Transcript 4.2 Cross Product

Engineering Mechanics:
Statics in SI Units, 12e
4
Force System Resultants
Copyright © 2010 Pearson Education South Asia Pte Ltd
Chapter Outline
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
Moment of a Force – Scalar Formation
Cross Product
Moment of Force – Vector Formulation
Principle of Moments
Moment of a Force about a Specified Axis
Moment of a Couple
Simplification of a Force and Couple System
Further Simplification of a Force and Couple System
Reduction of a Simple Distributed Loading
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4.1 Moment of a Force – Scalar Formation
•
•
Moment of a force about a point or axis – a measure
of the tendency of the force to cause a body to rotate
about the point or axis
Torque – tendency of rotation caused by Fx or simple
moment (Mo) z
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4.1 Moment of a Force – Scalar Formation
Magnitude
•
For magnitude of MO,
MO = Fd (Nm)
where d = perpendicular distance
from O to its line of action of force
Direction
•
Direction using “right hand rule”
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4.1 Moment of a Force – Scalar Formation
Resultant Moment
•
Resultant moment, MRo = moments of all the forces
MRo = ∑Fd
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Example 4.1
For each case, determine the moment of the force about
point O.
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Solution
( a )M o  (100 N )( 2 m )  200 N .m (CW )
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Solution
( b )M o  ( 50 N )( 0 . 75 m )  37 . 5 N .m (CW )
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Solution

( c )M o  ( 40 N )( 4 m  2 cos 30 m )  229 N .m (CW )
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Solution

( d )M o  ( 60 N )( 1 sin 45 m )  42 . 4 N .m (CCW )
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Solution
( e )M o  ( 7 kN )( 4 m  1m )  21 . 0 kN .m (CCW )
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TEST (4-1)
• Determine the resultant moment produced by the forces
about point O.
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4.2 Cross Product
• Cross product of two vectors A and B yields C, which
is written as
C=AXB
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4.2 Cross Product
Magnitude
• Magnitude of C is the product of
the magnitudes of A and B
• For angle θ, 0° ≤ θ ≤ 180°
C = AB sinθ
Direction
• Vector C has a direction that is perpendicular to the
plane containing A and B such that C is specified by
the right hand rule
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4.2 Cross Product
• Expressing vector C when
magnitude and direction are known
C = A X B = (AB sinθ)uC
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4.2 Cross Product
Laws of Operations
1. Commutative law is not valid
AXB≠BXA
Rather,
AXB=-BXA
• Cross product A X B yields a
vector opposite in direction to C
B X A = -C
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4.2 Cross Product
Laws of Operations
2. Multiplication by a Scalar
a( A X B ) = (aA) X B = A X (aB) = ( A X B )a
3. Distributive Law
AX(B+D)=(AXB)+(AXD)
• Proper order of the cross product must be maintained
since they are not commutative
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4.2 Cross Product
• Cartesian vector formulation
ii  0
i j  k
ik  j
jxi=
jxj =
jxk=
kxi=
kxj =
kxk=
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4.2 Cross Product
• A = Ax i + Ay j + Az k
• B = Bx i + By j + Bz k
• AxB=
• (Ay Bz- Az By) i - (Ax Bz- Az Bx) j + (Ax By- Ay Bx) k
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4.2 Cross Product
Cartesian Vector Formulation
• Use C = AB sinθ on pair of Cartesian unit vectors
• A more compact determinant in the form as
 
A  B  Ax

j

k
Ay
Az
Bx
By
Bz

i
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4.2 Cross Product
• Each component can be determined using 2  2
determinants.
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4.3 Moment of Force - Vector Formulation
• Moment of force F about point O can be expressed
using cross product
MO = r X F
Magnitude
• For magnitude of cross product,
MO = rF sinθ
• Treat r as a sliding vector. Since d = r sinθ,
MO = rF sinθ = F (rsinθ) = Fd
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4.3 Moment of Force - Vector Formulation
Direction
• Direction and sense of MO are determined by righthand rule
*Note:
- “curl” of the fingers indicates the sense of rotation
- Maintain proper order of r and F since cross product
is not commutative
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4.3 Moment of Force - Vector Formulation
Principle of Transmissibility
• For force F applied at any point A, moment created
about O is MO = rA x F
• F has the properties of a sliding vector, thus
M O = r1 X F = r2 X F = r3 X F
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4.3 Moment of Force - Vector Formulation
Cartesian Vector Formulation
• For force expressed in Cartesian form,

 
M O  r X F  rx

j

k
ry
rz
Fx
Fy
Fz

i
• With the determinant expended,
MO = (ryFz - rzFy) i
- (rxFz - rzFx) j
+ (rxFy - ryFx) k
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4.3 Moment of Force - Vector Formulation
Resultant Moment of a System of Forces
• Resultant moment of forces about point O can be
determined by vector addition
MRo = ∑(r x F)
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Example 4.4
Two forces act on the rod. Determine the resultant
moment they create about the flange at O. Express the
result as a Cartesian vector.
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Solution
Position vectors are directed from point O to each force
as shown.
These vectors are
rA  5 j  m
rB  4 i  5 j  2 k  m
The resultant moment about O is

MO 

 r  F   r
A
 F  rB  F
i
j
k
i
j
k
0
5
0  4
5
2
 60
40
20
40
 30
80
 30 i  40 j  60 k  kN  m
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TEST (4-2 ~ 4-3)
• Determine the moment of force F about point O. Express
the resultant as a Cartesian vector.
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4.4 Principles of Moments
• Also known as Varignon’s Theorem
“Moment of a force about a point is equal to the sum of
the moments of the forces’ components about the point”
• Since F = F1 + F2,
MO = r X F
= r X (F1 + F2)
= r X F1 + r X F2
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Example 4.5
Determine the moment of the force about point O.
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Solution
The moment arm d can be found from trigonometry,
d  3  sin 75   2 . 898 m
Thus,
M O  Fd  5 2 . 898   14 . 5 kN  m
Since the force tends to rotate or orbit clockwise about point O, the
moment is directed into the page.
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Solution II
Considering counterclockwise moments as positive, and
applying the principle of moments, we have
(CCW)＋MO ＝－Fxdy－Fydx
＝ – (5 cos45 kN) (3 sin30 m) – (5 sin45 kN) (3 cos30 m)
＝ – 14.5 kN‧m
＝ 14.5 kN‧m (CW)
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Solution III
• The x and y axes can be set parallel and perpendicular
to the rod’s axis. Here Fx produces no moment about
point O since its line of action passes through this point.
Therefore,
•
• (CCW)＋MO ＝－Fydx
•
＝－(5 sin75 kN)(3 m)
•
＝－14.5 kN‧m
•
＝14.5 kN‧m (CW)
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TEST (4-4)
• Determine the moment of the force about point O.
F4-6
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4.5 Moment of a Force about a Specified Axis
• For moment of a force about a point, the moment and
its axis is always perpendicular to the plane
• A scalar or vector analysis is used to find the
component of the moment along a specified axis that
passes through the point
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4.5 Moment of a Force about a Specified Axis
Scalar Analysis
• According to the right-hand rule, My is directed along
the positive y axis
• For any axis, the moment is
M a  Fd a
• Force will not contribute a moment
if force line of action is parallel or
passes through the axis
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4.5 Moment of a Force about a Specified Axis
Vector Analysis
• For magnitude of MA,
MA = MOcosθ = MO·ua
where ua = unit vector
• In determinant form,
u ax




M a  u ax  ( r X F )  rx
u ay
u az
ry
rz
Fx
Fy
Fz
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Example 4.8
Determine the moment produced by the force F which
tends to rotate the rod about the AB axis.
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Solution
Unit vector defines the direction of the AB axis of the rod,
where 
uB 
rB

0 . 4 i  0 . 2 j 
0 .4  0 .2
2
rB
 0 . 8944 i  0 . 4472 j
2
For simplicity, choose rD
rD  0 . 6 i m
The force is
F   300 k  N
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Solution -- Example 4.8
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TEST (4-5)
• Determine the magnitude of the moment of the force F =
{ 300 i – 200 j + 150 k } N about the OA axis.
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4.6 Moment of a Couple
• Couple
– two parallel forces
– same magnitude but opposite direction
– separated by perpendicular distance d
• Resultant force = 0
• Tendency to rotate in specified direction
• Couple moment = sum of moments of both couple
forces about any arbitrary point
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4.6 Moment of a Couple
Scalar Formulation
• Magnitude of couple moment
M = Fd
• Direction and sense are determined by right hand rule
• M acts perpendicular to plane containing the forces
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4.6 Moment of a Couple
Vector Formulation
•
For couple moment,
M=rXF
• If moments are taken about point A,
moment of –F is zero about this
point
• r is crossed with the force to which
it is directed
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4.6 Moment of a Couple
Equivalent Couples
• 2 couples are equivalent if they produce the same
moment
• Forces of equal couples lie on the same plane or plane
parallel to one another
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4.6 Moment of a Couple
Resultant Couple Moment
• Couple moments are free vectors and may be applied
to any point P and added vectorially
• For resultant moment of two couples at point P,
M R = M1 + M2
• For more than 2 moments,
MR = ∑(r X F)
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Example 4.12
Determine the couple moment acting on the pipe.
Segment AB is directed 30° below the x–y plane.
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SOLUTION I (VECTOR ANALYSIS)
Take moment about point O,
M = rA X (-250k) + rB X (250k)
= (0.8j) X (-250k) + (0.66cos30ºi
+ 0.8j – 0.6sin30ºk) X (250k)
= {-130j}N.cm
Take moment about point A
M = rAB X (250k)
= (0.6cos30°i – 0.6sin30°k)
X (250k)
= {-130j}N.cm
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SOLUTION II (SCALAR ANALYSIS)
Take moment about point A or B,
M = Fd = 250N(0.5196m)
= 129.9N.cm
Apply right hand rule, M acts in the –j direction
M = {-130j}N.cm
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TEST (4-6)
• Determine the resultant couple moment acting on the
pipe assembly.
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4.7 Simplification of a Force and Couple System
• An equivalent system is when the external effects are
the same as those caused by the original force and
couple moment system
• External effects of a system is the translating and
rotating motion of the body
• Or refers to the reactive forces at the supports if the
body is held fixed
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4.7 Simplification of a Force and Couple System
• Equivalent resultant force acting at point O and a
resultant couple moment is expressed as
FR 
M R
F
 M
O
O

M
• If force system lies in the x–y plane
and couple moments are
perpendicular to this plane,
 FR x
F
F    F
M    M
R

y
y
R
x
O
O

M
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Example 4.16
A structural member is subjected to a couple moment M
and forces F1 and F2. Replace this system with an
equivalent resultant force and couple moment acting at its
base, point O.
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Solution
Express the forces and couple moments as Cartesian
vectors.


F1  {  800 k } N


 rCB 

F2  ( 300 N ) u CB  ( 300 N )   
 rCB 


  0 . 15 i  0 . 1 j 


 300 
  {  249 . 6 i  166 . 4 j } N
2
2
 ( 0 . 15 )  ( 0 . 1) 


4
3
M   500   j  500   k  {  400 j  300 k } N .m
5
5
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Solution
Force Summation.


FR   F ;






F R  F1  F 2   800 k  249 . 6 i  166 . 4 j



 {  249 . 6 i  166 . 4 j  800 k } N

M
Ro



  

  M C   M O  M  rC X F1  rB X F 2

i




 (  400 j  300 k )  (1k ) X (  800 k )   0 . 15
 249 . 6

j

k
0 .1
1
166 . 4
0



 {  166 i  650 j  300 k } N .m
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TEST (4-7)
• Replace the loading system by an equivalent resultant
force and couple moment acting at point O.
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4.8 Further Simplification of a Force and Couple System
Concurrent Force System
•
A concurrent force system is where lines of action of
all the forces intersect at a common point O
FR 
F
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4.8 Further Simplification of a Force and Couple System
Coplanar Force System
•
•
Lines of action of all the forces lie in the same plane
Resultant force of this system also lies in this plane
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4.8 Further Simplification of a Force and Couple System
Parallel Force System
•
•
Consists of forces that are all parallel to the z axis
Resultant force at point O must also be parallel to this
axis
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4.8 Further Simplification of a Force and Couple System
Reduction to a Wrench
•
•
3-D force and couple moment system have an
equivalent resultant force acting at point O
Resultant couple moment not perpendicular to one
another
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Example 4.18
The jib crane is subjected to three coplanar forces.
Replace this loading by an equivalent resultant force and
specify where the resultant’s line of action intersects the
column AB and boom BC.
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Solution
Force Summation
  F Rx   F x ;
F Rx
3
  2 . 5 kN    1 . 75 kN
5
  3 . 25 kN  3 . 25 kN 
  F Ry   F y ;
F Ry
4
  2 . 5 N    0 . 6 kN
5
  2 . 60 kN  2 . 60 N 
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Solution
For magnitude of resultant force,
FR  ( FRx )2  ( FRy )2  (3.25)2  (2.60)2
 4.16kN
For direction of resultant force,
F 
1 2.60 
q  tan 1 Ry ÷

tan

÷
÷


F
3
.
25
 Rx 
 38.7
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Solution
Moment Summation
 Summation of moments about point A,
M
RA
 M A;
3 . 25 kN ( y )  2 . 60 kN ( 0 )
 1 . 75 kn (1m )  0 . 6 kN ( 0 . 6 m )
3
4
 2 . 50 kN   ( 2 . 2 m )  2 . 50 kN   (1 . 6 m )
5
5
y  0 . 458 m
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Solution
Moment Summation
 Principle of Transmissibility
M
RA
 M A ;
3 . 25 kN ( 2 . 2 m )  2 . 60 kN ( x )
 1 . 75 kn (1m )  0 . 6 kN ( 0 . 6 m )
3
4
 2 . 50 kN   ( 2 . 2 m )  2 . 50 kN   (1 . 6 m )
5
5
x  2 . 177 m
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TEST (4-8)
• Replace the loading system by an equivalent single
resultant force and specify the x and y coordinates of its
line of action.
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4.9 Reduction of a Simple Distributed Loading
• Large surface area of a body may be subjected to
distributed loadings
• Loadings on the surface is defined as pressure
• Pressure is measured in Pascal (Pa): 1 Pa = 1N/m2
Uniform Loading Along a Single Axis
• Most common type of distributed
loading is uniform along a
single axis
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4.9 Reduction of a Simple Distributed Loading
Magnitude of Resultant Force
• Magnitude of dF is determined from differential area dA
under the loading curve.
• For length L,
FR 
 w  x dx   dA
L
 A
A
• Magnitude of the resultant force is equal to the total
area A under the loading diagram.
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4.9 Reduction of a Simple Distributed Loading
Location of Resultant Force
• MR = ∑MO
• dF produces a moment of xdF = x w(x) dx about O
• For the entire plate,
M
 M O
Ro
x FR 
 xw ( x ) dx
L
• Solving for x
x 
 xw ( x ) dx
L
 w ( x ) dx
L

 xdA
A
 dA
A
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Example 4.21
Determine the magnitude and location of the equivalent
resultant force acting on the shaft.
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Solution
For the colored differential area element,
dA  wdx  60 x dx
2
For resultant force
FR   F ;
2
FR 
 dA 
 60 x dx
A
0
2
2
x 
 23 03 
 60    60 


3
3
3
 0


3
 160 N
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Solution
For location of line of action,
2
2
x 
 xdA
A
 dA

 x ( 60 x
2
) dx
0
160

 x4 
60  
 4 0
160
 24 04 
60 


4
4



160
A
 1 .5 m
Checking,
A
ab

2 m ( 240 N / m )
3
x 
3
4
a
 160
3
3
( 2 m )  1 .5 m
4
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TEST (4-9)
• Determine the resultant force and specify where it acts
on the beam measured from A.
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END OF CHAPTER 4
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