Download 2D-Resultants

2D-Resultants
Objectives
• To explain through examples how to replace a
number of forces acting on a rigid body by a
simple resultant force without altering the
external effects.
• To explain through examples how to specify a 3Dforce : (i) by two points on the line of action of the
force; and (ii) by two angles that orient the line of
action of the force.
• To illustrate the concept of projection and also
explain how to find out the moments about a
point and a line in three dimension force systems.
Resultants
• The resultant of a system of forces is the simplest force
combination which can replace the original forces without
altering the external effects of the system on the rigid body to
which the forces are applied.

F2

F2

F1

F1

F3

F3

R
Resultant using force polygon
• The resultant force R is obtained in magnitude and direction
by adding the forces head-to-tail in any sequence i.e. forming
the force polygon.

F2

F1
y

F3

F1

F2


F3

R
x
Resultant Calculation
For any system of coplanar forces resultant R can be calculated using:
In Vector form :
   


R  F1  F2  F3 ..........Fn  F
In Scalar form :
Rx  Fx
R y  Fy
R
Rx 2  Ry 2
Direction of the resultant with reference to x - axis.
tan  
Ry
Rx
   tan
1
Ry
Rx
Algebraic Method
• We can use algebra to obtain the resultant force and its line of action as
follows:
• 1. Choose a convenient reference point and move all forces to that point
(b). M1, M2, and M3 are the couples resulting from the transfer of forces
F1, F2, and F3 from their respective original lines of action to lines of
action through point O.
• 2. Add all forces at O to form the resultant force R (c), and resultant couple
MO (c).
• 3. Find the line of action of R by requiring R to have a moment of MO
about point O (d). Note that Σ(Fd) in (a) is equal to Rd (d).
Principle of Moments
• The moment of the resultant force about any point O equals
the sum of the moments of the original forces of the system
about the same point.
F2
O
F3
F1
Moment of the resultant force R about O  M o  Rd
Sum of the moments of the original forces about the same point  M  ( Fd )
According to principle of moments : Rd  ( Fd )
Problem
For the loaded system shown in the figure:
1. Replace the three forces and two couples by a single force R and moment M
about point O.
2. Determine the direction of R.
3. Sketch the resultant force R that represents the force-couple system alone
and find its intersection with the x- and y-axes.
100 N
y
25 mm
25 mm
20 N
50 mm
600
10 mm
20 mm
110 N
O
x
10 mm
5 Nm
30 mm
A
2 Nm
20 mm
1
 Rx   Fx  110  20 cos 60o  100 N 
100 N

 Ry   Fy  100  20 sin 60  117.3 N 
25 mm
y
25 mm
20 N
50 mm
o
600
10 mm
20 mm
110 N
R  Rx2  Ry2  (100)2  (117.3)2  154.1 N
O
x
10 mm
5 Nm
30 mm
A
2 Nm
 20 
 25 
CCW () M O  110  
  100  

20 mm
 1000 
 1000 
 50 
 30 
o
(20 sin 60o )  
  (20 cos 60 )  
  2  5  2.734 N.m (CCW)
 1000 
 1000 
2-
 Ry 
  117.3 
0
  tan    tan 1 
  49.6
 100 
 Rx 
1
3-
y
M O  Rd  d 
M O 2.734

 0.0177 m  17.7 mm
R
154.1
154.1 N
49.60
x
O
x  intercept, x 
2.734 Nm
MO
2.734

 0.0233 m  23.3 mm
Ry
 117.3
23.3 mm
M
2.734
y  intercept, y   O  
 0.0273 m  27.3 mm
Rx
100
y
154.1 N
x
49.60
O
17.7 mm
27.3 mm
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-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
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
(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-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
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
Moment in 3D
Moment in 3D
• In two-dimensional analyses it is often convenient to determine a moment
magnitude by scalar multiplication using the moment-arm rule. In three
dimensions, however, the determination of the perpendicular distance
between a point or line and the line of action of the force can be a tedious
computation. A vector approach with cross-product multiplication then
becomes advantageous.
Moment: Vector Definition
• In some two-dimensional and many of the three-dimensional problems to
follow, it is convenient to use a vector approach for moment calculations.
  
M  r F
O

r  Position v ector that runs from the moment
reference point A to any po int the line of action of F


MA r
d
O
Direction : The right - hand rule is used to identify the direction.
Thus the moment of F about O-O may be identified as a vector
pointing in the direction of thumb, with the fingers curled in the
direction of the tendency to rotate.

F

Evaluating the Cross Product



r  rx i  ry j  rz k



F  Fxi  Fy j  Fz k
  
M rF



i
j k

M  rx ry rz
Fx
Fy
Fz




M  M xi  M y j  M z k
Moment about an Arbitrary Axis


 
If Moment about a point O is M o ( r  F ), we have to find moment M  about a
line  (passing through the point O).

  
M   ( M o .n )n

  
M   (r  F .n )n
rx

M   M   Fx
ry
rz
Fy
Fz




 ,  ,  : direction cosines of the unit vecto r n
Varignon’s Theorem in Three Dimensions
The sum of the moments of a system of concurrent forces about a given
point equals the moment of their sum about the same point.

     
  
r  F1  r  F2  r  F3  .....  r  ( F1  F2  F3  .....)


 r  F

  
 

 M O  (r  F )  r  F  r  R
Problem-1
The turnbuckle is tightened until the tension in the cable AB equals 20 kN
(Figure).
(i) Determine the moment of the tension T about point O.
(ii) Determine the moment of the tension T about point B.
z
(iii)Calculate the moment of the tension T about line OC.
Note that points B, C and D lie in the x-y plane.
A
T=20 kN
30 m
D
x
O
6m
E
B 5m C 8m
y
Solution (i)


r  30k





T  20n AB  20[li  mj  nk ]



 (13  0)i  (6  0) j  (0  30)k 



 20 
  7.8i  3.6 j  18k
2
2
2
 (13  0)  (6  0)  (0  30) 






 
M O  r  T  234 j  108i  M O  108i  234 j kN.m
z
A
(0, 0, 30)
T=20 kN
Solution (ii)
As the line of action of 20 kN force is passing
through B, the force will not produce any
moment about point B.
That is MB = 0
30 m
D
x
O
6m
(0, 0, 0)
E
B 5m C 8m
(13, 6, 0)
y
(8, 6, 0)
Solution (iii)




nOC  [li  mj  nk ]



 (8  0)i  (6  0) j  (0  0)k 



nOC  
 0.8i  0.6 j

 (8  0) 2  (6  0) 2  (0  0) 2 
 




M OC  M O .nOC  (108i  234 j ).(0.8i  0.6 j )  53.6 kN.m

  


M OC  ( M O .nOC )nOC  53.6(0.8i  0.6 j )



 M OC  42.9i  32.2 j kN.m
z
A
(0, 0, 30)
T=20 kN
30 m
D
x
O
6m
(0, 0, 0)
E
B 5m C 8m
(13, 6, 0)
y
(8, 6, 0)