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Where is unit vector in the direction of force
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3D-Force System
RECTANGULAR COMPONENTS
• Many problems in mechanics require analysis in three dimensions, and for such problems it is often necessary to resolve a force into its three mutually perpendicular components.
x
jFy
F
x
iFx
kFz
yz
z
y
y
xiF x
cos
jF y
cos
yx
F
zzyyxx FFFFFF cos;cos;cos
kFjFiFF
kFjFiFF
zyx
Zyx
)cos()cos()cos(
)( knjmilFF
F
n
ml
z
y
x of cosinesDirection
cos
coscos
222zyx FFFF
1 :Note 222 nml
• Where is unit vector in the direction of force
x
jFy
F
x
iFx
kFz
yz
z
y
)( knjmilFF
)( FnFF
knjmilnF
Fn
Specification of a force vector• (a) Specification by two points on the line of action of the
force.• (b) Specification by two angles which orient the line of action
of the force.
• a) Two points:
• b) Two angles:
)sinsincoscos(cos kjiF
kFjFiFF zyx
Problem-1
x
D
C
O
z
y
A
B
30 mT=24 kN
6m
5m
18 m
The turnbuckle is tightened until the tension in the cable AB equals 24 kN. Express the tension T acting on point A as a vector
A=A(0, 18, 30) B=B(6, 13, 0)
kN 22.2387.364.4
22.2387.364.4)968.0161.0194.0(24
968.0161.0194.0
968.031
300 ;161.031
1813 ;194.031
06m 31)300()1813(0)-(6points obetween tw Distance
; 222
kjiT
kjikjinTT
kjiknjmiln
nml
knjmilnnTT
AB
AB
ABAB
x
D
C
O
z
y
A
B
30 mT=24 kN
6m
5m
18 m
Problem-2
kN 10
y
x
z
045
030
Consider a force as shown in the Figure. The magnitude of this force is 10 kN. Express it as a vector.
kN 00.512.612.6 kjiF
kFjFiFF zyx
kN 0.530sin10
kN 66.830cos100
0
z
xy
F
F
kN 10
y
x
z
045
030
xF
yF
xyF
zF
kN 12.645sin66.845sin
kN 12.645cos66.845cos00
00
xyy
xyx
FF
FF
(Orthogonal) ProjectionF
n
nFnAF direction- in theor BC lineon of ProjectionA
BC
ABCn line ofdirection in ther unit vecto
nnmlknjmiln r unit vecto of cosinesdirection theare ,, Here where,
ABzzn
AByym
ABxxl )( ;)( ;)( 121212
n
),,( 111 zyxA),,( 222 zyxB
212
212
212 )()()( zzyyxxAB
Problem-1
u
v
F = 100 N
O45
15
For the shown force:a. Determine the magnitudes of the projection of the force F = 100 N onto the
u and v axes.b. Determine the magnitudes of the components of force F along the u and v
axes.
• Projections of the force onto u and v axes
• Components of the force along u and v
axes
u
v
100 N
O45
15u
v
100 N
O45
15
N 6.9615cos100
N 7.7045cos100
proj
proj
v
u
F
F
projcomp
projcomp
compcomp
N 6.81
N 9.29120sin
10045sin15sin
vv
uu
vu
FF
FF
FF
Note
Rectangular components of a force along the two chosen perpendicular axes, and projection of the force onto the same axes are the same.
N 1003.646.76:
N 3.6440sin100
N 6.7640cos100
2222
projcomp
projcomp
xx
yy
xx
FFFCheck
FF
FF
Ox
y
100 N
40
Problem-2
y
z
A
O
4 m
3 m
6 m
N 176141106 kjiF
x
. line thealong ) find ..( projection of form vector theDetermine
Figure. in theshown as axes theof Oorigin at the applied is forceA
OAFeiF
x-y-zF
OAOA
384.081.7
)03(
;512.081.7
)04( ;768.081.7
)06(m 81.7)03()04()06(
)()()(222
212
212
212
n
ml
OA
zzyyxxLOA
y
z
A
O4 m
3 m
6 mx
OAF
N 176141106 kjiF
kjin
knjmiln
OA
OA
OA
384.0512.0768.0
:is line thealongr unit vecto Therefore,
N 18.221)384.0512.0768.0).(176141106(. kjikjinFF OAOA
N 93.8424.11387.169
)384.0512.0768.0(18.221).(
kjiF
kjinFnnFF
OA
OAOAOAOAOAOA
y
z
A
O4 m
3 m
6 mx
OAF
N 176141106 kjiF
Problem-3A force with a magnitude of 100 N is applied at the origin O of the axes x-y-z as shown. The line of action of force passes through a point A. Determine the projection Fxy of 100N force on the x-y plane.
x
y
z
A
O
4 m
3 m
6 m
N 100
923.0810.7211.7
)03()04()06(
)00()04()06(cos
222
222
xy
N 3.92)923.0(100cos xyxy FF
(6,4,3)
(6,4,0)
(0,0,0)
x
y
z
A
O
4 m
3 m
6 m
N 100F
xyθB
xyF
Alternative Solution
N ˆ4.38ˆ2.51ˆ8.76810.7
ˆ3ˆ4ˆ6100
)03()04()06(
ˆ)03(ˆ)04(ˆ)06(100)ˆˆˆ(222
kjikjiF
kjiknjmilFF
jikjin
kjiknjmiln
OB
OB
ˆ554.0ˆ832.0211.7
ˆ0ˆ4ˆ6
)00()04()06(
ˆ)00(ˆ)04(ˆ)06(ˆˆˆ222
N 3.92
N 3.9204.38554.02.51832.08.76
xy
OBxy
F
nFF
(6,4,3)
(6,4,0)
(0,0,0)
x
y
z
A
O
4 m
3 m
6 m
N 100F
xyθB
xyF
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