ME 221 Statics
LECTURE #4
Sections: 3.1 - 3.6
ME 221
Lecture 5
1
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
2
Chapter 3
Rigid Bodies; Moments
• Consider rigid bodies rather than particles
– Necessary to properly model problems
•
•
•
•
•
Moment of a force about a point
Moment of a force about an axis
Moment of a couple
Equivalent force couple systems
Problems
ME 221
Lecture 5
3
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
4
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
5
Moment
• A force acting at a distance is a moment
M
M
A
O
d
F
d is the perpendicular
distance from F’s line
of action to O
Defn. of moment: M = F • d
• Transmissibility tells us the moment is the
same about O or A
ME 221
Lecture 5
6
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
7
Cross Product
The cross product of two vectors results in a
vector perpendicular to both.
AxB

A  B  A B sin  nˆ
B
A

The right-hand rule
decides the direction
of the vector.
n^ =
ME 221
B
A
AxB=-BxA
BxA
AxB
AxB
Lecture 5
8
Base Vector Cross Product
Base vector cross products give us a means for
evaluating the cross product in components.
ˆi  ˆi  0 ; ˆj  ˆi  kˆ ; kˆ  ˆi  ˆj
ˆi  ˆj  kˆ ; ˆj  ˆj  0 ; kˆ  ˆj  ˆi
ˆi  kˆ  ˆj ; ˆj  kˆ  ˆi ; kˆ  kˆ  0
Here is how to remember all of this:
ME 221
ĵ
ĵ
î + k̂
î - k̂
Lecture 5
9
General Component Cross Product
Consider the cross product of two vectors
 A ˆi  A ˆj  A kˆ    B ˆi  B ˆj  B kˆ 
x
y
z
x
y
z
 Ax By kˆ  Ax Bz ˆj  Ay Bxkˆ  Ay Bz ˆi  Az Bx ˆj  AzBy ˆi
Or, matrix determinate gives a convenient
calculation
ME 221
ˆi
A  B  Ax
ˆj
Ay
kˆ
Az
Bx
By
Bz
Lecture 5
10
ˆi
A  B  Ax
ˆj
Ay
Bx
By
ˆi
kˆ
Az A  B  Ax
Bx
Bz
ˆj
Ay
kˆ
Az
ˆi
+B  Ax
A
ˆj
Ay
kˆ
Az
By
Bz
Bx
By
Bz
= (AyBz-AzBy) i - (AxBz-AzBx) j + (AxBy-AyBx)k
ME 221
Lecture 5
11
Example Problems
If: A = 5i + 3j & B = 3i + 6j
Determine:
• A·B
• The angle between A and B
• AxB
• BxA
ME 221
Lecture 5
12
ME 221
Lecture 5
13
Vector Moment Definition
The moment about point O of a force acting at point A is:
F
O
A
MO = rA/O x F
rA/O
Compute the cross product with whichever
method you prefer.
ME 221
Lecture 5
14
200 N
Example
Method # 1
O.4
0.2
60 o
tan 60°=0.2m/x
x=0.115m
sin 60°=d/0.285m
60 o
0.285
A
x
d
d = 0.247 m
MA =200N *0.247m= 49.4 Nm
ME 221
Lecture 5
15
200 sin 60
200 N
Method # 2
O.4
60 o
200 cos60
0.2
A
+
M =200N (sin 60)(0.4m)- 200N (cos 60)(0.2m)
= 49.4 Nm
Note: Right-hand rule applies to moments
ME 221
Lecture 5
16
200 N
Method # 3
O.4
0.2
60 o
r
A
F=200N cos 60 i + 200N sin 60 j
r =0.4 i + 0.2 j
^i
j^
0.2
MA= 0.4
200cos60 200sin60
ME 221
^k
0 =200 (sin 60)(0.4) - 200 (cos 60)(0.2)
0 = 49.4 Nm
Lecture 5
17
200 N
Method # 4
O.4
60 o
0.2
r =0.285 i
A
F=200N cos 60 i + 200N sin 60 j
r =0.285 i
i
j
k
0
0 = 49.4 Nm
MA= 0.285
200cos60 200sin60 0
ME 221
Lecture 5
18
Moment of a Force about an Axis
y ^
n
|Mn| =MA·n^
A
^
=n·(rB/A x F )
Same as the projection of MA
along n
rAB=rB/A
F
B
O
x
z
nx
r
|Mn|= B/Ax
Fx
ME 221
ny
r B/A y
Fy
nz
r B/Az
Fz
Lecture 5
19
Resolve the vector MA into MA
two vectors one parallel and
one perpendicular to n.
y Mp
Mn
A
Mn=|Mn|n^
n^
F
rAB=rB/A B
O
Mp = MA - Mn
x
z
^
^
=n x [(r B/A x F) x n]
ME 221
Lecture 5
20
Moment of a Couple
Let F1 = -F2
B
y
|C|=|F1| d
ME 221
d
rAB=rB/A
F2
Mo=rA x F2+ rB x F1
=(rB - rA ) x F1
=rAB x F1= C
F1
A
rB
rA
O
x
The Moment of two equal
and opposite forces is called
a couple
z
Lecture 5
21
Moment of a Couple (continued)
• 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
ME 221
Lecture 5
22
Moment of a Couple (continued)
• 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
ME 221
Lecture 5
23
Example
ME 221
Lecture 5
24