Problem 1:
Determine the moment created by the
force F = {50i + 100j – 50k} N acting at
D, about each of the joints at B and C.
Solution:
B(0,1.25,0 ) ; C(0,0,0) ; D(0.75,1.2 5,-0.3)
F  50 i  100 j 50 k N
Position v ector :
rBD  (0.75  0) i  (1.25  1.25) j ( 0.3  0) k m
 rBD  0.75 i  0.3 k m
rCD  (0.75  0) i  (1.25  0) j ( 0.3  0) k m
 rCD  0.75 i  1.25 j 0.3 k m
F  50 i  100 j 50 k N
rBD  0.75 i  0.3 k m
rCD  0.75 i  1.25 j 0.3 k m
Moment of force F about point B is :
M B  rBD F
i
 M B  0.75
50
j
k
0
 0.3
100
 50
(0)( 50)  ( 0.3)(100)i  (0.75)( 50)  ( 0.3)(50) j
 MB  
 N.m
 (0.75)(100)  (0)(50)k

 M B  30 i  22.5 j 75 k N.m
F  50 i  100 j 50 k N
rBD  0.75 i  0.3 k m
rCD  0.75 i  1.25 j 0.3 k m
Moment of force F about point C is :
M C  rCD  F
i
j
k
 M C  0.75 1.25  0.3
50
100
 50
(1.25)( 50)  ( 0.3)(100)i  (0.75)( 50)  ( 0.3)(50) j
 MC  

 (0.75)(100)  (1.25)(50)k

 M C   32.5 i  22.5 j 12.5 k N.m
Problem 2:
Determine Smallest force F that must be
applied to the rope, when held in the
direction shown, in order to cause the
pole to break at its base O. This
requires a moment of M = 900 N.m to
be developed at O.
Solution:
A(0,0,10.5 ) ; B(4,-3,1.5)
M  900 N.m
Position v ector :
rAB  4 i  3 j 9 k m
rAB  4 2  ( 3) 2  ( 9) 2  10.30 m
FF
4 i  3 j 9 km
rAB
F
rAB
10.30 m
 F  F 0.39 i  0.29 j 0.87 k m
A(0,0,10.5 ) ; B(4,-3,1.5)
M  900 N.m
F  F 0.39 i  0.29 j 0.87 k m
M O  rOA  F
i
j
k
 MO  F 0
0
10.5
0.39  0.29  0.87
(0)( 0.87 )  (10.5)( 0.29)i 


 M O  F  (0)( 0.87 )  (10.5)( 0.39) j N.m
 (0)( 0.29)  (0)( 0.39)k 


 M O  F 3.045 i  4.095 jN.m
M  900 N.m
M O  F 3.045 i  4.095 jN.m
F
MO
(3.045)  ( 4.095)
2
2

900
5.10
 F  176.5 N
Problem 3:
A 20-N horizontal force is applied
perpendicular to the handle of the socket
wrench. Determine the magnitude and
direction of the moment created by this
force about point O.
Solution:
rOA  0.2 sin15 i  0.2 cos15 j 0.075 km
 rOA  0.0518 i  0.1932 j 0.075 km
F   20 cos 15 i  20 sin15 j N
 F   19.32 i  5.18 j N
rO
A
Look at the red arrow on xy plane:
15o
F   20 cos 15 i  20 sin15 j N
 F   19.32 i  5.18 j N
x
15o
F
rOA  0.0518 i  0.1932 j 0.075 k m
F   19.32 i  5.18 j N
M O  rA  F
i
j
k
 M O  0.0518 0.1932 0.075
 19.32 5.18
0
(0.1932)( 0)  (0.075)(5.18)i



 M O   (0.0518)( 0)  (0.075)( 19.32) j
 N.m
 (0.0518)(5.18)  (0.1932)( 19.32)k 


 M O   0.39 i  1.45 j 4.00 k N.m
M O   0.39 i  1.45 j 4.00 k N.m
 M O  ( 0.39) 2  ( 1.45) 2  ( 4.00) 2  4.27 N.m
  0.39 
α  cos 1 
  95.2
 4.27 
  1.45 
β  cos 1 
  109.9
 4.27 
 4.00 
γ  cos 1 
  20.5
 4.27 
y
(90o-15o)
Problem 4:
The force F acts at the end of the angle
bracket shown. Determine the moment of
the force about point O.
It is noted that in this problem the scalar
analysis provides a more convenient
method for analysis that the vector
analysis.
Solution I (Scalar analysis)
The force is resolved into its x and y components .
Moments of the components are computed about point O :
 M O  400 sin 30(0.2)  400 cos 30(0.4)
 M O  98.6 N.m  98.6 N.m (clockwise )
Solution II (Vector analysis)
Using a Cartesian vector approach, the force and position v ectors
shown can be represente d as :
r  0.4 i  0.2 j m
F  400 sin 30 i  400 cos 30 j N
 F  200 i  346.4 j N
M O  r F
i
 M O  0.4
j
 0.2
k
0
200  346.4 0
 M O  0 i  0 j 98.6 k
 M O   98.6 k  N.m
As it is clear, it is recommended to use the scalar analysis for 2-dimensional problems,
and the vector method for 3-dimensional problems.
Problem 5:
A force F = {6i –2j +1k} kN produces a
moment of MO = {4i + 5j –14k} kN.m
about the origin of coordinates, point O.
If the force acts at a point having an x
coordinate of x=1 m, determine the y
and z coordinates.
Solution:
P(1, y, z)
F  6 i  2 j 1k  kN
M O  4 i  5 j 14 k  kN.m
M O  r F
i
j
k
 4 i  5 j 14 k  1 y z
6 2 1
 4 i  5 j 14 k  ( y  2 z ) i  (1  6 z ) j ( 2  6 y ) k
Equate i, j, and k coefficien ts
4  y  2z
5  6z 1
 14  6 y  2
(1)
( 2)
(3)
P(1, y, z)
4  y  2z
5  6z 1
 14  6 y  2
(1)
( 2)
(3)
( 2)  6 z  6
 z 1m
Substitute in (1)
 y  4  2z  4  2
 y 2m
 P(1,2,1)