Moment 2
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(0.75,1.25,-0.3)
F = {50 i + 100 j− 50 k }N
Position vector :
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
j
k
⇒ M B = 0.75 0 − 0.3
50 100 − 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 vector :
rAB = {4 i − 3 j− 9 k}m
rAB = 4 2 + ( −3) 2 + ( −9) 2 = 10.30 m
F=F
rAB
{4 i − 3 j− 9 k}m
=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
⇒ MO = F 0
0
k
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) 2 + ( 4.095) 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 sin 15° i + 0.2 cos15° j+ 0.075 k }m
⇒ rOA = {0.0518 i + 0.1932 j+ 0.075 k }m
F = {− 20 cos15° i + 20 sin 15° j}N
⇒ F = {−19.32 i + 5.18 j}N
rO
A
Look at the red arrow on xy plane:
15o
F = {− 20 cos15° i + 20 sin 15° j}N
⇒ F = {−19.32 i + 5.18 j}N
x
15o
F
(90o-15o)
y
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 
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 vectors
shown can be represented 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
j
k
⇒ M O = 0.4
− 0.2 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 coefficients
(1)
4 = y + 2z
( 2)
5 = 6z −1
− 14 = −6 y − 2 (3)
P(1, y, z)
4 = y + 2z
(1)
( 2)
5 = 6z −1
− 14 = −6 y − 2 (3)
( 2) ⇒ 6 z = 6
⇒ z =1m
Substitute in (1)
⇒ y = 4 − 2z = 4 − 2
⇒ y = 2m
⇒ P(1,2,1)