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
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