lecture 8- 1page

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Vector Mechanics for Engineers: StaticsE i gh t h

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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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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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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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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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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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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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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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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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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?



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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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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



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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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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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Mixed Triple Product of Three Vectors

• Mixed triple product of three vectors,

resultscalar=×• 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

×•−=×•−=×•−=

×•=×•=×•

( )

( )

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

=

−+

−+−=×• rrr

• Evaluating the mixed triple product,

lecture 8- 1page

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