lecture 7- 1page

8/14/2019 Lecture 7- 1page

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Vector Mechanics for Engineers: StaticsE i gh t h

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MOMENT OF A FORCE

Moment of aforce





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3 - 2

APPLICATIONS

What is the net effect of the

two forces on the wheel?





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3 - 3

APPLICATIONS (continued)

What is the effect of the 30 N

force on the lug nut?





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3 - 4

MOMENT IN 2-D

The moment of a force about a point provides a measure of thetendency for rotation (sometimes called a torque).





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3 - 5

MOMENT IN 2-D (continued)

In the 2-D case, the magnitude of the moment is

Mo = F d

As shown, d is the perpendicular distance from point O to the

line of action of the force.

In 2-D, the direction of MO is either clockwise or

counter-clockwise depending on the tendency for rotation.





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MOMENT IN 2-D (continued)

For example, MO = F d and the

direction is counter-clockwise.

Often it is easier to determine MO by using the components of F

as shown.

Using this approach, MO = (FY a) – (FX b). Note the different

signs on the terms! The typical sign convention for a moment in2-D is that counter-clockwise is considered positive. We can

determine the direction of rotation by imagining the body pinned

at O and deciding which way the body would rotate because of

the force.

Fa b

d

O

a bO

F

F x

F y





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Moment of a Force About a Point

• A force vector is defined by its magnitude and

direction. Its effect on the rigid body also depends

on it point of application.

• The moment of F about O is defined as F r M O ×=

• The moment vector M O

is perpendicular to the

plane containing O and the force F.

• Any force F’ that has the same magnitude and

direction as F, is equivalent if it also has the same line

of action and therefore, produces the same moment.

• Magnitude of M O

measures the tendency of the force

to cause rotation of the body about an axis along M O.

The sense of the moment may be determined by theright-hand rule.

Fd rF M O == θ sin





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Moment of a Force About a Point

• 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. M O, the moment of the force about O is

perpendicular to the plane.

• If the force tends to rotate the structure clockwise, the

sense of the moment vector is out of the plane of the

structure and the magnitude of the moment is positive.

• If the force tends to rotate the structure counterclockwise,

the sense of the moment vector is into the plane of the

structure and the magnitude of the moment is negative.





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Varignon’s Theorem

• The moment about a give point O of the

resultant of several concurrent forces is equal

to the sum of the moments of the various

moments about the same point O.

• Varigon’s Theorem makes it possible to

replace the direct determination of the

moment of a force F by the moments of two

or more component forces of F.

L

rr

rr

L

rrr

+×+×=++× 2121 F r F r F F r





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MOMENT IN 3-D

Moments in 3-D can be calculated using scalar (2-D) approach but

it can be difficult and time consuming. Thus, it is often easier to

use a mathematical approach called the vector cross product.

Using the vector cross product, M O

= r F .

Here r is the position vector from point O to any point on the line

of action of F

.





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Rectangular Components of the Moment of a Force

( ) ( ) ( )k yF xF j xF zF i zF yF

F F F

z y x

k ji

k M j M i M M

x y z x y z

z y x

z y xO

rrr

rrr

rrrr

−+−+−=

=

++=

The moment of F about O,

k F jF iF F

k z j yi xr F r M

z y xO rrrr

rrrrr

rr

++=

++=×= ,





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The moment of F about B,

F r M B A B

rr

r

×= /

( ) ( ) ( )

k F jF iF F

k z z j y yi x x

r r r

z y x

B A B A B A

B A B A

rrrr

rrr

rrr

++=

−+−+−=

−=/

( ) ( ) ( )

z y x

B A B A B A B

F F F

z z y y x x

k ji

M −−−=

rrr

r





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For two-dimensional structures,

z y

Z O

z yO

yF xF M M

k yF xF M

−=

=

−=rr

( ) ( )

( ) ( ) z B A y B A

Z O

z B A y B AO

F y yF x x

M M

k F y yF x x M

−−−=

=

−−−=rr





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Sample Problem 3.1

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 A which creates the same

moment,

c) smallest force at 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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Sample Problem 3.1

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. Since the force

tends to rotate the lever clockwise, the momentvector is into the plane of the paper.

( )

( )( )in.12lb100

in.1260cosin.24

=

=°=

=

O

O

M

d

Fd M

inlb1200 ⋅=O M





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Sample Problem 3.1

b) Horizontal force at A that produces the same

moment,

( )

( )

in.8.20

in.lb1200

in.8.20in.lb1200

in.8.2060sinin.24

⋅=

=⋅

=

=°=

F

F

Fd M

d

O

lb7.57=F





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Sample Problem 3.1

c) The smallest force A to produce the same moment

occurs when the perpendicular distance is a

maximum or when F is perpendicular to OA.

( )

in.42

in.lb1200

in.42in.lb1200

⋅=

=⋅

=

F

F

Fd M O

lb50=F





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Sample Problem 3.1

d) To determine the point of application of a 240 lb

force to produce the same moment,

( )

in.5cos60

in.5lb402

in.lb1200

lb240in.lb1200

=°

=⋅

=

=⋅

=

OB

d

d Fd M O

in.10=OB





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Sample Problem 3.1

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, or on the same

line of action. None of the forces is equivalent to the100 lb force.

lecture 7- 1page

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