me 221lecture 51 me 221 statics lecture #4 sections: 3.1 - 3.6
Post on 22-Dec-2015
215 views
TRANSCRIPT
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 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)