Chapter 3

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Namas ChandraIntroduction to Mechanical engineering

Hibbeler

Chapter 3-1

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Chapter 3: Force System Resultants

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Hibbeler

Chapter 3-2

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Cross Product

The Cross productof twovectors andA B

C A B

Magnitude:

C=ABsin

Direction:

C is perpendicular toboth A and B

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HibbelerChapter 3-3

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Laws of Operation for Cross Product

Commutative law is not valid

In fact

A B B A

A B B A

A B A B A B A B Scalar Multiplication

Distributive Law

A B D A B A D

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HibbelerChapter 3-4

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Cartesian Vector Formulation (sec 3.1)

Using general definition,

Magnitude: ( )( )(sin )

Direction: R.H. Rule

i j

i j

k

i j k

0

0

0

i j k i k j i i

j k i j i k j j

k i j k j i k k

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HibbelerChapter 3-5

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Cross Product of Two Vectors (sec 3.1)

x y z

x y z

A A i A j A k

B B i B j B k

( - )

- ( - )

( - )

x y z

x y z

y z z y

x z z x

x y y x

i j k

A B A A A

B B B

i A B A B

j A B A B

k A B A B

Let

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HibbelerChapter 3-6

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Moment Systems (sec 3.2)

The moment of a force about an axis

(sometimes represented as a pointin a

body) is the measure of the forces tendency

to rotate the body about the axis (or point).

The magnitude of the moment is:

Direction R.H. Rule

oM Fd

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HibbelerChapter 3-7

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Moment Systems of System of Forces (sec 3.2)

0

1, 2 3

1, 2 3

Consider a system of Forces andThey are at and from point 0.

CCW R

F F Fd d d

M Fd

It is customary to assume CCWas the positive direction.

Resultant Moment of four Forces:

50 (2 ) 60 (0) 20 (3sin 30 )40 (4 3cos30 )

334 =334 Nm(CW)

oRM N m N N mN m m

Nm

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HibbelerChapter 3-8

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Moment of a Force-Vector Formulation (sec 3.3)

0

The moment of a Force about a point O,

is the position vector of between O

and any point on the line of action of

F

M r F

r F

F

sin for any d,

:Note

r d

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HibbelerChapter 3-9

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Moment of a Force-Vector Formulation (sec 3.3)

Let

and

Then,

x y z

x y z

o x y z

x y z

F F i F j F k

r r i r j r k

i j k

M r F r r r

F F F

The axis of the moment isperpendicular to the plane that

contains both F and r

The axis passes through point O

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HibbelerChapter 3-10

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Moment of Force Systems-Vector Formulation(sec 3.3)

Let a system of forces act upon a

body. We like to compute the net

moment of all the forces about the

point O.Net moment is the sum of moment

of each force with separate

oR

F r

M r F

The moment will have three

components in , and

o

o

R

R x y z

M

x y zM M i M j M k

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EML 3004C

CCW ;

4 3

375 (11) (500) (5) (500) (0)5 5

160 cos 30 (0) 160 sin 30 (0.5)

6165 lb ft 6.16 kip ft C CW

B

B

RB

RB

R

M M

M

M

Problem 3-10 (page 84, Section 3.1-3.3)

3.10 Determine the resultant moment about point B on the

three forces acting on the beam.

Solution:

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EML 3004C


3.20 The cable exerts a 140-N force on the telephone pole as

shown. Determine the moment of this force at the baseAof

the pole. Solve the problem two ways, i.e., by using a position

vector from A to C, thenA toB.

Solution:

2 2 2

Position Vector:

r 6k m r 2i -3j m

Force Vector:

(2-0) i (-3-0) j (0-6) k F 140

(2 0) ( 3 0) (0-6)

40i - 60 j -12 0k N

AB AC

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EML 3004C



the pole. Solve the problem two ways, i.e., by using a position

vector from A to C, thenA toB.

Solution-Cont

360i 240j N m

Moment about point :

M r FA

A

Use r r

i j k

0 0 6

40 -60 -120

0 (-120) - (-60)(6) i - 0(-120)-40(6) j 0(-60)-40(0) k

AB

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the pole. Solve the problem two ways, i.e., by using a positionvector from A to C, thenA toB.

Solution-Cont

Use r r

i j k

M 2 -3 0

40 -60 -120 ( 3) (-120) - (-60)(0) i - 2(-120)-40(0) j 2(-60)-40(-3) k

360i 240j N m

AC

A

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EML 3004C

Section 3.1-3.3 (In-class Exercise)

If the man B exerts a force P=30 lbs on his rope,

determine the magnitude of F the man at C must exert

to prevent pole from tipping.

Solution:

Net moment should be zero

45

Assume CCW +ve.

30(cos45)(18) ( )(12

39

)

.8lb

F

F

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EML 3004C

Section 3.1-3.3 (In-class Challenge Exercise)

The foot segment is subjected to the pull of the two

plantar flexor muscle. Determine the moment of

each force about the point of contact A on the ground.

Solution:

1

20cos30(4.5) 20sin30(4)

118 (cw)

AM

lb

2

30cos30(4.0) 30sin 70(3.5)

140 (cw)

AM

lb

1 1

258 (cw)

A A AM M M

lb

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Namas ChandraIntroduction to Mechanical engineering HibbelerChapter 3-17

EML 3004C

3.4 Principle of Moments (sec 3.4)

The moment of a force is equal to the sum of the moment of theforces component about a point. (Varginons theorem 1654-1722)

0 1 2 1 2M r F r F F r F r F

ACable exerts F on pole with moment M .

F can be slided by the

Note A x y

principle of transmissibility

M F h F b F d

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3.5 Moment of a Force about a specified axis

0When the moment of a force F is computed using M =r F,

the axis is perpendicular to r and F.

If we need the moment about other axis still through O, we

can use either scalar or vector analysis.

Here we have F=20 N applied.

Though the typical equation gives

moment with respect to b-axis, we

require it through y-axis.

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3.5 Moment of a Force about a specified axis-2

Step 1. Find M about using cross produc

(0.3 0.4 ) ( 20 )

8 6 Nm

t.

o AM r F i j k

i j

Step 2. Find M about the given axis

( 0.8 6 )

6 Nm

=j.

y o A

Au

M M u i j j

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3.5 Moment of a Force about a specified axis-3

The two steps in the previous analysis can be combined with thedefinition of a scalar triple product. Since dot product is commutative

If and then

x y z

x y z

o a a o

a a

a a a x y z

x y z

a a a

x y z

x y z

M r F M u M

M u r F

i j k

u i u j u k r r r

F F F

u u ur r r

F F F

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3.6 Moment of a Couple-1

A couple is defined as two parallel forces with same magnitude and

opposite direction. Net force is zero, but rotates in specified direction.

is the moment of the couple.

Sum of the moment is same about any point.

Moment about O,

Since does not depend on O, the moment

is same at any point.

A B

B A

Couple moments

M r F r F

r r F r F

r

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3.6 Equivalent Couples -2

Two couples are equivalent if they produce the same

moment. The forces should be in the same or parallelplanes for two couple to be equivalent.

Couple moments are free vectors. They can be added

at any point P in the body.

1 2

1 2

There are two couples, with moments and+

FR

R

M MM M M

M r

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3.39The bracket is acted upon by a 600-N force atA.

Determine the moment of this force about theyaxis.

Solution:

Force Vector:

F 600 (cos 60 i cos 60 j cos45 k)300i 300j 424.26 k N

Position Vector:

r -0.1 i 0.15 k m

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3.39The bracket is acted upon by a 600-N force atA.

Determine the moment of this force about theyaxis.

Magnitude of the moment along axis:

j (r F)

0 1 0 -0.1 0 0.15

300 300 424.26

=0-1 (-0.1)(424.26)-(300)(0.15) 0

8

y

y

M

7.4 N m

In cartesian vector form

87

:

. 4 j N M my

Solution-Cont

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3.54 Two couples act on the frame. If d= 6 ft, determine the

resultant couple moment. Compute the result by resolving

each force intoxandycomponents and (a) finding the

moment of each couple (Eq. 3-14) and (b) summing the

moments of all the force components about pointA

1

2

1 2

(a)

100cos30 (6)

519.6 lb ft CW

4 (150)(4) 480 lb ft CCW5

519.6-480

39.6 lb ft CW

R

M

M

M M M

Solution:

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3.54 Two couples act on the frame. If d= 6 ft, determine the

resultant couple moment. Compute the result by resolving

each force intoxandycomponents and (a) finding themoment of each couple (Eq. 3-14) and (b) summing the

moments of all the force components about pointA

(b)

CCW ;

4 100cos30 (3) (150)(4) 100cos30 (9)5

39.6 lb ft

39.6 lb CW ft

R B

R

R

M M

M

M

Solution-Cont

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Solution:

a. Find the normal distance foreach case first.

Two couples act on the frame. If d = 4ft, find the resultant

couple moment by (a) direct method, and (b) resolving the x

and y components (take moment about A).

4540cos30(4) 60 (4)

53.4lb.ft (CW)

cM

3 34 45 5 5 5

40cos30(2) 40cos30(6)

60 (3) 60 (7) 60 (7) 60 (7)

53.4lb.ft (CW)

cM

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Solution:

The meshed gears are subjected to the couple moments shown.

Determine the magnitude of the resultant couple moment andspecify its coordinate direction angles.

1

2

1 2

50 N.m20cos 20sin 30 20cos 20cos30

20sin 20 N.m

= 9.397 16.276 6.840 N.m

9.397 16.276 6.840 50 N.m

= 9.397 16.276 56.840 N.m

R

M kM i j

k

i j k

M M M

i j k k

i j k

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Section 3.4-3.6 (In-class Challenge Exercise-2)

Solutioncontd.

2 2

1 9.37959.867

1 16.27659.867

1 56.84059.867

9.379 16.276 + 56.840 N.m

59.867 N.m=59.9 N.m

cos 99.0

cos 106.0

cos 18.3

RM

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3.7 Movement of a Force on a Rigid Body-1

A single force on a body can cause it to rotate (moment) and translate(force).

In the first example, the ruler

causes a force F and in addition

a moment M=Fd.

In the example, the ruler

causes a force F and NOADDITIONAL

moment.

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3.7 Movement of a Force on a Rigid Body-2

Extend this idea to a general 3-D case. Now, the force can be moved

Force now causes the force at any point 0 and then a couple.

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3.8 Resultant of a Force and Couple System-1

0

By applying the same concepts we have

x

y

z

o

R

R x

R y

R z

R

F F

F F

F F

F F

M M

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3.9 Further Reduction on Force/Couples-1

If resultant force and moment is known then it is possible to

reduce them to a single force at P. d=

o

o

R R

R

R

F M

M

F

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Concurrent Force Systems

Only equivalent force

Coplanar Force Systems A single force at d from point 0

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Parallel Force Systems

Here we have parallel forces and moments that are perpendicular.

Resultant moment (see b):

A single force =

oR C

R

M M r F

F F

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3.103 The weights of the various components of the truck

shown. Replace this system of forces by an equivalentresultant force and couple moment acting at pointA.

Force Summation:

;

3500 5500 1750

=10750lb 110.75lb

R y

R

F F

F

Moment Summation:

CCW ;

3500(20) 5500(6) 1750(2)

=99500lb ft 99.5 kip ft

RA A

RA

M M

M

Solution:

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Hibbeler

Chapter 3-37


3.93 The building slab is subjected to four parallel column

loadings. Determine the equivalent resultant force and specifyits location (x,y) on the slab.

;

30 50 40 20 140kN

M ;

-140 50(3) 30(11) 40(13)

M M ;

140 50(4) 20(

140kN

7.

10) 40(10)

14m

5.

R x

R

R xx

R yy

y

x

F F

F

M

y

x

71m

Solution:

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Hibbeler

Chapter 3-38

3.69 The gear is subjected to the two forces shown. Replace

these forces by an equivalent resultant force and couple

moment acting at point O.


Force Summation:

;

40 (2.25) 3sin 60 -0.40295 kN

41

;

9 (2.25) 3sin60 -1.0061 kN

41

Rx x

Rx

Ry y

Ry

F F

F

F F

F

Solution:

EML 3004C

3 69 Th i bj t d t th t f h R l

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Hibbeler

Chapter 3-39

3.69 The gear is subjected to the two forces shown. Replace

these forces by an equivalent resultant force and couple

moment acting at point O.

Solution-Cont

2 2

2 2

-1 -1

1.08 kN

68.2

( 0.40295) ( 1.0061)

1.0061 tan tan0.40295

Moment Summation:

CCW ;

Rx RyR

Ry

Rx

Ro o

F F F

FF

M M

0

0

3sin 60 (0.175cos 20 )

403cos60 (0.175sin 20 )+ (2.25)(0.17

0.901 kN

5)41

m CCW

RoM

EML 3004C

S ti 3 7 3 9 (I l E i )

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Namas Chandra

Introduction to Mechanical engineeringHibbeler

Chapter 3-40


Replace the force system by an equivalent force and couple

moment at the point A.

Solution:

1 2 3

300 100 400 100

100 50 500

400 300 650 N

R RF F F F F F

i j

k

i j k

EML 3004C

S ti 3 7 3 9 (I l E i 2)

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Namas Chandra


Chapter 3-41

Section 3.7-3.9 (In-class Exercise..2)

Solution (contd).

1 2 3

0 0 12 0 0 12

300

31

400 100 100 100 5

00 4800 N.m

0

0 1 120 0 500

R A

AB AB AB

M M

r F r F r F

i j k i j k

j

i j

i

k

EML 3004C

Section 3 7 3 9 (In class Challenge Exercise)

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Namas Chandra


Chapter 3-42


The weights of the various components of the truck are shown.

Replace this system by an equivalent resultant force and specifyits location from point A.

Solution: Equivalent force

1750 5500 3500

107 10.75Kip50 s

R y

y

F FF

lb

Location of force

10750( ) 3500(20) 5500(6) 1750(

9

2

2 f

)

. 6 t

AR ACCW M M

d

d

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Namas Chandra


Chapter 3-43

Chapter 3: Force SystemResultantsconcludes

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