ME 201 Engineering Mechanics: Statics

1
ME 201Engineering Mechanics Statics

• Chapter 4 Part B
• 4.5 Moment of a Force about a Specified Axis
• 4.6 Moment of a Couple

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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

3
Moment of a Force about a Specified Axis

• Remember
• MO r x F
• r is vector from O to line of action of F
• And
• defining a unit vector UA
• and perform a dot product operation
• MA UA (r x F)
• MA MA Ua

4
Example Problem
z

• Given
• F 700 N
• Find
• MAB

C
F
D
y
.6m
A
.3m
.4m
B
.2m
x
5
Example Problem Solution
z
C
F
D
y
.6m
A
.3m
.4m

• Given
• F 700 N
• Find
• MAB
• Solution

B
.2m
x
6
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

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Moment of a CoupleClass Exercise
Compute the Moments for the problem below

• ? F
• MA
• MB
• MC

F
C
d
A
x
x
x
B
F
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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
d
F

-
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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

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Example Problem

• Given
• F 400 N (couple)
• Find
• equiv couple at DE

F
D
E
.2 m
F
.1m
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Example Problem Solution
F
D
E
.2 m

• Given
• F 400 N (couple)
• Find
• equiv couple at DE
• Solution

F
.1m
F2
-F2
(cw rotation)
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Example Problem
F
5
3

• Given
• F 150 lb (couple)
• Find
• M

4
F
A
.1m
B
.3 m
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Example Problem Solution
F

• Given
• F 150 lb (couple)
• Find
• M
• Solution

5
3
4
F
A
.1m
B
.3 m
Breaking into components may be simpler than
finding perpendicular distance.
Note There are now 2 couples
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Example Problem Solution
F
5
3

• Given
• F 150 lb (couple)

4
F
A
.1m
B
.3 m
Fy
A
Fx
Fy
.1m
Fx
B
.3 m
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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 forces
• Think 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
F
r
y
F
x
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Example Problem

• Given
• F25 lb (couple)
• ? 30o
• Find
• Couple Moment

z
F
F
A
8
O
y
?
6
x
B
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Example Problem Solution
z
F
F
A
8
O
y
?
6
x
B
or
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Example Problem

• Given
• F1 200 N (couple)
• F2 100 N (couple)
• Find
• MR

z
E
.2m
5
3
A
F1
F2
4
D
O
.3m
y
F2
B
F1
.4m
C
x
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Example Problem Solution
z
E
.2m
5
3
A
F1
F2
4
D
O
.3m
y
F2

• Given
• F1 200 N (couple)
• F2 100 N (couple)
• Find
• MR
• Solution

B
F1
.4m
C
x