Transcript ge201-lecture_7.pptx

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.
z




F  Fx i  Fy j  FZ k



Fz k


F  ( F cos x )i  ( F cos y ) j  ( F cos z )k

z
Fx  F cos  x ; Fy  F cos  y ; Fz  F cos  z

Fx i
F  Fx  Fy2  Fz2
2
l  cos  x 


m  cos  y Direction cosines of F

n  cos  z 




F  F (li  mj  nk )
x
Note : l 2  m 2  n 2  1
F
y
y

Fy j
x

F
y

F cos y j 
y
x

F cos  x i
x




F  F (li  mj  nk )


 F  F (nF )
• Where nF is unit vector in the direction of force




nF  li  mj  nk
z

Fz k
z

Fx i
x
x

F
y

Fy j
y
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:




F  Fx i  Fy j  Fz k 



 F (cos  cos i  cos  sin j  sin k )
Problem-1
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
z
18 m
A
T=24 kN
30 m
D
O
6m
x
B
5m
C
y
z
18 m
A=A(0, 18, 30)
B=B(6, 13, 0)
A
T=24 kN
30 m


T  T n AB ;




n AB  li  mj  nk
D
Distance between tw o points  (6 - 0)  (13  18)  (0  30)  31 m
2
2
2
60
13  18
0  30
l
 0.194; m 
 0.161; n 
 0.968
31
31
31







n AB  li  mj  nk  0.194i  0.161 j  0.968k








T  T n AB  24(0.194i  0.161 j  0.968k )  4.64i  3.87 j  23.22k




 T  4.64i  3.87 j  23.22k kN
x
O
6m
B
5m
C
y
Problem-2
Consider a force as shown in the Figure. The magnitude of this force is 10 kN.
Express it as a vector.
z
10 kN
300
450
x
y
z
Fxy  10 cos 300  8.66 kN
Fz
Fz  10 sin 30  5.0 kN
0
Fx  Fxy cos 45  8.66 cos 45  6.12 kN
0
0
Fy  Fxy sin 450  8.66 sin 450  6.12 kN
Fx




F  Fx i  Fy j  Fz k 




 F  6.12i  6.12 j  5.00k kN
x
10 kN
Fy
300
450
Fxy
y
(Orthogonal) Projection

F
n

C
B
A
Projection of F on line ABC or in the n - direction  Fn

n  unit vector in the direction of line ABC


 
where, n  li  mj  nk
l

Here l , m, n are the direction cosines of unit vector n
( x2  x1 )
(y  y )
(z  z )
; m 2 1 ; n 2 1
AB
AB
AB
A( x1, y1, z1 )

n
B( x2 , y2 , z2 )
AB  ( x2  x1 ) 2  ( y2  y1 ) 2  ( z 2  z1 ) 2
Problem-1
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.
v
F = 100 N
15
O
45
u
• Projections of the force
onto u and v axes
• Components of the
force along u and v
v axes
v
100 N
15
100 N
15
O
45
O

45
u
u
Fu proj  100 cos 45  70.7 N
Fv proj  100 cos15  96.6 N
Fu comp

sin 15
 Fu comp
Fv comp
100
sin 45 sin 120
 29.9 N  Fu proj

 Fv comp  81.6 N  Fv proj
Note
Rectangular components of a force along the two chosen
perpendicular axes, and projection of the force onto the same
axes are the same.
y
100 N
40
O
x
Fx comp  100 cos 40  76.6 N  Fx proj
F 
y comp
 100 sin 40  64.3 N  Fy proj
Check : F  Fx  Fx  76.6 2  64.32  100 N
2
2
Problem-2
A force F is applied at the origin O of the axes x-y-z as shown in the Figure.
Determine the vector form of projection FOA (i.e. find F OA ) along the line OA.
z




F  106i  141 j  176k N
y
A
3m
O
4m
x
OA  L  ( x2  x1 ) 2  ( y2  y1 ) 2  ( z2  z1 ) 2
 OA  (6  0) 2  (4  0) 2  (3  0) 2  7.81 m
(6  0)
(4  0)
 0.768; m 
 0.512;
7.81
7.81
(3  0)
n
 0.384
7.81
l
z




F  106i  141 j  176k N
y
A
Therefore, unit vecto r along the line OA is :

FOA
n OA  l i  m j  n k
 n OA  0.768i  0.512 j  0.384k
3m
O
4m
x







FOA  F.nOA  (106i  141 j  176k ).(0.768i  0.512 j  0.384k )  221.18 N





 FOA  ( FOA.nOA )nOA  FOA nOA  221.18(0.768i  0.512 j  0.384k )




 FOA  169.87i  113.24 j  84.93k N
z




F  106i  141 j  176k N
y
A

FOA
3m
O
4m
x
Problem-3
A 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.
z
y
100 N A
3m
O
4m
x
z
cos  xy 
(6  0)  (4  0)  (0  0)
2
2
y
2
(6  0) 2  (4  0) 2  (3  0) 2

7.211
 0.923
7.810
Fxy  F cos  xy  100(0.923)  92.3 N
F  100 N A (6,4,3)
3m
θ xy
(0,0,0)
O
Fxy
(6,4,0)
B
4m
x
Alternative Solution
 (6  0)iˆ  (4  0) ˆj  (3  0)kˆ

ˆ
F  F (liˆ  mˆj  nk )  100
 (6  0) 2  (4  0) 2  (3  0) 2


 6iˆ  4 ˆj  3kˆ 
  76.8iˆ  51.2 ˆj  38.4kˆ N
 F  100

 7.810 


 z

y
F  100 N A (6,4,3)

(6  0)iˆ  (4  0) ˆj  (0  0)kˆ
ˆ
ˆ
ˆ
nOB  li  mj  nk 
(6  0) 2  (4  0) 2  (0  0) 2

6iˆ  4 ˆj  0kˆ
 nOB 
 0.832iˆ  0.554 ˆj
7.211
(0,0,0)
O
 
Fxy  F  nOB  76.8  0.832  51.2  0.554  38.4  0  92.3 N
 Fxy  92.3 N
3m
θ xy
Fxy
(6,4,0)
B
4m
x