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7/18/2019 Lecture 2 - Basic Concept of Moments
http://slidepdf.com/reader/full/lecture-2-basic-concept-of-moments 1/15
3/10/201
SCHOOL OF MATERIALS ENGINEERING,
UNIVERSITI MALAYSIA PERLIS
Taman Muhibah, Jejawi, 02600 Perlis
EBT 112 - STATICS
Lecture 2
Basic Concept of Moments
Juyana A Wahab
PPK Bahan (Taman Utara Jejawi)
[email protected]/ext:6296
Moment of a Force
• The ability of a force to cause a body to rotate is
measured by a quantity called the moment of the force.
7/18/2019 Lecture 2 - Basic Concept of Moments
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Moment of a Force – Scalar Formation
• The turning effect of the force is measured by the
product of F and d .
Moment of a Force – Scalar Formation
Direction of a moments.
• The two forces P and Q causes the lever to rotate about point O in
opposite direction.
– The force P causes a counterclockwise rotation (positive sign)
– The force Q causes a clockwise rotation (negative sign)
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Example 1
For each case, determine the moment of the force about
point O.
Example 1: Solution
Line of action is extended as a dashed line to establish
moment arm d.
Tendency to rotate is indicated and the orbit is shown as
a colored curl.
)(.200)2)(100()( CW mN mN M a o
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Example 1: Solution
)(.5.37)75.0)(50()( CW mN mN M b o
)(.229)30cos24)(40()( CW mN mmN M c o
Example 1: Solution
)(.4.42)45sin1)(60()( CCW mN mN M d o
)(.0.21)14)(7()( CCW mkN mmkN M e o
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Consider this situation..
What are the resultant effects on the person’shand when the force is applied in these two
different ways?
1) Resolve the 100 N force along x and y-axes.
2) Determine MO using a scalar analysis for the two force
components and then add those two moments together.
Example 2
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Example 2: Solution
Since this is a 2-D problem:
1) Resolve the 100 N force alongthe handle’s x and y axes.
2) Determine M A using a scalar
analysis.
Example 3
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Solution:
+ Fy = 100 sin 30° N
+ Fx = 100 cos 30° N
x
y
+ M A = { –(100 cos 30°)N (450 mm) – (100 sin 20°)N (125 mm)}
= – 43.2464 N·mm = 43.2 N·m (clockwise or CW)
Example 3: Solution
Exercise 1
• A force F of magnitude 60 N is applied to the gear.
Determine the moment of F about point O.
Ans. Mo = 5.64 N.m CW
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Example 4
A 500 N force is applied to the end of a lever pivoted at
point O. Determine the moment of the force about O if θ =
30°
Example 4: Solution
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Example 5
Determine the moment of the force about point O.
Neglecting the thickness and mass of the member.
Example 5: Solution
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Moment of a Force – Scalar Formation
Resultant Moment
• Resultant moment, MRo = moments of all the forces
MRo = ∑Fd
Example 6
• Figure shows a member which is subjected to two forces
at point A. Calculate the resultant moment produced by
the forces about point O.
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1
+ Fx = 450 (cos 60) – 700 (sin 30)
= – 125 N
+ Fy = – 450 (sin 60) – 300 – 700 (cos 30)
= – 1296 N
+ MRA = 450 (sin 60) (2) + 300 (6) + 700 (cos 30) (9) + 1500
= 9535 Nm
Now find the magnitude and direction of the resultant.
FRA = (1252 + 12962)1/2 = 1302 N and
= tan-1 (1296 /125) = 84.5°
Summing the force components:
Example 7: Solution
Moment of a couple
• Couple
– Two parallel forces that have the same magnitude
but opposite directions and separated by a
perpendicular distance d.
• Equivalent Couples
– 2 couples are equivalent if they produce the same moment
– Forces of equal couples lie on the same plane or plane parallelto one another
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Moments due to couples can be addedtogether using the same rules as adding any
vectors. The resultant couple moment, MR =
M1 + M2
The moment produced by a couple is called
couple moment having a magnitude of F*d.
The value of couple moment can be
determined by finding the sum of moments
of both couple forces about any point. The
moment of a couple is a free vector . It can
be moved anywhere on the body and have
the same external effect on the body.
APPLICATIONS
Would older vehicles without power steering need larger
or smaller steering wheels?
When you grip a vehicle’s steering wheel with both hands and
turn, a couple moment is applied to the wheel.
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Example 8
• Two couples act on the beam as shown. Determine the
magnitude of F so that the resultant couple moment is
300 N · m counterclockwise
Example 8: Solution
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1) Add the two couples to find the resultant couple.
2) Equate the net moment to 1.5 kNm clockwise to find F.
Example 9
Solution:
The net moment is equal to:
+ M = – F (0.9) + (2) (0.3)
= – 0.9 F + 0.6
– 1.5 kNm = – 0.9 F + 0.6
Solving for the unknown force F, we get
F = 2.33 kN
Example 9: Solution