lecture 8- 1page
TRANSCRIPT
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
3 - 1
Sample Problem 3.4
The rectangular plate is supported by
the brackets at A and B and by a wire
CD. Knowing that the tension in the
wire is 200 N, determine the moment
about A of the force exerted by the
wire at C .
SOLUTION:
The moment M A of the force F exerted
by the wire is obtained by evaluating
the vector product,
F r M AC A
r
r
r
×=
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Vector Mechanics for Engineers: StaticsE i gh t h
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Sample Problem 3.4
SOLUTION:
12896120
08.003.0
−−
=
k ji
M A
rrr
r
( ) ( ) ( )k ji M Ar
rrv
m N8.82m N8.82m N68.7 ⋅+⋅+⋅−=
( ) ( ) jir r r AC AC
rrrrr
m08.0m3.0 +=−=
F r M AC A
r
r
r
×=
( )
( ) ( ) ( ) ( )
( ) ( ) ( )k ji
k ji
r
r F F
DC
DC
rrr
rrr
r
rr
N128 N69 N120
m5.0
m32.0m0.24m3.0 N200
N200
−+−=
−+−=
== λ
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Vector Mechanics for Engineers: StaticsE i gh t h
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Given: A 400 N force is
applied to the frame
and θ = 20°.
Find: The moment of the
force at A.
Plan:
1) Resolve the force along x and y axes.
2) Determine MA using scalar analysis.
EXAMPLE 1
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Vector Mechanics for Engineers: StaticsE i gh t h
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EXAMPLE 1 (continued)
Solution
+ ↑ Fy = -400 cos 20° N
+ → Fx = -400 sin 20° N
+ MA = {(400 cos 20°)(2) + (400 sin 20°)(3)} N·m
= 1160 N·m
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Vector Mechanics for Engineers: StaticsE i gh t h
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EXAMPLE 2
Given: a = 3 in, b = 6 in and c = 2 in.
Find: Moment of F about point O.
Plan:
1) Find rOA.
2) Determine M O = rOA F .
Solution rOA = {3 i + 6 j – 0 k} in
i j k
3 6 0
3 2 -1
MO = = [{6(-1) – 0(2)} i – {3(-1) – 0(3)} j +
{3(2) – 6(3)} k] lb·in
= {-6 i + 3 j – 12 k} lb·in
o
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Vector Mechanics for Engineers: StaticsE i gh t h
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GROUP PROBLEM SOLVING
Given: A 40 N force is
applied to the wrench.
Find: The moment of the
force at O.
Plan: 1) Resolve the force
along x and y axes.
2) Determine MO using
scalar analysis.
Solution: + ↑ Fy = - 40 cos 20° N
+ → Fx = - 40 sin 20° N
+ MO = {-(40 cos 20°)(200) + (40 sin 20°)(30)}N·mm
= -7107 N·mm = - 7.11 N·m
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Vector Mechanics for Engineers: StaticsE i gh t h
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Scalar Product of Two Vectors
• The scalar product or dot product between
two vectors P and Q is defined as
( )resultscalar cosθ PQQP =• rr
• Scalar products:
- are commutative,
- are distributive,
- are not associative,
PQQPrrrr
•=•
2121 QPQPQQPrrrrrrr
•+•=+•
undefined =•• S QPrrr
• Scalar products with Cartesian unit components,
000111 =•=•=•=•=•=• ik k j jik k j jiir
rrvrr
rrrrrr
k Q jQiQk P jPiPQP z y x z y x
rrr
rrrrr
++•++=•
2222PPPPPP
QPQPQPQP
z y x
z z y y x x
=++=•
++=•rr
rr
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Vector Mechanics for Engineers: StaticsE i gh t h
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Scalar Product of Two Vectors: Applications
• Angle between two vectors:
PQ
QPQPQP
QPQPQPPQQP
z z y y x x
z z y y x x
++=
++==•
θ
θ
cos
cosrr
• Projection of a vector on a given axis:
OL
OL
PPQ
QP
PQQP
OLPPP
==•
=•
==
θ
θ
θ
cos
cos
alongof projectioncos
rr
rr
z z y y x x
OL
PPP
PP
θ θ θ
λ
coscoscos ++=
•= rr
• For an axis defined by a unit vector:
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Vector Mechanics for Engineers: StaticsE i gh t h
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APPLICATIONS
For this geometry, can you determine
angles between the pole and the cables?
For force F at Point A, what
component of it (F1) acts along the
pipe OA? What component (F2) acts
perpendicular to the pipe?
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: StaticsE i gh t h
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EXAMPLE
Given: The force acting on the pole
Find: The angle between the force
vector and the pole, and the
magnitude of the projection
of the force along the pole
OA.
Plan:
A
1. Get rOA
2. θ = cos-1{( F • rOA)/(F r OA)}
3. FOA = F • uOA or F cos θ
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Vector Mechanics for Engineers: StaticsE i gh t h
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EXAMPLE (continued)
A
rOA = {2 i + 2 j – 1 k} m
r OA = (22 + 22 + 12)1/2 = 3 m
F = {2 i + 4 j + 10 k}kN
F = (22 + 42 + 102)1/2 = 10.95 kN
θ = cos-1{( F • rOA)/(F r OA)}
θ = cos-1 {2/(10.95 * 3)} = 86.5°
uOA = rOA/r OA = {(2/3) i + (2/3) j – (1/3) k}
FOA = F • uOA = (2)(2/3) + (4)(2/3) + (10)(-1/3) = 0.667 kN
Or FOA = F cos θ = 10.95 cos(86.51°) = 0.667 kN
F • rOA = (2)(2) + (4)(2) + (10)(-1) = 2 kN·m
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: StaticsE i gh t h
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GROUP PROBLEM SOLVING
Given: The force acting on the pole.
Find: The angle between the force
vector and the pole, and the
magnitude of the projection of
the force along the pole AO.
Plan:
1. Get r AO
2. θ = cos-1{( F • r AO)/(F r AO)}
3. FOA = F • u AO or F cos θ
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Vector Mechanics for Engineers: StaticsE i gh t h
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r AO = {-3 i + 2 j – 6 k} ft.
r AO = (32 + 22 + 62)1/2 = 7 ft.
F = {-20 i + 50 j – 10 k}lb
F = (202 + 502 + 102)1/2 = 54.77 lb
GROUP PROBLEM SOLVING (continued)
θ = cos-1{( F • r AO)/(F r AO)}
θ = cos-1 {220/(54.77 × 7)} = 55.0°
F • r AO = (-20)(-3) + (50)(2) + (-10)(-6) = 220 lb·ft
u AO = r AO/r AO = {(-3/7) i + (2/7) j – (6/7) k}
FAO = F • u AO = (-20)(-3/7) + (50)(2/7) + (-10)(-6/7) = 31.4 lb
Or FAO = F cos θ = 54.77 cos(55.0°) = 31.4 lb
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Vector Mechanics for Engineers: StaticsE i gh t h
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Mixed Triple Product of Three Vectors
• Mixed triple product of three vectors,
resultscalar=ו QPS rrr
• The six mixed triple products formed from S, P, and
Q have equal magnitudes but not the same sign,
( ) ( ) ( )S PQQS PPQS
PS QS QPQPS rrrrrrrr
rrrrrrrrr
ו−=ו−=ו−=
ו=ו=ו
( )
( )
z y x
z y x
z y x
x y y x z
z x x z y y z z y x
QQQ
PPP
S S S
QPQPS
QPQPS QPQPS QPS
=
−+
−+−=ו rrr
• Evaluating the mixed triple product,