mechancis slides

1. 11/26/2013
1
Vectors
• Vector classifications:
- Fixed or bound vectors have well defined points of
application that cannot be changed without affecting
an analysis.
- Free vectors may be freely moved in space without
changingtheir effect on an analysis.
• Equal vectors have the same magnitude and direction.
• Negative vector of a given vector has the same magnitude
and the opposite direction.
Engineers Mechanics- Review of Vector Algebra/Applications
Addition of Vectors
• Parallelogramrule for vector addition
• Triangle rule for vector addition
B
B
C
C
QPR
BPQQPR


 cos2222
• Law of cosines,
• Law of sines,
P
C
R
B
Q
A sinsinsin

• Vector addition is commutative,
PQQP


• Vector subtraction
Engineers Mechanics- Review of Vector Algebra/Applications

2. 11/26/2013
2
Addition of Vectors
• Addition of three or more vectors through
repeated application of the triangle rule
• The polygon rule for the addition of three or
more vectors.
• Vector addition is associative,
   SQPSQPSQP


• Multiplication of a vector by a scalar
Engineers Mechanics- Review of Vector Algebra/Applications
Engineers Mechanics- Review of Vector Algebra/Applications
Addition of Vectors
A quantity which has magnitude and
direction, but doesn’t follow parallelogram
law, cannot be a vector.
Can you name such a quantity?
Thinkin terms of commutative property!!
Answer: Finite Rotation

3. 11/26/2013
3
Application: Resultant of Several Concurrent Forces
• Concurrent forces: set of forces which all
pass through the same point.
A set of concurrent forces applied to a
particle may be replaced by a single
resultantforce which is the vector sum of the
applied forces.
• Vector force components: two or more force
vectors which, together, have the same effect
as a single force vector.
Engineers Mechanics- Review of Vector Algebra/Applications
Rectangular Components of a Vector
cosFFx 
• Vector components may be expressed as products of
the unit vectors with the scalar magnitudes of the
vector components.
• Fx and Fy are referred to as the scalar components of
jFiFF yx


F

• May resolve a vector into perpendicular components so
that the resulting parallelogram is a rectangle.
are referred to as rectangular vector components and
yx FFF


yx FF

and
• Define perpendicular unit vectors which are
parallel to the x and y axes.
ji

and
sinFFy 
Engineers Mechanics- Review of Vector Algebra/Applications

4. 11/26/2013
4
Application: Addition of Concurrent Forces
SQPR


• Wish to find the resultantof 3 or more
concurrentforces,
   jSQPiSQP
jSiSjQiQjPiPjRiR
yyyxxx
yxyxyxyx




• Resolve each force into rectangularcomponents


x
xxxx
F
SQPR
• The scalar components of the resultant are
equal to the sum of the corresponding scalar
components of the given forces.


y
yyyy
F
SQPR
x
y
yx
R
R
RRR 122
tan
 
• To find the resultant magnitude and direction,
Engineers Mechanics- Review of Vector Algebra/Applications
Example
Two structural members A and B are bolted to
a bracket as shown. Knowing that both
members are in compression and that the force
is 20 kN in member A and 30 kN in member B,
determine, using trigonometry, the magnitude
and direction of the resultant of the forces
applied to the bracket by members A and B.
SOLUTION KEY
o Construct the force triangle
and apply the sine and
cosine rules.
Engineers Mechanics- Review of Vector Algebra/Applications
Example of a timber truss joint

5. 11/26/2013
5
SOLUTION
Engineers Mechanics- Review of Vector Algebra/Applications
Example
A collar that can slide on a vertical rod is subjected to
the three forces shown. Determine (a) the value of the
angle for which the resultant of the three forces is
horizontal, (b) the corresponding magnitude of the
resultant.
SOLUTION KEY
o Since the resultant(R) is to be
horizontal, sum of the vertical
comp. of the forces, i.e., Ry = 0. Example of an Umbrella
Engineers Mechanics- Review of Vector Algebra/Applications
Example

6. 11/26/2013
6
1 - 11
0y yR F  
   90 lb 70 lb sin 130 lb cos 0   
   13 cos 7 sin 9  
 2
13 1 sin 7 sin 9   
     2 2
169 1 sin 49 sin 126 sin 81     
   2
218 sin 126 sin 88 0   
sin 0.40899 
 1.24
(a) Since R is to be horizontal, Ry = 0
Then,
Squaring both sides:
Solving by quadratic formula:
or,
SOLUTION
Engineers Mechanics- Review of Vector Algebra/Applications
Example
1 - 12
117.0 lbR  or,
Engineers Mechanics- Review of Vector Algebra/Applications
Example-2

7. 11/26/2013
7
Rectangular Components of a Vector in Space
1coscoscos 222
 zyx 
• Components of the vector F

 
kji
F
kjiF
kFjFiFF
FFFFFF
zyx
zyx
zyx
zzyyxx








coscoscos
coscoscos
coscoscos





• is a unit vector along the line of action of
and are the direction
cosines for
F

F



zyx  cosand,cos,cos
222
zyx FFFF 
Engineers Mechanics- Review of Vector Algebra/Applications
Application: Rectangular Components of a Force Vector in Space
222
zyx dddd 
The magnitude of the force vector is
F and the direction of the force is
defined by the location of two
points,
   222111 ,,and,, zyxNzyxM
 
d
Fd
F
d
Fd
F
d
Fd
F
kdjdid
d
FF
zzdyydxxd
kdjdid
NMd
z
z
y
y
x
x
zyx
zyx
zyx










1
andjoiningvector
121212


d
d
d
d
d
d z
z
y
y
x
x   cos;cos;cos
Engineers Mechanics- Review of Vector Algebra/Applications

8. 11/26/2013
8
Example
2 - 15
The tension in the guy wire is 2500 N.
Determine:
a) components Fx, Fy, Fz of the force
acting on the bolt at A,
b) the angles qx, qy, qz defining the
direction of the force (the direction
cosines)
SOLUTION:
• Based on the relative locations of the
points A and B, determine the unit
vector pointing from A towards B.
• Apply the unit vector to determine the
components of the force acting on A.
• Noting that the components of the unit
vector are the direction cosines for the
vector, calculate the corresponding
angles.
Engineers Mechanics- Review of Vector Algebra/Applications
Example
2 - 16
SOLUTION:
• Determine the unit vector pointing from A
towards B.
     
     
m394
m30m80m40
m30m80m40
222
.
AB
AB


 kji
• Determine the components of the force.
  
     kji
kji
λF
N795N2120N1060
318084804240N2500



...
F
kji
kjiλ
318084804240
394
30
394
80
394
40
...
...


















 

Engineers Mechanics- Review of Vector Algebra/Applications

9. 11/26/2013
9
Example
2 - 17
• Noting that the components of the unit vector are
the direction cosines for the vector, calculate the
correspondingangles.
kji
kjiλ
318084804240
coscoscos
...
zyx

 



5.71
0.32
1.115



z
y
x



Engineers Mechanics- Review of Vector Algebra/Applications
2 - 18
What are the components of the
force in the wire at point B? Can
you find it without doing any
calculations?
SOLUTION:
• Since the force in the guy wire must be
the same throughout its length, the force
at B (and acting toward A) must be the
same magnitude but opposite in
direction to the force at A.
     kji
FF
N795N2120N1060 
 ABBA
FBA
FAB
Engineers Mechanics- Review of Vector Algebra/Applications
Example

10. 11/26/2013
10
Scalar Product of Two Vectors
• The scalar product or dot product between
two vectors P and Q is defined as
 resultscalarcosPQQP 

• Scalar products:
- are commutative,
- are distributive,
- are not associative,
PQQP


  2121 QPQPQQP


  undefined SQP

• Scalar products with Cartesian unit components,
000111  ikkjjikkjjii

   kQjQiQkPjPiPQP zyxzyx


2222
PPPPPP
QPQPQPQP
zyx
zzyyxx




Engineers Mechanics- Review of Vector Algebra/Applications
Applications: Scalar Product of Two Force Vectors
• Angle between two force vectors:
PQ
QPQPQP
QPQPQPPQQP
zzyyxx
zzyyxx





cos
cos

• Projection of a force vector on a given axis:
OL
OL
PP
Q
QP
PQQP
OLPPP







cos
cos
alongofprojectioncos


zzyyxx
OL
PPP
PP


coscoscos 


- For an axis defined by a unit vector ():
Q
Q




Note:
Engineers Mechanics- Review of Vector Algebra/Applications

11. 11/26/2013
11
Vector Product of Two Vectors
• Vector product of two vectors P and Q is defined
as the vector V which satisfies the following
conditions:
1. Line of action of V is perpendicular to plane
containingP and Q.
2. Magnitude of V is
3. Direction of V is obtained from the right-hand
rule.
sinQPV 
• Vector products:
- are not commutative,
- are distributive,
- are not associative,
 QPPQ 
  2121 QPQPQQP 
   SQPSQP 
Engineers Mechanics- Review of Vector Algebra/Applications
Vector Products: Rectangular Components
• Vector products of Cartesian unit vectors,
0
0
0



kkikjjki
ijkjjkji
jikkijii



• Vector products in terms of rectangular
coordinates
   kQjQiQkPjPiPV zyxzyx


   
 kQPQP
jQPQPiQPQP
xyyx
zxxzyzzy




zyx
zyx
QQQ
PPP
kji


Engineers Mechanics- Review of Vector Algebra/Applications

12. 11/26/2013
12
Application: Moment of a Force About a Point
• Moment of a force produces a “turning action” on a
rigid body
• The moment of a force F about O is defined as
FrMO


Engineers Mechanics- Review of Vector Algebra/Applications
r is the position vector of A from O
• The moment vector MO is perpendicular to the
plane containingO and the force F.
• Magnitude of MO measures the tendency of the force
to cause rotation of the body about an axis along MO.
d is the perpendicular distance of the line of action
of Force F from O. The sense of the moment may be
determinedby the right-handrule.
FdrFMO  sin
Moment of a Force About a Point
3 - 24
• Two-dimensional structures have length and breadth but
negligible depth and are subjected to forces contained
only in the plane of the structure.
• The plane of the structure contains the point O and the
force F. MO, the moment of the force about O, is
perpendicularto the plane.
• If the force tends to rotate the structure clockwise, the
sense of the moment vector is out of the plane of the
structure and the magnitude of the moment is positive.
• If the force tends to rotate the structure counterclockwise,
the sense of the moment vector is into the plane of the
structure and the magnitude of the moment is negative.
Engineers Mechanics- Review of Vector Algebra/Applications

13. 11/26/2013
13
Engineers Mechanics- Review of Vector Algebra/Applications
Examples: Application of moments
Varignon’s Theorem
• The moment about a give point O of the
resultantof several concurrent forces is equal
to the sum of the moments of the various
forces about the same point O.
• Varignon’sTheorem makes it possible to
replace the direct determination of the
moment of a force F by the moments of two
or more component forces of F.
    2121 FrFrFFr
Engineers Mechanics- Review of Vector Algebra/Applications

14. 11/26/2013
14
Application: Rectangular Components of the Moment of a Force about origin
     kyFxFjxFzFizFyF
FFF
zyx
kji
kMjMiMM
xyzxyz
zyx
zyxO






The moment of force F applied at A about O,
kFjFiFF
kzjyixrFrM
zyx
O



 ,
Engineers Mechanics- Review of Vector Algebra/Applications
Application: Rectangular Components of the Moment of a Force about an
Arbitrary Point
The moment of force F applied at A about B,
FrM BAB

 /
     
kFjFiFF
kzzjyyixx
rrr
zyx
BABABA
BABA





/
     
zyx
BABABAB
FFF
zzyyxx
kji
M 


Engineers Mechanics- Review of Vector Algebra/Applications

15. 11/26/2013
15
Rectangular Components of the Moment of a Force
For two-dimensional structures,
 
xy
ZO
xyO
yFxF
MM
yFxF


 kM
    
    xBAyBA
ZB
xBAyBAB
FyyFxx
MM
FyyFxx


 kM
Engineers Mechanics- Review of Vector Algebra/Applications
Example
A 500-N vertical force is applied to the end of a lever
which is attached to a shaft (not shown) at O.
Determine:
a) the moment about O,
b) the horizontalforce at A which creates the same
moment,
c) the smallest force at A which produces the same
moment,
d) the location for a 1200-N vertical force to
produce the same moment
0.6 m
500 N
Engineers Mechanics- Review of Vector Algebra/Applications

16. 11/26/2013
16
Example
a) Moment about O is equal to the product of the
force and the perpendicular distance between the
line of action of the force and O. Since the force
tends to rotate the lever clockwise, the moment
vector is into the plane of the paper which, by our
sign convention, would be negative or
counterclockwise.
500 N
0.6 m
 
   N.m150m0.3N500
m3060cosm60



O
O
M
..d
FdM
mN150 OM
Engineers Mechanics- Review of Vector Algebra/Applications
Example
Why must the direction of this F be to the right?
b) Horizontal force at A that produces the same
moment,
0.6 m
 
 
m520
mN150
m520mN150
m52060sinm60
.
F
.F
FdM
..d
O





N5288.F
Engineers Mechanics- Review of Vector Algebra/Applications

17. 11/26/2013
17
Example
What is the smallest force at A which produces
the same moment?
F(min) ?
0.6 m
c) The smallest force at A to produce the same
moment occurs when the perpendicular distance is
a maximum or when F is perpendicular to OA.
 
m60
mN150
m60mN150
.
F
.F
FdMO




N250F 30o
Engineers Mechanics- Review of Vector Algebra/Applications
Example
d) To determine the point of application of a 1200 N
force to produce the same moment,
 






60cos
m1250
N0120
mN150
N1200mN150
OBd
.d
d
FdMO
m250.OB 
1200 N
Engineers Mechanics- Review of Vector Algebra/Applications

18. 11/26/2013
18
A 36-N force is applied to a wrench to
tighten a showerhead. Knowing that the
centerline of the wrench is parallel to the x
axis. Determine the moment of the force
about A.
Example
SOLUTION KEY
o Find out the position vector of the
point ‘C’ .
o Get the components of the applied
force (F) along X, Y and Z directions.
Engineers Mechanics- Review of Vector Algebra/Applications
Example
/A C A M r F
     / 215 mm 50 mm 140 mmC A   r i j k
 36 N cos45 sin12xF      36 N sin 45yF   
 36 N cos45 cos12zF    
     5.2926 N 25.456 N 24.900 N    F i j k
where
, ,
0.215 0.050 0.140 N m
5.2926 25.456 24.900
A   
  
i j k
M
     4.8088 N m 4.6125 N m 5.7377 N m     i j k
and
     4.81 N m 4.61 N m 5.74 N mA      M i j k 
SOLUTION
Engineers Mechanics- Review of Vector Algebra/Applications

19. 11/26/2013
19
Mixed Triple Product of Three Vectors
• Mixed triple product of three vectors,
  resultscalar QPS

• The six mixed triple products formed from S, P, and
Q have equal magnitudes but not the same sign,
     
     SPQQSPPQS
PSQSQPQPS




     
 
zyx
zyx
zyx
xyyxz
zxxzyyzzyx
QQQ
PPP
SSS
QPQPS
QPQPSQPQPSQPS




• Evaluating the mixed triple product,
P
QS
= volume of parallelepiped
Engineers Mechanics- Review of Vector Algebra/Applications
Application: Moment of a Force About an Axis Passing Through the Origin
kzjyixr


• Moment MO of a force F applied at the point A
about a point O,
FrMO


• Scalar moment MOL about an axis OL is the
projection of the moment vector MO onto the
axis,
 FrMM OOL

 
• Moments of F about the coordinate axes
(components of )
xyz
zxy
yzx
yFxFM
xFzFM
zFyFM



Engineers Mechanics- Review of Vector Algebra/Applications
OM


20. 11/26/2013
20
Application: Moment of a Force About an Arbitrary Axis
• Moment of a force about an arbitraryaxis,
 
BABA
BA
BBL
rrr
Fr
MM








• The result is independent of the location
of point B along the given axis!!
Engineers Mechanics- Review of Vector Algebra/Applications
Example
Engineers Mechanics- Review of Vector Algebra/Applications
The frame ACD is hinged at A and D and is supported by a cable that
passes through a ring at B and is attached to hooks at G and H. Knowing
that the tension in the cable is 1125 N, determine the moment about the
diagonal AD of the force exerted on the frame by portion BH of the cable.
SOLUTION KEY
o Find unit vector along AD.
o Find the force vector BH.
o Choose either point A or D , get the
position vector AB or DB
 /AD AD B A BHM   λ r T

21. 11/26/2013
21
Example
Engineers Mechanics- Review of Vector Algebra/Applications
 /AD AD B A BHM   λ r T
   
   2 2
0.8 m 0.6 m
0.8 0.6
0.8 m 0.6 m
AD

  
 
i k
λ i k
 / 0.4 mB A r i
 
     
     
2 2 2
0.3 m 0.6 m 0.6 m
1125 N
0.3 0.6 0.6 m
BH BH
BH
T
BH
    
  
i j k
T

0.8 0 0.6
0.4 0 0 180 N m
375 750 750
ADM

   

180.0 N mADM    
Moment of a Couple – Couple Moment
• Two forces F and -F having the same magnitude,
parallel lines of action, and opposite sense are said
to form a couple.
• Moment of the couple,
 
 
CFdrFM
CFr
Frr
FrFrM
BA
BA




sin



Special Notation
• The moment vector of the couple is
independent of the choice of the origin of the
coordinate axes, i.e., it is a free vector that can
be applied at any point with the same effect.
Engineers Mechanics- Review of Vector Algebra/Applications

22. 11/26/2013
22
Moment of a Couple
Two couples will have equal moments if
• 2211 dFdF 
• the two couples lie in parallel planes, and
• the two couples have the same sense or
the tendency to cause rotation in the same
direction.
Engineers Mechanics- Review of Vector Algebra/Applications
Addition of Couples
• Consider two intersectingplanes P1 and
P2 with each containinga couple
222
111
planein
planein
PFrM
PFrM




• Resultants of the vectors also form a
couple
 21 FFrRrM


• Sum of two couples is also a couple that is equal
to the vector sum of the two couples
21
21
MM
FrFrM




Engineers Mechanics- Review of Vector Algebra/Applications

23. 11/26/2013
23
Summary – Couple
• A couple can be represented by a vector with magnitude
and direction equal to the moment of the couple.
• Couple vectors obey the law of addition of vectors.
• Couple vectors are free vectors, i.e., the point of application
is not significant.
• Couple vectors may be resolved into component vectors.
Engineers Mechanics- Review of Vector Algebra/Applications