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Engineering 36
Chp 4: Intro to
Moments
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-07_Moments_Intro.ppt
Moment (Torque) Described
 In Physics and Engineering a MOMENT
is a measure of TWISTING Power
 The MAGNITUDE of a Moment is the
PRODUCT of a Lever Arm Distance
and an Intensity
• The “Intensity” can be a Force, an Electric
Charge, an Area, a Mass, or other
• In Engineering Mechanics
the Intensity takes the
form of a Force
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-07_Moments_Intro.ppt
Moment Described
 In General, MOMENTS are VECTOR
Quantities with Magnitude (see previous
slide) and Direction
 The Direction of the Moment
Vector is determined by the
Right Hand Rule
• Wrap Fingers in the Direction of
ROTATION (or tendency to rotate),
then THUMB points in the
Direction of the Moment Vector
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-07_Moments_Intro.ppt
Moment Center
 The MOMENT
CENTER is equivalent
to the PIVOT POINT
about which Rotation
would occur upon
application of a Force
whose Line of Action is
OFFSET from the Pivot
Point
 distance from the
Pivot to Force Loa
Moment Center (MC)
or Pivot Point
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-07_Moments_Intro.ppt
Position Vector
 The Position Vector
runs from the
Moment Center to
ANY Point on the
LoA of the Force
• Often times the most Convenient Point on
the LoA is the Point of Application (PoA)
– i.e., Pos. Vector runs from the Pivot to the PoA
 The position Vector Contains within it
the “LeverArm” part of the Moment Calc
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-07_Moments_Intro.ppt
Picking the Position Vector, r
 Often the Most
Convenient Position
Vector is that which
runs From the Pivot to
the Point of Application
 OTHER Pts on the
Force LoA may be
more easily determined
and are thus More
Convenient
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r runs from the
Pivot to the Point
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-07_Moments_Intro.ppt
Shortest Position Vector
 The Shortest vector
is that which is  to
the Force LoA
 The Mag of the
Shortest r is called
the Perpendicular
Distance, d:
d  r min
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 distance
yields Shortest r
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-07_Moments_Intro.ppt
Moment Magnitude
 As Noted Previously the
Magnitude of a Moment is
related to the product of

• The Position Vector, r
• The Force, F
 Mathematically
M  M  r  F sin   r sin  F  d  F
• Thus knowing F & d allows Calc of the Moment
magnitude, but NOT its SENSE (direction)
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-07_Moments_Intro.ppt
Moment Sense
 TWO-DIMENSIONAL
STRUCTURES
Have Length And Breadth But
Negligible Depth And Are
Subjected To Forces Contained
In The PLANE Of The Structure
 The Plane Of The Structure
Contains The Point O And The
Force F. MO, The Moment Of
The Force About O Is
Perpendicular To The Plane.
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-07_Moments_Intro.ppt
Moment Sense/Direction
 If The Force Tends To Rotate The
Structure COUNTER-clockwise,
The Sense Of The Moment
Vector Is OUT Of The Structure
Plane
• SIGN{MO} → POSITIVE
 If The Force Tends To Rotate The
Structure CLOCKWISE, The
Sense Of The Moment Vector Is
INTO The Structure Plane
• SIGN{MO} → NEGATIVE
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-07_Moments_Intro.ppt
Moment Direction by Rt Hand Rule
1. Point Fingers in r Direction
2. Curl Fingers Toward +F Direction
3. THUMB Points in the Direction of M
 k̂
 k̂
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HINT:
Put r & F
Tail-to-Tail
F
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-07_Moments_Intro.ppt
r
Moments Point in ALL Directions
 Since r and F
can be
arbitrarily
oriented
relative to the
CoOrd Axes,
then M will
also be
arbitrarily
Oriented
• Confirm These using your
own Right Hand
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-07_Moments_Intro.ppt
Moment Units  Force • Dist
 Discern the UNITS for Moments from
M  M  r  F sin   r sin  F  d  F
Units M   Distance Force
 Typical Units
• Ft-lbs
• In-lbs
• N-m
• N-mm
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-07_Moments_Intro.ppt
Example: Moment Calculation

A 100-lb Vertical Force Is Applied To
The End Of A Lever Which Is Attached
To a Shaft At O. DETERMINE
a) Moment About O
b) Horizontal Force At Pt-A Which Creates
The SAME Moment
c) Smallest Force At Pt-A Which Produces
The SAME Moment
d) Location For a 240-lb Vertical Force To
Produce The SAME Moment
e) Whether Any Of The Forces From b, c, and
d is EQUIVALENT To The ORIGINAL
Force
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-07_Moments_Intro.ppt
Example M Calc – Soln (a)
 Moment About O Is Equal To The Product
Of The Force And The
PERPENDICULAR DISTANCE Between
The Line Of Action Of The Force And O
 The Force Tends To Rotate The Lever
CLOCKWISE, Thus The Moment Vector
points INTO The Plane Of The Paper
• The Moment Vector Qty is thus NEGATIVE
M O  M O  Fd
d  24 in.  cos 60  12 in.
M O  1200 lb  in
M O  100 lb 12 in. 
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-07_Moments_Intro.ppt
Example M Calc – Soln (b)
 Horizontal Force at A That Produces
The Same Moment
d  24 in. sin 60  20.8 in.
60
M O  Fd
1200 lb  in.  F 20.8 in. 
1200 lb  in.
F
20.8 in.
F  57.7 lb
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-07_Moments_Intro.ppt
Example M Calc – Soln (c)
 The Smallest Force at A To Produce
The Same Moment Occurs When The
Perpendicular Distance is a Maximum
• i.e., When F Is Perpendicular To OA
M O  Fd
1200 lb  in.  F 24 in. 
1200 lb  in.
F
24 in.
F  50 lb
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-07_Moments_Intro.ppt
Example M Calc – Soln (d)
 To Determine The Point Of Application
Of A 240 lb Vertical Force To Produce
The Same Moment
M O  Fd
1200 lb  in.  240 lb d
1200 lb  in.
d
 5 in.
240 lb
OB cos60  d  5 in.
or OB  5 in.   cos60
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OB  10 in.
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-07_Moments_Intro.ppt
Example M Calc – Soln (e)
 Although Each Of The Forces In Parts b),
c), and d) Produces The Same Moment
As The 100 lb Force, NONE Are of The
Same MAGNITUDE And SENSE (Line of
Action) as the original pull
 NONE Of The Forces Is Equivalent To
The 100 lb force
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-07_Moments_Intro.ppt
WhiteBoard Work
Let’s Work
Problem
4.21
In order to pull out the nail at B, the force F exerted on the
handle of the hammer must produce a clockwise moment
of 500 in∙lb. about point A. Determine the required
magnitude of force F.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-07_Moments_Intro.ppt
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-07_Moments_Intro.ppt
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-07_Moments_Intro.ppt
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-07_Moments_Intro.ppt
Engineering 36
Appendix
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-36_Lec-07_Moments_Intro.ppt