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Lecture 5 – Rigid bodies: Moment 3D
What you will learn for today?
1. Moment of a force about a point in 3D
2. Moment of a force about an axis in 3D
• Angle between two vectors
• Nearest distance between two vectors
3. Example and exercise
By: Ts. Dr. Muhammad Hanif Ramlee
Credit to: Prof. Dato’ Ir. Dr. Mohammed Rafiq Abdul Kadir
Moment of a force about a point (3D)
Vector identity:
Cross Product (×)
Q

P
P
Q
Q
P
PQ
P
Q
PQ
Q
P










sin
sin
RIGHT HAND RULE
{ + k }
{ - k }
Moment of a force about a point (3D)
Definition of Moment about a point
F
r
M 
 where r = position vector
Start – the point where moment is taken
End – the point where the force acts
k
F
j
F
i
F
F
k
r
j
r
i
r
r
z
y
x
z
y
x






   
k
F
j
F
i
F
k
r
j
r
i
r
M z
y
x
z
y
x 





Moment of a force about a point (3D)
In 3D analysis, r and F are resolved into x, y, and z components.
sin  = 0 or 1
j
k
i
k
j
i
i
i






 0
i
k
j
j
j
k
i
j







0
0







k
k
i
j
k
j
i
k
           
     k
F
r
F
r
j
F
r
F
r
i
F
r
F
r
M
i
F
r
j
F
r
i
F
r
k
F
r
j
F
r
k
F
r
M
x
y
y
x
z
x
x
z
y
z
z
y
y
z
x
z
z
y
x
y
z
x
y
x












Moment of a force about a point (3D)
The equation can also be solved using determinant
x
x
F
r
i
y
y
F
r
j
z
z
F
r
k
           
     k
F
r
F
r
j
F
r
F
r
i
F
r
F
r
M
j
F
r
i
F
r
k
F
r
k
F
r
j
F
r
i
F
r
M
x
y
y
x
z
x
x
z
y
z
z
y
z
x
y
z
x
y
y
x
x
z
z
y












x
x
F
r
i
y
y
F
r
j
Example 1
b) Determine the shortest distance between point A and line of action of the force
a) Determine the moment of force F = 3000N about a point A.
x
y
z
D
2.2m
2.0m
0.4m
O
C
A
1.2m
Example 1
Nk
Nj
Ni
F
k
j
i
F
k
d
d
j
d
d
i
d
d
F
F
F
F
CD
CD
z
y
x
CD
CD
CD
CD
CD
2200
2000
400
3
2
.
2
3
0
.
2
3
4
.
0
3000


























 
Solutions for (a)
x
y
z
D
2.2m
2.0m
0.4m
O
C
A
1.2m
CD
AD
CD
AC
A F
r
F
r
M 



Example 1
   
       
       
Nmk
Nmj
Nmi
M
Nmi
Nmj
Nmi
Nmk
M
i
N
m
j
N
m
i
N
m
k
N
m
M
Nk
Nj
Ni
mk
mj
M
F
r
M
A
A
A
A
CD
AC
A
800
480
6800
2400
480
4400
800
2000
2
.
1
400
2
.
1
2200
0
.
2
400
0
.
2
2200
2000
400
2
.
1
0
.
2























r can be chosen from either rAC or rAD.
mk
mi
r
mk
mj
r
AD
AC
4
.
3
4
.
0
2
.
1
0
.
2





lets choose rAC
Example 1
 
m
F
M
d
Nm
M
M
M
M
M
M
F
M
d
d
F
M
A
A
A
z
y
x
A
A
A
288
.
2
3000
7
.
6863
7
.
6863
800
480
6800 2
2
2
2
2
2















Solutions for (b)
Moment of a force about a point (3D)
Vector identity:
Dot Product (·), is a scalar.
Q

P
P
Q
Q
P
PQ
P
Q
PQ
Q
P










cos
cos
Direction
not
associated
Moment of a force about a point (3D)
In 3D analysis, vectors are resolved into x, y, and z components.
cos  = 1 or 0
0
0
1






k
i
j
i
i
i
0
1
0






k
j
j
j
i
j
1
0
0






k
k
j
k
i
k
Moment of a force about a point (3D)
The dot product is used to determine:
The angle between two vectors
The moment of a force about an axis
The perpendicular / 90° / nearest distance between two
vectors (the line of action)
The angle between two vectors
    
cos
PQ
k
Q
j
Q
i
Q
k
P
j
P
i
P z
y
x
z
y
x 






cos
PQ
Q
P 

k
Q
j
Q
i
Q
Q
k
P
j
P
i
P
P
z
y
x
z
y
x






Q

P
2
2
2
z
y
x P
P
P
P 


2
2
2
z
y
x Q
Q
Q
Q 


  
cos
PQ
Q
P
Q
P
Q
P z
z
y
y
x
x 


 
PQ
Q
P
Q
P
Q
P z
z
y
y
x
x 


 
cos
Example 2
Determine the angle between vectors P = 6i +6j – 7k and Q = -6i +33j -30k
 
PQ
Q
P
Q
P
Q
P z
z
y
y
x
x 



cos
         372
30
7
33
6
6
6 







 z
z
y
y
x
x Q
P
Q
P
Q
P
  11
7
6
6
2
2
2
2
2
2







 z
y
x P
P
P
P
    45
30
33
6
2
2
2
2
2
2








 z
y
x Q
Q
Q
Q
  

38
.
41
7515
.
0
45
11
372
cos





Moment of a force about an axis
B
AB
AB
A
AB
AB
M
M
or
M
M






int
po
axis
axis M
M 
 
definition
application
where
MAB = moment of force F about axis AB
AB = unit vector from A to B
MA or MB = moment of force F about point A or point B
Example 3
Determine the moment of force F = 3000N about the axis AB.
x
y
z
2.2m
2.0m
0.4m
2.4m
2.4m
1.2m
C
D
A
B
Example 3
A
AB
AB M
M 
  B
AB
AB M
M 
 
int
po
axis
axis M
M 
 
Solution
       
CD
BD
AB
CD
BC
AB
CD
AD
AB
CD
AC
AB
AB F
r
F
r
F
r
F
r
M 










 



Observe that AB and FCD are common to all equations, the difference lies in the
position vector, r. Unless otherwise stated, it is advisable to choose the ‘simplest’ r.
Example 3
Solution
k
j
i
k
j
i
k
d
d
j
d
d
i
d
d
k
j
i
Z
Y
X
ABz
ABy
ABx
AB

































 














3
1
3
2
3
2
6
.
3
2
.
1
6
.
3
4
.
2
6
.
3
4
.
2




Example 3
Solution
Nk
Nj
Ni
F
k
j
i
F
k
d
d
j
d
d
i
d
d
F
F
F
F
Z
Y
X
CD
CD
CD
2200
2000
400
3
2
.
2
3
0
.
2
3
4
.
0
3000






















 






 










 
Example 3
Solution
mk
mj
mi
r
mj
mi
r
mk
mi
r
mk
mj
r
BD
BC
AD
AC
2
.
2
4
.
2
8
.
2
4
.
4
4
.
2
4
.
3
4
.
0
2
.
1
0
.
2












Example 3
 
CD
AC
axis
axis F
r
M 

 
Solution
   
 
   
   
   
   
 
 
Nm
Nmk
Nmj
Nmi
k
j
i
Nmi
Nmj
Nmi
Nmk
k
j
i
i
N
m
j
N
m
i
N
m
k
N
m
k
j
i
Nk
Nj
Ni
mk
mj
k
j
i
5120
3
800
3
960
3
13600
800
480
6800
3
1
3
2
3
2
2400
480
4400
800
3
1
3
2
3
2
2000
2
.
1
400
2
.
1
2200
0
.
2
400
0
.
2
3
1
3
2
3
2
2200
2000
400
2
.
1
0
.
2
3
1
3
2
3
2


















































































Nearest distance between two vectors
2
2
parallel
lar
Perpendicu F
F
F 

d
F
M lar
Perpendicu
axis 
F
F AB
parallel 
 
The component of a vector (F) that is parallel to an axis (AB) is given by
The perpendicular component of the vector is
From the definition of moment about an axis
Where d = perpendicular distance
lar
Perpendicu
axis
F
M
d 

Example 4
Determine the shortest distance between the line of action of force
F = -400Ni – 2000Nj +2200Nk and the line AB.
x
y
z
2.2m
2.0m
0.4m
2.4m
2.4m
1.2m
C
D
A
B
Example 4
Nm
M
M AB
axis 5120


 
     
m
F
M
d
N
F
N
N
N
N
Nk
Nj
Ni
k
j
i
F
F
F
F
F
lar
perpendicu
axis
lar
perpendicu
AB
parallel
parallel
lar
Perpendicu
133
.
2
2400
5120
2400
3
5400
3000
3
5400
2200
3
1
2000
3
2
400
3
2
2200
2000
400
3
1
3
2
3
2
2
2
2
2










































Solution
N
F 3000

Exercise 1
x
y
z
0.15m
0.25m
0.1m
A
B
O
F=150N
0.3m
C
Determine the moment of the F = 150N force about the diagonal BA.
Answer:
MAB = 14.52 Nm

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dynamics and static for the advanced Moments 3D.ppt

  • 1. Lecture 5 – Rigid bodies: Moment 3D What you will learn for today? 1. Moment of a force about a point in 3D 2. Moment of a force about an axis in 3D • Angle between two vectors • Nearest distance between two vectors 3. Example and exercise By: Ts. Dr. Muhammad Hanif Ramlee Credit to: Prof. Dato’ Ir. Dr. Mohammed Rafiq Abdul Kadir
  • 2. Moment of a force about a point (3D) Vector identity: Cross Product (×) Q  P P Q Q P PQ P Q PQ Q P           sin sin RIGHT HAND RULE { + k } { - k }
  • 3. Moment of a force about a point (3D) Definition of Moment about a point F r M   where r = position vector Start – the point where moment is taken End – the point where the force acts k F j F i F F k r j r i r r z y x z y x           k F j F i F k r j r i r M z y x z y x      
  • 4. Moment of a force about a point (3D) In 3D analysis, r and F are resolved into x, y, and z components. sin  = 0 or 1 j k i k j i i i        0 i k j j j k i j        0 0        k k i j k j i k                  k F r F r j F r F r i F r F r M i F r j F r i F r k F r j F r k F r M x y y x z x x z y z z y y z x z z y x y z x y x            
  • 5. Moment of a force about a point (3D) The equation can also be solved using determinant x x F r i y y F r j z z F r k                  k F r F r j F r F r i F r F r M j F r i F r k F r k F r j F r i F r M x y y x z x x z y z z y z x y z x y y x x z z y             x x F r i y y F r j
  • 6. Example 1 b) Determine the shortest distance between point A and line of action of the force a) Determine the moment of force F = 3000N about a point A. x y z D 2.2m 2.0m 0.4m O C A 1.2m
  • 8. Example 1                     Nmk Nmj Nmi M Nmi Nmj Nmi Nmk M i N m j N m i N m k N m M Nk Nj Ni mk mj M F r M A A A A CD AC A 800 480 6800 2400 480 4400 800 2000 2 . 1 400 2 . 1 2200 0 . 2 400 0 . 2 2200 2000 400 2 . 1 0 . 2                        r can be chosen from either rAC or rAD. mk mi r mk mj r AD AC 4 . 3 4 . 0 2 . 1 0 . 2      lets choose rAC
  • 9. Example 1   m F M d Nm M M M M M M F M d d F M A A A z y x A A A 288 . 2 3000 7 . 6863 7 . 6863 800 480 6800 2 2 2 2 2 2                Solutions for (b)
  • 10. Moment of a force about a point (3D) Vector identity: Dot Product (·), is a scalar. Q  P P Q Q P PQ P Q PQ Q P           cos cos Direction not associated
  • 11. Moment of a force about a point (3D) In 3D analysis, vectors are resolved into x, y, and z components. cos  = 1 or 0 0 0 1       k i j i i i 0 1 0       k j j j i j 1 0 0       k k j k i k
  • 12. Moment of a force about a point (3D) The dot product is used to determine: The angle between two vectors The moment of a force about an axis The perpendicular / 90° / nearest distance between two vectors (the line of action)
  • 13. The angle between two vectors      cos PQ k Q j Q i Q k P j P i P z y x z y x        cos PQ Q P   k Q j Q i Q Q k P j P i P P z y x z y x       Q  P 2 2 2 z y x P P P P    2 2 2 z y x Q Q Q Q       cos PQ Q P Q P Q P z z y y x x      PQ Q P Q P Q P z z y y x x      cos
  • 14. Example 2 Determine the angle between vectors P = 6i +6j – 7k and Q = -6i +33j -30k   PQ Q P Q P Q P z z y y x x     cos          372 30 7 33 6 6 6          z z y y x x Q P Q P Q P   11 7 6 6 2 2 2 2 2 2         z y x P P P P     45 30 33 6 2 2 2 2 2 2          z y x Q Q Q Q     38 . 41 7515 . 0 45 11 372 cos     
  • 15. Moment of a force about an axis B AB AB A AB AB M M or M M       int po axis axis M M    definition application where MAB = moment of force F about axis AB AB = unit vector from A to B MA or MB = moment of force F about point A or point B
  • 16. Example 3 Determine the moment of force F = 3000N about the axis AB. x y z 2.2m 2.0m 0.4m 2.4m 2.4m 1.2m C D A B
  • 17. Example 3 A AB AB M M    B AB AB M M    int po axis axis M M    Solution         CD BD AB CD BC AB CD AD AB CD AC AB AB F r F r F r F r M                 Observe that AB and FCD are common to all equations, the difference lies in the position vector, r. Unless otherwise stated, it is advisable to choose the ‘simplest’ r.
  • 21. Example 3   CD AC axis axis F r M     Solution                           Nm Nmk Nmj Nmi k j i Nmi Nmj Nmi Nmk k j i i N m j N m i N m k N m k j i Nk Nj Ni mk mj k j i 5120 3 800 3 960 3 13600 800 480 6800 3 1 3 2 3 2 2400 480 4400 800 3 1 3 2 3 2 2000 2 . 1 400 2 . 1 2200 0 . 2 400 0 . 2 3 1 3 2 3 2 2200 2000 400 2 . 1 0 . 2 3 1 3 2 3 2                                                                                  
  • 22. Nearest distance between two vectors 2 2 parallel lar Perpendicu F F F   d F M lar Perpendicu axis  F F AB parallel    The component of a vector (F) that is parallel to an axis (AB) is given by The perpendicular component of the vector is From the definition of moment about an axis Where d = perpendicular distance lar Perpendicu axis F M d  
  • 23. Example 4 Determine the shortest distance between the line of action of force F = -400Ni – 2000Nj +2200Nk and the line AB. x y z 2.2m 2.0m 0.4m 2.4m 2.4m 1.2m C D A B
  • 24. Example 4 Nm M M AB axis 5120           m F M d N F N N N N Nk Nj Ni k j i F F F F F lar perpendicu axis lar perpendicu AB parallel parallel lar Perpendicu 133 . 2 2400 5120 2400 3 5400 3000 3 5400 2200 3 1 2000 3 2 400 3 2 2200 2000 400 3 1 3 2 3 2 2 2 2 2                                           Solution N F 3000 
  • 25. Exercise 1 x y z 0.15m 0.25m 0.1m A B O F=150N 0.3m C Determine the moment of the F = 150N force about the diagonal BA. Answer: MAB = 14.52 Nm