Engineering Mechanics: Statics: Statics

bracket P, , A 3 bracket 3 b ac et

: bracket bracket

: : ,

2

(sliding vector) (sliding vector) , , ( )

,

.

3

, ,

4

5

FFF yx += :

i jFFF yx +=

, F

:

sin cos FFFF yx ==

yyx F

FFFF 122 tan

,

=+=

:

xF

6

:

7

yx FFF 111 +=

FFF += yx FFF 222 +=

yx RRFFR +=+= 21

=+==+= xxxx

FFFRFFFR 21

=+= yyyy FFFR 21

8

51 6.22125tan 1 =

=

( )cos=x PP

( )

( ) 999622180sin260sin

2406.22180cos260

=+=

==+=

y PP ( )

N 100 , N 240

9.996.22180sin260

==

=+=

yx PP

9

tPt

5

3030

t t

o60=

6.22125tan 1 =

=

Pn

30n

Pn

10

( )15

15

( )( )

( ) 932015sin6315sin414.015sin6.115sin55.115cos6.115cos

===

======

NNFFFF

y

x15

15 ( )( ) 48.315cos6.315cos

932.015sin6.315sin===

===NN

NN

y

x

89.348.3414.0618.0932.055.1

22

=+=+===+=

yyy

xxx

NFRNFR

0.81618089.3tantan

94.389.3618.0

11

22

=

=

=

=+=

yRR

R

o0.81 , kN 94.3R

618.0

==

xR

11

Sol 1)

TT150

57.26200100tan 1

=

=

87.36200150tan 1 =

=

( ) 57116180 +

kN 15.2 57266.1

8736i== PP

( ) 57.116180 =+=

kN 3.20 5726sin

6157116sin

57.2687.36sin

== T .

..

T

12

100

Sol 2)

TT8736150tan

57.26200100tan

1

1

=

=

=

=

sin610sin6.1sin

87.36200

tan

==

=

=

PRy

15.2sin

sin6.1==

P

13

45

Bn

15(a)

45

C45

30

Pn

Pt15

R=90Nt

B

Pt

(b)B

t

60

15

60

15

A

n45

30Pn

Pt

14

45R=90N

( t) (t ) (moment) or (torque)

(M) d : Nm, lbft

FdM =

(CCW) : + (CW) : -

FdFrM ==

=sin

FrM r

15

V i Varignon

= RrMo( )+==

+=+=QrPrRrM

QPR Q)(PrRr

o

Q

qQpPRdM o +==

o

qQpo

16

d1

T

d1

W55

d2 d3O

FGA

17

d

18

F

15B

D

EF

60

45

A

CR=90N

19

F

15B

D

EF

60

45

A

CR=90N

20

(couple)(couple)

FadaFM += )(

FdM =

21

FrrF)rFrM BABA =+= )((

r)rr( FrM BA

BABA

== Q

O

M M

22

Fd

23

(f l t )- (force-couple system)

,

24

0

25

FFFFR

( ) ( )

=+++=

22

FFFFR 321

( ) ( )

+=== yxyyxx FFRFRFR 22

==

x

y

x

y

FF

RR 11 tantan

26

Step 1

O O

27

S 2 ( ) Step 2 ( - ) O R . Mo.

step 3 R O M Mo .

28

= FdM o=FR

RdM o =

29

30

O O (Varignon (Varignon )

=FR

( )o FdMM ==

oMRd =

31

couple. a is Oat System The

couple. a is Oat System The

45( ) mmN 21700

252545sin100180=

++== FdMN700

32

20=20(a) At point A(b)At point O Force

For what value of would the results of parts (a) and (b) be identical?

MAACouple

SlidingSlidingVector

33

=20

Force

MA

(a) (b) =0 =180 .

34

Couple

R

35

R

MA

kN644.142.170cos2 =+==FR

kN159.1532.170sin2

kN644.15

2.170cos2

=

==

+

yy

xx

FR

FR

kN j 159.1 i 644.1

system couple-force

+=R

( ) ( )34

5.05.170sin215.070cos25

++=

A

yy

MCCW mkN 22.2 =AM

( ) ( )

CCWmkN22.2

5.05.1532.115.0

542.1

=

+

36

CCW mkN 22.2

37

38

The rolling rear wheel of a gfront-wheel drive automobile which is accelerating to the right is subjected to the five forces and one moment shown. the forces Ax=240N and Ay=2000N are forces transmitted from the axle to the wheel, F=160N is thethe wheel, F 160N is the friction force exerted by the road surface on the tire, N=2400N is the normal reaction force exerted by thereaction force exerted by the road surface, and W=400N is the weight of the wheel/tire unit. The couple M=3N.m is the bearing friction momentthe bearing friction moment. Determine and locate the resultant of the system.

39

24001604002402000 ++== jijijFR

N.m 57)375.0(1603 N 80

===

Mi

A

m712057 ===

=Md

MRd

A

A

mm 712or m 712.0

m 712.080

=

===

dR

d

A80 N

dR

MA=57 N.m

40

cos cos

FFFF

yy

xx

==

cosFF zzy

=

222 FFFF zyx ++=

)coscoscos( kjiFkjiF

FFFF xxx

++=++=

runit vecto : , ,

)coscoscos(

kji kjiF

whereF zyx ++=

41

F F

coscoscos === lll

1

cos ,cos ,cos222 =++

===

nml

lll zyx

F

nF F F=kjin nmlF ++=

42

AB

( ) ( ) ( )121212 zzyyxxFABABFF F

++

==

kji

nF

( ) ( ) ( )( ) ( ) ( )212212212

121212 zzyyxx

zzyyxxF++

++=

kji

43

The 70-mThe 70 m microwave transmission tower t a s ss o to eis steadied by three guy cables as shown. Cable AB carries a tension of 12kN E th12kN. Express the corresponding force on point B asforce on point B as a vector.

44

602535 kji

nT

= ABT

kN75.906.469.5602535

60253512222

kji

kjiT

=

++=

kN 75.906.469.5 kji

45

The cable BCThe cable BC carries a tension of 750N Write this750N. Write this tension as a force T acting on point B inacting on point B in terms of the unit vectors i j and kvectors i, j, and k. The elbow at A forms a right anbleforms a right anble.

46

B B ( )

)693.0,4.0,6.1()30cos8.0,30sin8.0,6.1(,,

== ooBBB zyx

BC( ) ( ) ( )kji 6930214070610 +++=BC

nBC kjiTnT

zyx

BC

TTTT

++==

( ) ( ) ( )kji

kji507.01.16.1

693.02.14.07.06.10++=

+++=BC

nBC

T

j zyx

kjin507.01.16.1507.01.16.1

222 ++

++=BC

kji 253.0548.0797.0 ++= ( )

N 5.189411598253.0548.0797.0750

kjikjinT

++=++== BCT

47