1 The scalar product or dot product between two vectors P and Q is defined as Scalar products: -are...
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Transcript of 1 The scalar product or dot product between two vectors P and Q is defined as Scalar products: -are...
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• The scalar product or dot product between two vectors P and Q is defined as
resultscalarcosPQQP
• 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
3.9 SCALAR PRODUCT OF TWO VECTORS
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• Angle between two vectors:
PQQPQPQP
QPQPQPPQQP
zzyyxx
zzyyxx
cos
cos
• Projection of a vector on a given axis:
OL
OL
PPQ
QP
PQQP
OLPPP
cos
cos
along of projection cos
zzyyxx
OL
PPPPP
coscoscos
• For an axis defined by a unit vector:
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• 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,
FdrFMFr
Frr
FrFrM
BA
BA
sin
• 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.
3.12 MOMENT OF A COUPLE
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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.
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A couple is formed by any two parallel forces acting on the same plane with equal magnitude and opposite directions.
The moment of a couple is a vector whose line of action is perpendicular to the plane of the couple and whose direction can be determined by the right-hand rule.
MOMENT OF A COUPLE
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APPLICATIONS
A torque or moment of 12 N · m is required to rotate the wheel. Which one of the two grips of the wheel above will require less force to rotate the wheel?
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APPLICATIONS (continued)
The crossbar lug wrench is being used to loosen a lug net. What is the effect of changing dimensions a, b, or c on the force that must be applied?
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F
-F
Couple
The forces will result in rotation of the body
0F
OM ≠ 0
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Couple Moment• If F is the magnitude of the forces forming the
couple and d is the perpendicular distance between the lines of actions of the forces, then the magnitude of the couple-moment is:
• M=dF
F
F
A
B
dd-b
b
M = dF = (d – b)F + bF
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MOMENT OF A COUPLE
A couple is defined as two parallel forces with the same magnitude but opposite in direction separated by a perpendicular distance d.
The moment of a couple is defined as
MO = F d (using a scalar analysis) or as
MO = r F (using a vector analysis).
Here r is any position vector from the line of action of –F to the line of action of F.
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Couple Moment
• The couple-moment is known as a free vector, meaning that it can be moved anywhere in space without changing its meaning, provided that its magnitude and direction are kept intact (couple has same moment about every point in space).
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100mm
100mm
150mm
100mm
135 N
135 N
90 N
90 N
100mm
135 N
135 N
3.14 Equivalent Couples
Figure shows three couples which act successively on the same rectangular box. As seen in the preceding section, the only motion a couple can impart to a rigid body is a rotation.Since each of the three couples shown has the same moment M ( same direction and same magnitudeM = 135 N. m ), we can expect the three couples to have the same effect on the box.
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• Consider two intersecting planes P1 and P2 with each containing a couple
222
111
planein
planein
PFrM
PFrM
• Resultants of the vectors also form a couple
21 FFrRrM
• By Varigon’s theorem
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21
MM
FrFrM
• Sum of two couples is also a couple that is equal to the vector sum of the two couples
3.14 ADDITION OF COUPLES
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• 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.
3.15 COUPLES CAN BE REPRESENTED BY VECTORS
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• Force vector F can not be simply moved to O without modifying its action on the body.
• Attaching equal and opposite force vectors at O produces no net effect on the body.
• The three forces may be replaced by an equivalent force vector and couple vector, i.e, a force-couple system.
3.16 Resolution of a given force into a force at O and A Couple
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• Moving F from A to a different point O’ requires the addition of a different couple vector MO’
FrM O
'
• The moments of F about O and O’ are related,
FsM
FsFrFsrFrM
O
O
''
• Moving the force-couple system from O to O’ requires the addition of the moment of the force at O about O’.
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Sample problem 3.6
Determine the components of the single couple equivalent to the two couples shown.
135 N
135 N
90 N
90 N
x
300mm
225mm
175mm
225mmz
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x
z
y
Mx = -(60.75 N.m)i
My = +(27.0 N.m)j
Mz = +(20.25 N.m)k
Our computations will be simplified if we attach two equaland opposite 90N forces at A.
This enables us to replace the original 90N force couple by twonew 90N force couples, one of which lies in the zx plane and the other in a plane parallel to the xy plane.
The three couples shown in the adjoining sketch can be represented by three couple vectors Mx, My and Mz directed along the coordinate axes. The corresponding moments are
Mx = - (135N) (0.45m) = - 60.75N.m
My = +( 90 N) (0.3 m) = + 27N.m
Mz = +( 90 N) (0.225m) = + 20.25N.m
90 N
90 N
90 N
90 N
135 N
135 N
225mm
225mm
300mm
175mm
y
z
Solution
These thre moments represent the components of the single couple M equivalent to the given. We write
M = -(60.75 N.m)i + (27.0 N.m)j + (20.25 N.m)k
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• Replace the couple and force shown by an equivalent single force applied to the lever.
• Determine the distance from the shaft to the point of application of this force.
Sample problem 3.7
400 N
260 N
260 N
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(i) Step 1
The given force and couple are replaced by an equivalent force-couple system at O. The force F = - (400N)j is moved to O, and a moment couple is added to the system, which is the moment about point O of the force in its original position.
MO = x F
= (0.150mi + 0.260mj) x (-400N)j
= -60k
MO = - (60 Nm)k
OBr
Solution
400 N
400 N
200 N
F = - (400N)j
- (24 N.m)k - (24 N.m)k - (60 N.m)k
- (400 N.)j
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Example 2 (i) Step 2
This couple added to couple moment of –(24Nm)k produced due to two 200N forces, thus the couple moment is now –(84Nm)k. We then apply F at point C.
- (84 Nm)k = x F
= [(OC) cos 60i + (OC) sin 60j] x (-400)j
= - (OC) cos 60 (400 N)k
OC cos 60 = 0.210 m
= 210 mm
OC
OC = 420 mm
- (400 N) j
- (84 N.m) k- (400 N) j
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The effect of a couple does not depend on its location, thus the couple moment –(24 Nm)k can be moved to B.
-(24 Nm)k = x F
= - BC cos 60(400N) k
BC
We conclude that
(BC) cos 60˚ = 0.060m = 60mm BC =120mm
OC = OB + BC = 300 mm + 120 mm
OC = 420 mm
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• A system of forces may be replaced by a collection of force-couple systems acting a given point O
• The force and couple vectors may be combined into a resultant force vector and a resultant couple vector,
FrMFR RO
• The force-couple system at O may be moved to O’ with the addition of the moment of R about O’ ,
RsMM RO
RO
'
• Two systems of forces are equivalent if they can be reduced to the same force-couple system.
3.17 Reduction of A System of Forces To One Force and One Couple
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Equivalent System
MOVING A FORCE ITS LINE OF ACTIONON
Sliding VECTOR
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MOVING A FORCE OF ITS LINE OF ACTION OFF
Free VECTOR
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Resultant of a Force and Couple System
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3.20 Further Reduction of a Force and Couple System
If R and MRP are perpendicular to each other, the force-couple system at P can be further reduced to a single resultant force.
Will be the case for system consisting either :
(a) concurrent force (b) coplanar force (c) parallel forces
(a) concurrent force
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(b) Coplanar force systems
(c) Parallel force systems
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The moment of the resultant force about the grip
the moment of all the forces about the grip
FRd = F1d1 + F2d2 + F3d3
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Sample problems 3.8
A 4.80m long beam is subjected to the forces shown. Reduce the given system of forces to a) an equivalent force couple system at Ab) single force or resultant
150N 600N 100N 250N
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use FRd = F1d1 + F2d2 + F3d3 ;
solution
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= - 600(1.6m) + 100(2.8m) – 250(4.8)
= - 1880 Nm
+ FR = 150 - 600 + 100 - 250 = - 600 N = 600 N
MA = F1d1 + F2d2 + F3d3
F1 F2 F3
solution150N 600N 100N 250N
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A B
d
FR = 600 [N]
FRd=MA (600) d = 1880 Nm d = 1880/600 = 3.13 m
Thus, the single force equivalent to the given system;
FR = 600 [N] , d = 3.13 [m]
solution
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Example Three stage lights are
mounted on a pipe as shown. The lights at A and B each weigh 16.4 N, while the one at C weighs 14 N.
(a)If d = 0.625 m, determine the distance from D to the line of action of the resultant of the weights of the three lights.
(b)Determine the value of d so that the resultant of the weights passes through the midpoint of the pipe.
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Solution
16.4 N 16.4 N 14 N
DA B C =
D
L R
2.1 m
0.85 m d0.25 m
For equivalence, F : - 16.4 – 16.4 – 14 = - R
R = 46.8 N : - (0.25)(16.4) – (1.10)(16.4) – (1.10 + d)(14) = - L (46.8)
37.54 + 14d = 46.8 L »Eq 1
DM
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Example 3 (cont)
(a)For d = 0.625 m, substitute the value of d in Eq 1,
37.54 + 14 (0.625) = 46.8 L
L = 0.9891 m
The resultant passes through a point 0.9891 m to the right of D
(b) For L = 1.05 m, substitute the value of L in Eq. 1,
37.54 + 14d = 46.8 (1.05)
d = 0.8275 m
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THE END