr
F
F
Example Problem
Given:F=25 lb (couple)θ = 30o
Find:Couple Moment
z
x
y8” F
F
O
6”
Aθ
B
Example Problem Solution
FrM
BFrM ABA ABAB FrM
z
x
y8” F
F
O
6”
Aθ
B
or
kikir
3196.530sin630cos6
AB
lbkFB 25
inlbA jkkiM 130253196.5
Example Problem
Given:F1 = 200 N (couple)
F2 = 100 N (couple)
Find:MR
z
x
y
.4mF1
F1O.3m
A
BC
D
E.2m
F2
F23
45
Example Problem Solution
Given:F1 = 200 N (couple)F2 = 100 N (couple)
Find:MR
Solution: BFrM ABA
z
x
y
.4mF1
F1O.3m
A
BC
D
E.2m
F2
F23
45
Nm
C
kjkji
kjiM
1612)6080(2.
)3.(4.
3.4.1002.22
NmA
ijkM
602003.
DCDC FrM
NmR kjiM 161260 CAR MMM
