Statics chapter 2

1. Chapter 3
Force System Resultants

2. 3-1 MECHANICS

3. 3-1 MECHANICS
Objective:
1. Understand different definitions of statics and mechanics of
materials.

4. 3-2 Cross Product
• The Cross product of two
vectors
Magnitude:
C=AB sin
Direction:
C is perpendicular to
both A and B
C A B
 

5. 3-2 Cross Product
Commutative law is not valid
In fact
A B B A
A B B A
  
   
       
A B A B A B A B
   
      
Scalar Multiplication
Distributive Law
 
A B D A B A D
     
Laws of Operation for Cross Product

6. 3-2 Cross Product
Using general definition,
ˆ ˆ
Magnitude: ( )( )(sin )
ˆ
Direction: R.H. Rule
ˆ
ˆ ˆ
i j
i j
k
i j k



 
0
0
0
i j k i k j i i
j k i j i k j j
k i j k j i k k
      
      
      
Cartesian Vector Formulation

7. 3-2 Cross Product
x y z
x y z
A A i A j A k
B B i B j B k
  
  
( - )
- ( - )
( - )
x y z
x y z
y z z y
x z z x
x y y x
i j k
A B A A A
B B B
i A B A B
j A B A B
k A B A B
 


Let
Cross Product of Two Vectors

8. 3-2 Cross Product
Moment Systems
The moment of a force about an axis
(sometimes represented as a point in a
body) is the measure of the force’s tendency
to rotate the body about the axis (or point).
The magnitude of the moment is:
Direction R.H. Rule
o
M Fd



9. Moment Systems of System of Forces
0
1, 2 3
1, 2 3
Consider a system of Forces and
They are at and from point 0.
CCW R
F F F
d d d
M Fd
  
It is customary to assume CCW
as the positive direction.
Resultant Moment of four Forces:
50 (2 ) 60 (0) 20 (3sin30 )
40 (4 3cos30 )
334 =334 Nm(CW)
o
R
M N m N N m
N m m
Nm
   
 
 
3-2 Cross Product

10. 3-3 Moment of a Force-Vector Formulation
0
The moment of a Force about a point O,
is the position vector of between O
and any point on the line of action of
F
M r F
r F
F
 
sin for any d,
:
Note
r d
 


11. 3-3 Moment of a Force-Vector Formulation
Let
and
Then,
x y z
x y z
o x y z
x y z
F F i F j F k
r r i r j r k
i j k
M r F r r r
F F F
  
  
  
The axis of the moment is
perpendicular to the plane that
contains both F and r
The axis passes through point O

12. 3-3 Moment of a Force-Vector Formulation
Net moment is the sum of moment
of each force with separate
o
R
F r
M r F
 

The moment will have three
components in , and
o
o
R
R x y z
M
x y z
M M i M j M k
  
Let a system of forces act upon a
body. We like to compute the net
moment of all the forces about the
point O.

13. Problem 3-10 (Page 84, Section 3.1-3.3)
3.10 Determine the resultant moment about point B on the
three forces acting on the beam.
Solution:
CCW ;
4 3
375 (11) (500) (5) (500) (0)
5 5
160 cos 30 (0) 160 sin 30 (0.5)
6165 lb ft 6.16 kip ft CCW
B
B
RB
RB
R
M M
M
M
 
 


   
  


14. Problem 3-20 (Page 86, Section 3.1-3.3)
3.20 The cable exerts a 140-N force on the telephone pole as
shown. Determine the moment of this force at the base A of
the pole. Solve the problem two ways, i.e., by using a position
vector from A to C, then A to B.
Solution:
   
 
2 2 2
Position Vector:
r 6k m r 2i -3j m
Force Vector:
(2-0) i (-3-0) j (0-6) k
F 140
(2 0) ( 3 0) (0-6)
40i - 60 j -120k N
AB AC
 
 
 
  
 
    
 


15. Problem 3-20 (Page 86, Section 3.1-3.3)
3.20 The cable exerts a 140-N force on the telephone pole as
shown. Determine the moment of this force at the base A of
the pole. Solve the problem two ways, i.e., by using a position
vector from A to C, then A to B.
Solution-Con’t
Moment about point :
M r F
A
A
 
     
Use r r
i j k
0 0 6
40 -60 -120
0 (-120) - (-60)(6) i - 0(-120)-40(6) j 0(-60)-40(0) k
AB


 
  
360i 240j N m
 

16. Problem 3-20 (Page 86, Section 3.1-3.3)
3.20 The cable exerts a 140-N force on the telephone pole as
shown. Determine the moment of this force at the base A of
the pole. Solve the problem two ways, i.e., by using a position
vector from A to C, then A to B.
Solution-Con’t
     
 
Use r r
i j k
M 2 -3 0
40 -60 -120
( 3) (-120) - (-60)(0) i - 2(-120)-40(0) j 2(-60)-40(-3)
360i 240j N m
AC
A
 


  


17. Section 3.1-3.3 (In-class Exercise)
If the man B exerts a force P=30 lbs on his rope,
determine the magnitude of F the man at C must exert
to prevent pole from tipping.
Solution:
Net moment should be zero
4
5
Assume CCW +ve.
30(cos45)(18) ( )(12
39
)
.8lb
F
F




18. Section 3.1-3.3 (In-class Challenge Exercise)
The foot segment is subjected to the pull of the two
plantar flexor muscle. Determine the moment of
each force about the point of contact A on the ground.
Solution:
 1
20cos30(4.5) 20sin30(4)
118 (cw)
A
M
lb
  
 
 2
30cos30(4.0) 30sin70(3.5)
140 (cw)
A
M
lb
  
 
     
1 1
258 (cw)
A A A
M M M
lb
 
 

19. 3-4 Principle of Moments
The moment of a force is equal to the sum of the moment of the
force’s component about a point. (Varginon’s theorem 1654-1722)
 
0 1 2 1 2
M r F r F F r F r F
        
A
Cable exerts F on pole with moment M .
F can be slided by the
Note A x y
principle of transmissibility
M F h F b F d
  

20. 3-5 Moment of a force about a specified axis
0
When the moment of a force F is computed using M =r F,
the axis is perpendicular to r and F.

If we need the moment about other axis still through O, we
can use either scalar or vector analysis.
Here we have F=20 N applied.
Though the typical equation gives
moment with respect to b-axis, we
require it through y-axis.

21. 3-5 MOMENT OF A FORCE ABOUT A SPECIFIED AXIS-2
 
Step 1. Find M about using cross produc
(0.3 0.4 ) ( 20 )
8 6 Nm
t.
o A
M r F i j k
i j
     
   ˆ
Step 2. Find M about the given axis
ˆ ( 0.8 6 )
6 Nm
=j.
y o A
A
u
M M u i j j
   


22. 3-5 Moment of a force about a specified axis- 3
The two steps in the previous analysis can be combined with the
definition of a scalar triple product. Since dot product is commutative
 
 
ˆ
If and then
ˆ
x y z
x y z
o a a o
a a
a a a x y z
x y z
a a a
x y z
x y z
M r F M u M
M u r F
i j k
u i u j u k r r r
F F F
u u u
r r r
F F F
  
 
  


23. 3-5 Moment of a Couple- 1
A couple is defined as two parallel forces with same magnitude and
opposite direction. Net force is zero, but rotates in specified direction.
 
 
is the moment of the couple.
Sum of the moment is same about any point.
Moment about O,
Since does not depend on O, the moment
is same at any point.
A B
B A
Couplemoments
M r F r F
r r F r F
r
    
    

24. 3-5 Moment of a Couple- 2
Two couples are equivalent if they produce the same
moment. The forces should be in the same or parallel
planes for two couple to be equivalent.
Couple moments are free vectors. They can be added
at any point P in the body.
 
1 2
1 2
There are two couples, with moments and
+
F
R
R
M M
M M M
M r

 


25. Problem 3-39 (Page 95, Section 3.4-3.6)
3.39The bracket is acted upon by a 600-N force at A.
Determine the moment of this force about the y axis.
Solution:
 
 
Force Vector:
F 600 (cos 60 i cos 60 j cos45 k)
300i 300j 424.26 k N
Position Vector:
r -0.1 i 0.15 k m
     
  
 

26. Problem 3-39 (Page 95, Section 3.4-3.6)
3.39The bracket is acted upon by a 600-N force at A.
Determine the moment of this force about the y axis.
 
Magnitude of the moment along axis:
ˆ
j (r F)
0 1 0
-0.1 0 0.15
300 300 424.26
=0-1 (-0.1)(424.26)-(300)(0.15) 0
8
y
y
M   



 
7.4 N m
In cartesian vector form
87
:
.4j N
M m
y 


Solution-Con’t