ME 221 Statics
LECTURE #9
Sections: 3.1 - 3.3
ME 221
Lecture 9
1
Exam #1 Results
Average Score: ?
Scores posted on Angel
Solution to be posted on Angel today
See syllabus for regrade policy
Homework #4
• Chapter 3 problems:
–
–
–
–
ME 221
1, 4, 8, 11, 17, 25, 26, 28, 35 & 40
To be solved using hand calculations
May check work using MathCAD, Matlab, etc.
Due Friday, September 26
Lecture 9
3
Chapter 3
Rigid Bodies; Moments
• Consider rigid bodies rather than particles
– Necessary to properly model problems
• Moment of a force
• Problems
ME 221
Lecture 9
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 9
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 9
6
Moment
• A force acting at a distance is a moment
M
M
d is the perpendicular
distance from F’s line
of action to O
A
O
d
F
Defn. of moment: M = Fd
• Transmissibility tells us the moment is the
same about O or A
ME 221
Lecture 9
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
• Example problems
ME 221
Lecture 9
8
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 9
9
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 9
10
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
ˆj
kˆ
A  B  Ax
Ay
Az
Bx
By
Bz
Lecture 9
11
ˆi
ˆj
kˆ
ˆi
ˆj
kˆ
ˆj
kˆ
A  B  Ax
Ay
Ay
Az
A+
 B  Ax
Ay
Az
Bx
By
Az A  B  Ax
Bx
Bz
ˆi
By
Bz
Bx
By
Bz
= (AyBz-AzBy) i - (AxBz-AzBx) j + (AxBy-AyBx)k
ME 221
Lecture 9
12
Problems
A = 5i + 3j
B = 3i + 6j
Find
• A·B
• The angle between A and B
• AxB
• BxA
ME 221
Lecture 9
13