Problems (on Moment)
1. The rod on the power control mechanism for a business jet is subjected to a force of
80 N. Determine the moment of this force about the bearing at A.
+
M A  80
200.15 sin 60   80
200.15 cos 60 

cos



sin


Fx
M A  7.71 N  m
(counterclockwise - ccw)
Fy
Problems (on Moment)
2. In order to raise the lamp post from the position
shown, force F is applied to the cable. If F=200 N,
determine the moment produced by F about point A.
Geometry:
200sinq
q
6m
B
200cosq
6m
C
3m
A
3m
Cosine law:
BC  32  6 2  236  cos105
2
Sine law:
7.37
3


sin 105 sin q
q  23.15o
+
M A  200 sin 23.156   471.768 N  m
(counterclockwise - ccw)
BC  7.37 m
Problems (on Moment)
3. Calculate the moment Mo of the 250 N force about the base point O of the robot.
20o
Fx = 250sin40 N
Fy = 250cos40 N
400 mm
A
60o
50o 40o
30o
40o
Fy
500 mm
300 mm
O
+
0.50.4sin30+0.3sin40
Fx
0.4cos30+0.3cos40
M o  250 cos 400.4 cos 30  0.3 cos 40   250 sin 400.5  0.4 sin 30  0.3 sin 40 
M o  189.55 N  m
(counterclockwise - ccw)
4. The towline exerts a force of P=4 kN at the end of the 20-m-long crane boom. If x=25 m, determine
the position of the boom so that this force creates a maximum moment about point O. What is this
B
moment?
q
1.5 m
O
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑚𝑜𝑚𝑒𝑛𝑡, OB  AB


25 sin   1.5 cos   20 / : cos 
25 tan   1.52  400 sec 2 
625 tan 2   75 tan   2.25  400 1  tan 2 

 225 tan 2   75 tan   397.75  0
75  752  4225 397.75

2225
tan   m
225m  75m  397.75  0

m1, 2
m1, 2 tan   
1.5067

q  90  56.426  33.574o
2
 1.1733

M o max  420  80 kN  m
4 sin  25  4 cos  1.5  80 
25 tan   1.5  20 sec 
25 m
  56.426o
A
5. The 120 N force is applied as shown to one end of the curved wrench. If a=30o, calculate
the moment of F about the center O of the bolt. Determine the value of a which would
maximize the moment about O; state the value of this maximum moment.
If a=30o, MO=?
o
F = 120 N 30
A
+
(70+150+70)=290 mm
(25+70+70+25)
=190 mm
O
M o  120
cos
3070  150  70   120
sin 
3025  70  70  25



 



Fy
290
Fx
M o  41500 N  mm  41.5 N  m
190
(clockwise - cw)
Determine the value of a which would maximize the moment about O. MOmax=?
F = 120 N a
A
O
(70+150+70)
=290 mm
(25+70+70+25)
=190 mm
a
For MOmax, F must be perpendicular to OA.
 190 
o
a  arctan
  33.2
 290 
M o  120 190 2  290 2  41603.85 N  mm  41.6 N  m
 M o  F cos a0.29   F sin a0.19 
 dM

o
0

 F sin a0.29   F cos a0.19  

 da


0.19
0.29
a  33.2o
tan a 






Problems (on Moment)
6. The force exerted by the hydraulic cylinder AB on the door is 3 kN (constant) directed
from A to B. Determine the moment of this force about point O (Mo) as a function of q.
Also determine Mo for q =50o.
40 cm
O
q
150 cm
A
B
20 cm
Problems (on Moment)
7. The spring-loaded follower A bears
against the circular portion of the cam
until the lobe of the cam lifts the
plunger. The force required to lift the
plunger is proportional to its vertical
movement h from its lowest position.
For design purposes determine the angle
q for which the moment of the contact
force on the cam about the bearing O is
a maximum. In the enlarged view of the
contact, neglect the small distance
between the actual contact point B and
the end C of the lobe.
Problems (on Moment)
8.
The
pipe
assembly
is
subjected to an 80 N force.
Determine the moment of this
force about point A.
Also determine the minimum
distance from point A to the line
of action of force 𝐹.
Force in vector form
Fxy  80 cos 30  69.28 N
Fx  Fxy sin 40  69.28 sin 40  44.53 N
Fy  Fxy cos 40  69.28 cos 40  53.07 N
Fz  80 sin 30  40 N




F  44.53i  53.07 j  40k
𝒓𝑪/𝑨
Position vector from point A to point C





rC / A  0.4 j  0.3i  0.2k  0.25i




rC / A  0.55i  0.4 j  0.2k
Fx
Fxy
Moment of force 𝑭 about point A




M A  rC / A  F

Fy

Fz







M A  0.55i  0.4 j  0.2k  44.53i  53.07 j  40k











M A  29.19k  22 j  17.81k  16i  8.91 j  10.61i M A  5.39i  13.09 j  11.38k
minimum distance from point A to the line of action of force 𝑭
d: minimum distance



M A  rC / A  F

 


M A  rC / A  F  rC / A F sin q

d  rC / A sin q


MA  d F



 F

M A  5.39i  13.09 j  11.38k
Magnitude of the moment

M A  5.39 2  13.09 2  11.382  18.16 N  m
18.16  d 80 
𝒅 = 𝟎. 𝟐𝟐𝟕 𝒎
q
𝒓𝑪/𝑨
Problems (on Moment)
9. Concrete pump is subjected to force 𝐹 acting on point C as shown in the figure.
Arms OA, AB and BC lie in the same plane which makes 35o with xz plane.
a) Determine the moment of force 𝐹 about point 𝑂.
b) Determine the moment of force 𝐹 about line 𝑂𝐴.
35o
10. The spring which connects point B of the disk and
point C on the vertical surface is under a tension of 500 N.
Write this tension as it acts on point B as a force vector
and determine the moment Mz of this force about the shaft
axis OA.
𝑭spring=Fspring𝒏C/B
𝑭spring
Coordinates :
𝐵 −150𝑠𝑖𝑛30, 150𝑐𝑜𝑠30, 600 𝑩 −𝟕𝟓, 𝟏𝟐𝟗. 𝟗, 𝟔𝟎𝟎
𝑪(𝟑𝟓𝟎, 𝟑𝟎𝟎, 𝟎)







425i  170.1 j  600k
Fspring  500
 281.57i  112.7 j  397.5k
754.69



Position vector from point O to point C (It may be
rC / O  0.35i  0.3 j
written the position vector from point O to point B)



Moment of force 𝑭𝑠𝑝𝑟𝑖𝑛𝑔 about point O M O  rC / O  Fspring









M O  0.35i  0.3 j  281.57i  112.7 j  397.5k  119.25i  139.25 j  45.02k

Moment about the shaft axis M OA

 
 
 M O  nOA  M O  k  45.02 N  m
11. Strut AB of the 1 meter diameter hatch door exerts a force of 450 N on point B. Determine
the moment of this force about point O. Line OB of the hatch door lies in the yz plane.
z
𝐴 0.5𝑠𝑖𝑛30, 0.5 + 0.5𝑐𝑜𝑠30, 0
𝐵 0, 1𝑐𝑜𝑠30, 1𝑠𝑖𝑛30
𝑨 𝟎. 𝟐𝟓, 𝟎. 𝟗𝟑𝟑, 𝟎
𝑩(𝟎, 𝟎. 𝟖𝟔𝟔, 𝟎. 𝟓)
z
B
O
𝑭
30o
0.5 m
30o
y
A
x

F  450



0  0.25i  0.866  0.933 j  0.5  0k
0.252  0.067 2  0.52



 200.89i  53.84 j  401.79k
z




F  200.89i  53.84 j  401.79k
Position vector from point O to point B (It may be
written the position vector from point O to point A)



rB / O  0.866 j  0.5k



M O  rB / O  F

B
𝑭
 O







M O  0.866 j  0.5k   200.89i  53.84 j  401.79k
x




M O  374.86i  100.45 j  179.97 k
! Obtain the same result using position vector 𝒓𝑨/𝑶






rA / O  0.5 j  0.5 cos 30 j  0.5i  0.5i  0.933 j
0.5 m
𝑀𝑂 = 𝑀𝑂 =
𝑀𝑂
𝑑=
= 0.951 𝑚
𝐹
30o
A



M O  rA / O  F
Minimum distance from point O to line AB, d:
𝐹 = 450 𝑁 ,
y
30o
374.862 + 100.452 + 179.972 = 427.84 𝑁. 𝑚
Problems (on Moment)
12. Concentrated force is acting perpendicular to the crank arm BC at point C. For
the position q=70°, what is Mx , the moment of about the x axis? At this instant,
My =20 N·m and Mz =37.5 N·m.
P
100 mm
y
B

C
O
200 mm
q
z
150 mm
A
x
q=70°, My =20 N·m and Mz =37.5 N·m Mx=?
P
P



P   P sin j  P cos k
y

B, C

B
C
200 mm
z
70o
y
200 mm
O
O, A
q
z
A





rC / O  r  250i  200 sin 70 j  200 cos 70k




r  250i  187.94 j  68.4k

 
Mo  r  P






M o  250i  187.94 j  68.4k   P sin j  P cos k





M o  250 P sin k  250 P cos j  187.94 P cos i  68.4 P sin i



x
q=70°, My =20 N·m and Mz =37.5 N·m Mx=?





M o  250 P sin k  250 P cos j  187.94 P cos i  68.4 P sin i
M y  250 P cos   20 103 N  mm
1
M z  250 P sin   37.5 103 N  mm 2
2
1
 250 P sin   37.5 103

 250 P cos 
 20 103

tan   1.875
  61.93
P  170 N
M x  187.94 P cos   68.4 P sin   25294 N  mm  25.294 N  m
Problems (on Moment)
z
13. Rectangular plate ADCF is held
A (0, 5, 3) m
in equilibrium by string AB which
D (0, 2, 2)
y
has
a
tension
of
T=14
kN.
Determine,
F
O
a) The magnitude of the moment of
C (4, 6, 0) m
about the axis DC,
b) The perpendicular distance
B (6, 3, 0) m
x
between lines AB and DC.
z
A (0, 5, 3) m
D (0, 2, 2)
y
O
C (4, 6, 0) m




 6i  2 j  3k 

T  TnT  14 

7






T  12i  4 j  6k



 4i  4 j  2k  2  2  1 

nDC  
 i  j k
6
3
3

 3
a) The magnitude of the moment of about the axis DC
B (6, 3, 0) m
x
M DC  ?



 



M C  rB / C  T  2i  3 j  12i  4 j  6k



 18i  12 j  28k
  2  2  1 
 


M DC  M C  nDC  18i  12 j  28k   i  j  k 
3
3 
3
2
2
 1
 18   12   28  
3
3
 3
M DC  12  8-9.33  10.67 kN  m




z
b) The perpendicular distance between lines AB and DC
A (0, 5, 3) m
D (0, 2, 2)
y
O
C (4, 6, 0) m
B (6, 3, 0) m




T  12i  4 j  6k ,
 
 
T  nDC  T nDC cosq

nDC

 
2i  2 j  k

3
 2   2   1  22
 12   4   6   
 3  3  3 3
22
22
 14(1) cos q , cos q 
 0.52
3
42
q  58.67 
x
Theparallel component of a force to a line does not generate moment with respect to that line.
If T is resolved into two components parallel and normal to line DC, then
M DC  T sin θ  d
M DC
10.67
, d

 0.89 m
T sin θ 14 sin 58.67

The magnitude of moment MDC is equal to the normal component of T to line DC and the
perpendicular distance between T and line DC.
14. The arms AB and BC of a desk lamp lie in a vertical plane that forms an angle of 30o with xy
plane. To reposition the light, a force of magnitude 8 N is applied as shown. Determine,
a) the moment of the force about point A,
b) the moment of the force about the axis of arm AB,
c) the angle between the line of action of force and line AB,
d) the perpendicular distance between the line AB and line of action of force.
Direction CD is parallel to the z-axis and
lies in a parallel plane to the horizontal
plane.
45o
y
AB  450 mm , BC  325 mm
D
50o
45o
20o
C
A
150 mm
x
30o
z




F  Fx i  Fy j  Fz k
Fx  8Cos 45Sin20  1.93 N 

Fy  8Sin45  5.66 N


Fz  8Cos 45Cos 20  5.32 N 
a)




F  1.93i  5.66 j  5.32k
MA ?




rC / A  0.45Sin45  0.325Sin50 Cos 30i  0.45Cos 45  0.325Cos 50  j  0.45Sin45  0.325Sin50 Sin30k




rC / A  0.491i  0.11 j  0.2835k



M A  rC / A  F







M A  0.491i  0.11 j  0.2835k   1.93i  5.66 j  5.32k



 2.19i  3.16 j  2.57 k

 

AB  450 mm , BC  325 mm
b) M AB  ?
 
M AB  M A  n AB




rAB  0.45Sin45Cos 30i  0.45Cos 45 j  0.45Sin45Sin30k




rAB  0.275i  0.318 j  0.159k




rAB

n AB    0.61i  0.71 j  0.35k
rAB


AB  450 mm , BC  325 mm







M AB  2.19i  3.16 j  2.57 k  0.61i  0.71 j  0.35k  1.81 Nm







M AB   1.81 0.61i  0.71 j  0.35k  1.1i  1.28 j  0.64k



c) F  n AB  F n AB Cosq










 1.93i  5.66 j  5.32k  0.61i  0.71 j  0.35k  81Cos q
 3.3  8Cos q
q  114.63
d) If the force is resolved into two components
parallel and normal to line AB, only normal component
generate a moment. The parallel component of a force to a line does not generate moment with respect to
that line.
M AB  FSinqd
d
M AB
1.81

 0.249 m
FSinq 8Sin114.63