ME 221 Lecture 5 1

ME 221 Statics

LECTURE #4

Sections: 3.1 - 3.6

ME 221 Lecture 5 2

Announcements• HW #2 due Friday 5/28

Ch 2: 23, 29, 32, 37, 47, 50, 61, 82, 105, 113

Ch 3: 1, 8, 11, 25, 35

• Quiz #3 on Friday, 5/28

• Exam #1 on Wednesday, June 2

ME 221 Lecture 5 3

Chapter 3Rigid Bodies; Moments

• Consider rigid bodies rather than particles– Necessary to properly model problems

• Moment of a force about a point

• Problems

• Moment of a force about an axis

• Moment of a couple

• Equivalent force couple systems

ME 221 Lecture 5 4

Rigid Bodies• The point of application of a force is very

important in how the object responds

F

F

• We must represent true geometry in a FBD and apply forces where they act.

ME 221 Lecture 5 5

Transmissibility• A force can be replaced by an equal

magnitude force provided it has the same line of action and does not disturb equilibrium

B

A

ME 221 Lecture 5 6

Moment• A force acting at a distance is a moment

• Transmissibility tells us the moment is the same about O or A

F

d

M

O

M

A d is the perpendiculardistance from F’s lineof action to O

Defn. of moment: M = F • d

ME 221 Lecture 5 7

Vector Product; Moment of Force

• Define vector cross product– trig definition

– component definition

• cross product of base vectors

• Moment in terms of cross product

ME 221 Lecture 5 8

Cross ProductThe cross product of two vectors results in a vector perpendicular to both.

ˆsin A B A B nB

A

A x B

The right-hand rule decides the direction of the vector.

B x A

A

B A x B = - B x A

n =AxBAxB

^

ME 221 Lecture 5 9

Base Vector Cross ProductBase vector cross products give us a means for

evaluating the cross product in components.

ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ; ;

ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ; ;

ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ; ;

i i 0 j i k k i j

i j k j j 0 k j i

i k j j k i k k 0

Here is how to remember all of this:

i

j

k+ i k-

j

ME 221 Lecture 5 10

General Component Cross Product

Consider the cross product of two vectors

ˆ ˆ ˆ ˆˆ ˆx y z x y zA A A B B B i j k i j k

ˆx yA B k ˆ

x zA B j ˆy xA B k ˆ

y zA B i ˆz xA B j ˆ AzBy

i

Or, matrix determinate gives a convenient calculation

ˆ ˆ ˆ

x y z

x y z

A A A

B B B

i j k

A B

ME 221 Lecture 5 11

ˆ ˆ ˆ

x y z

x y z

A A A

B B B

i j k

A B

ˆ ˆ ˆ

x y z

x y z

A A A

B B B

i j k

A B -

ˆ ˆ ˆ

x y z

x y z

A A A

B B B

i j k

A B +

= (AyBz-AzBy) i - (AxBz-AzBx) j + (AxBy-AyBx)k

ME 221 Lecture 5 12

Example ProblemsIf: A = 5i + 3j & B = 3i + 6j

Determine:

• A·B

• The angle between A and B

• AxB

• BxA

ME 221 Lecture 5 13

ME 221 Lecture 5 14

Vector Moment DefinitionThe moment about point O of a force acting at point A is:

MO = rA/O x F

Compute the cross product with whichever method you prefer.

rA/OA

O

F

ME 221 Lecture 5 15

A

O.4

0.2

200 N

60 o

x

60 o

d

0.285tan 60°=0.2m/x

x=0.115m

sin 60°=d/0.285m

d = 0.247 m

MA =200N *0.247m= 49.4 Nm

ExampleMethod # 1

ME 221 Lecture 5 16

A

O.4

0.2

200 N

60 o

200 cos60

200 sin 60

M+ =200N (sin 60)(0.4m)- 200N (cos 60)(0.2m)= 49.4 Nm

Method # 2

Note: Right-hand rule applies to moments

ME 221 Lecture 5 17

A

O.4

0.2

200 N

60 o

Method # 3

r

F=200N cos 60 i + 200N sin 60 j

r =0.4 i + 0.2 j

0.4 0.2 0

200cos60 200sin60 0MA= =200 (sin 60)(0.4) - 200 (cos 60)(0.2)

= 49.4 Nm

i j k

^ ^ ^

ME 221 Lecture 5 18

A

O.4

0.2

200 N

60 o

Method # 4

r =0.285 i

F=200N cos 60 i + 200N sin 60 j

r =0.285 i i j k 0.285 0 0

200cos60 200sin60 0MA= = 49.4 Nm

ME 221 Lecture 5 19

Moment of a Force about an Axis

x

y

z

O

F

B

n

ArAB=rB/A

|Mn| =MA·n^

=n·(rB/A x F )^

Same as the projection of MA along n

|Mn|=

nx ny nzrB/Ax r B/A y r B/Az

F x F y F z

ME 221 Lecture 5 20

x

y

z

O

F

B

n

ArAB=rB/A

MA

Mn

Mp

Resolve the vector MA into two vectors one parallel and one perpendicular to n.

Mn=|Mn|n^

Mp = MA - Mn

=n x [(r B/A x F) x n]^ ^

ME 221 Lecture 5 21

Moment of a Couple

x

y

z

O

F2

B

rA

F1

rAB=rB/A

rB

A

Let F1 = -F2

d

Mo=rA x F2+ rB x F1

=(rB - rA ) x F1

=rAB x F1= C

The Moment of two equal and opposite forces is called a couple

|C|=|F1| d

ME 221 Lecture 5 22

• The two equal and opposite forces form a couple (no net force, pure moment)

• The moment depends only on the relative positions of the two forces and not on their position with respect to the origin of coordinates

Moment of a Couple (continued)

ME 221 Lecture 5 23

• Since the moment is independent of the origin, it can be treated as a free vector, meaning that it is the same at any point in space

• The two parallel forces define a plane, and the moment of the couple is perpendicular to that plane

Moment of a Couple (continued)

ME 221 Lecture 5 24

Example