Lecture 2 - Basic Concept of Moments

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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.



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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 .


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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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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+ 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

Lecture 2 - Basic Concept of Moments

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