ENGI 1313 Mechanics I Lecture 11: 2D and 3D Particle Equilibrium Shawn Kenny, Ph.D., P.Eng. Assistant Professor Faculty of Engineering and Applied Science Memorial University of.
Download ReportTranscript ENGI 1313 Mechanics I Lecture 11: 2D and 3D Particle Equilibrium Shawn Kenny, Ph.D., P.Eng. Assistant Professor Faculty of Engineering and Applied Science Memorial University of.
Slide 1
ENGI 1313 Mechanics I
Lecture 11:
2D and 3D Particle Equilibrium
Shawn Kenny, Ph.D., P.Eng.
Assistant Professor
Faculty of Engineering and Applied Science
Memorial University of Newfoundland
[email protected]
Slide 2
Chapter 3 Objectives
to introduce the concept of the free-body
diagram for a particle.
to show how to solve particle equilibrium
problems using the equations of
equilibrium
2
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 11
Slide 3
Lecture 11 Objectives
3
to further examine and apply Chapter 3
objectives in 2D and 3D space
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 11
Slide 4
Note on Free Body Diagram
F2
Force Sense and Solution
Negative sign indicates
the force sense is
opposite that shown
on the FBD
F
y
0
F1= mg
F1 F 2 0
F1 F 2
F2
F1= mg
4
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 11
Slide 5
Omit
Ch.3
Spring
Problems
5
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 11
Slide 6
Example 11-01
6
Each cord can sustain a
maximum tension of 200 N
Determine the largest weight
of the sack that can be
supported. Also, determine
θ of cord DC for equilibrium.
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 11
Slide 7
Example 11-01 (cont.)
Where to Start?
F
x
0
F
y
0
FBE
30
B
45
C
FBC
FCB
Point B
2 Equations
3 Unknowns
FAC
FAB
45
A
60
© 2007 S. Kenny, Ph.D., P.Eng.
FAH
60
H
FAH
7
FCA
Point C
2 Equations
3 Unknowns
FBA
Point A
2 Equations
3 Unknowns
FCD
Point H
1 Equation
2 Unknowns
but….
Newton’s 3rd Law
W = mg
ENGI 1313 Statics I – Lecture 11
Slide 8
Example 11-01 (cont.)
FBD at Point H
What Cord Will Have the
Maximum Tension?
Educated guess
Experience
Theoretical approach
FAH
H
• Assume W = 1N
• Maximum cord tension
8
200 N
© 2007 S. Kenny, Ph.D., P.Eng.
W = mg
F
y
0
F AH W m g
ENGI 1313 Statics I – Lecture 11
Slide 9
Example 11-01 (cont.)
FBD at Point A
F
0
x
F AB F AC
F AB cos 45
cos 60
cos 45
F AC cos 60
0
0 . 7071 F AC
F AB 0 . 7071 F AC 0 . 5176 N
F
y
0 F AB sin 45
0 . 7071 F AC sin 45
F AC sin 60
F AC sin 60
0 . 5 F AC 0 . 866 F AC 1 0
F AH 0
FAC
FAB
45
© 2007 S. Kenny, Ph.D., P.Eng.
60
W 0
FAH = W = 1N
F AC 0 . 7321 N
9
A
ENGI 1313 Statics I – Lecture 11
Slide 10
Example 11-01 (cont.)
FBD at Point B
F
x
0
FBC FBA cos 45
FBC FBA cos 45 FBE cos 30
FBE cos 30
F
y
0
0 . 5176 N sin 45
FBA sin 45
FBE sin 30
0
0
FBC 0 . 732 N cos 30 0 . 5176 N cos 45
0 . 268 N
FBE sin 30 0
0
FBE
30
F BE 0 . 7320 N
B
45
FBC
FBA = FAB = 0.5176N
10
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 11
Slide 11
Example 11-01 (cont.)
FBD at Point C
F
0
x
FCD cos FCA cos 60
FCB 0
FCD cos 0 . 7321 N cos 60 0 . 268 N 0 . 6341 N
F
y
0
FCA sin 60
FCD sin 0 . 7321 N sin 60
F
F
y
x
tan
11
1
FCD sin
FCD cos
FCD sin 0
0 . 6340 N
FCD
0 . 6340 N
sin
0 . 8966 N
0 . 6340 N
0 . 6341 N
0 . 6340 N
45
0 . 6341 N
© 2007 S. Kenny, Ph.D., P.Eng.
FCB = FBC = 0.268N FCD
C
60
FCA = FAC = 0.7321N
ENGI 1313 Statics I – Lecture 11
Slide 12
Example 11-01 (cont.)
Cord Forces
Analysis summary unit force
F AH 1
F AB 0 . 5176
F
0 . 7321
AC
F
F BE 0 . 7320
F
0 . 268
BC
F 0 . 8966
CD
12
N
Maximum force
• 200 N
F AH 1
F AB 0 . 5176
F
0 . 7321
AC
F 200
F BE 0 . 7320
F 0 . 268
BC
F 0 . 8966
CD
© 2007 S. Kenny, Ph.D., P.Eng.
F AH
F AB
F
N AC
F BE
F
BC
F
CD
ENGI 1313 Statics I – Lecture 11
200
104
146
N
146
53 . 6
179
Slide 13
Example 11-01 (cont.)
Use of Vector Algebra in
Mathematical Software
to Solve Mechanics Problems
Mathcad
• www.mathcad.com
Engineering calculations
This discussion on the use of Mathcad is
just for knowledge
It is not part of any course requirement
13
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 11
Slide 14
Example 11-01 (cont.)
Mathcad Solution
14
Set-up equilibrium
equations
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 11
Slide 15
Example 11-01 (cont.)
Mathcad Solution
15
Uses a command Find to
solves a system of linear
equations
This system of linear
equations is
based on the FBD
analysis that defines the
equilibrium equations (Fx
and Fy)
The Find command
function requires an
initial guess or estimate of
the forces and angle ()
to start the mathematical
search of the solution
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 11
Slide 16
Example 11-01 (cont.)
Mathcad Solution
16
Solve system of equations
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 11
Slide 17
Particle Equilibrium in 3D
Newton’s 1st Law
F F1 F2 F3 0
Cartesian Vector
Scalar components = 0
3 Equations
Solve for at most 3 unknowns
17
© 2007 S. Kenny, Ph.D., P.Eng.
F x ˆi
F y ˆj
F
x
0
F
y
0
F
z
0
ENGI 1313 Statics I – Lecture 11
Fz kˆ 0
Slide 18
Comprehension Quiz 11-01
In 3-D, the direction of a force is known but not
the force magnitude, how many unknowns
corresponding to that force remain?
18
A) One
B) Two
C) Three
D) Four
Answer: A
H int :
F F uˆ
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 11
Slide 19
Comprehension Quiz 11-02
In 3-D, when you don’t know either the direction
or magnitude of a force, how many unknowns do
you have corresponding to that force?
19
A) One
B) Two
C) Three
D) Four
Answer: C
H int :
© 2007 S. Kenny, Ph.D., P.Eng.
F F uˆ
uˆ cos ˆi cos ˆj cos kˆ
ENGI 1313 Statics I – Lecture 11
Slide 20
Comprehension Quiz 11-03
Four forces act at point A and the system
is in equilibrium. Select the correct force
vector F4 to balance the system. z
A)
B)
C)
D)
20
F4
F4
F4
F4
20 ˆi 10 ˆj 10 kˆ N
10 ˆi 20 ˆj 10 kˆ N
20 ˆi 10 ˆj 10 kˆ N
none of the above
F3 = 10 N
A
y
F1 = 20 N
x
Answer: D
© 2007 S. Kenny, Ph.D., P.Eng.
F2 = 10 N
ENGI 1313 Statics I – Lecture 11
Slide 21
Classification of Textbook Problems
Hibbeler (2007)
21
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 11
Slide 22
References
Hibbeler (2007)
http://wps.prenhall.com/esm_hibbeler_eng
mech_1
22
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 11
ENGI 1313 Mechanics I
Lecture 11:
2D and 3D Particle Equilibrium
Shawn Kenny, Ph.D., P.Eng.
Assistant Professor
Faculty of Engineering and Applied Science
Memorial University of Newfoundland
[email protected]
Slide 2
Chapter 3 Objectives
to introduce the concept of the free-body
diagram for a particle.
to show how to solve particle equilibrium
problems using the equations of
equilibrium
2
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 11
Slide 3
Lecture 11 Objectives
3
to further examine and apply Chapter 3
objectives in 2D and 3D space
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 11
Slide 4
Note on Free Body Diagram
F2
Force Sense and Solution
Negative sign indicates
the force sense is
opposite that shown
on the FBD
F
y
0
F1= mg
F1 F 2 0
F1 F 2
F2
F1= mg
4
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 11
Slide 5
Omit
Ch.3
Spring
Problems
5
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 11
Slide 6
Example 11-01
6
Each cord can sustain a
maximum tension of 200 N
Determine the largest weight
of the sack that can be
supported. Also, determine
θ of cord DC for equilibrium.
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 11
Slide 7
Example 11-01 (cont.)
Where to Start?
F
x
0
F
y
0
FBE
30
B
45
C
FBC
FCB
Point B
2 Equations
3 Unknowns
FAC
FAB
45
A
60
© 2007 S. Kenny, Ph.D., P.Eng.
FAH
60
H
FAH
7
FCA
Point C
2 Equations
3 Unknowns
FBA
Point A
2 Equations
3 Unknowns
FCD
Point H
1 Equation
2 Unknowns
but….
Newton’s 3rd Law
W = mg
ENGI 1313 Statics I – Lecture 11
Slide 8
Example 11-01 (cont.)
FBD at Point H
What Cord Will Have the
Maximum Tension?
Educated guess
Experience
Theoretical approach
FAH
H
• Assume W = 1N
• Maximum cord tension
8
200 N
© 2007 S. Kenny, Ph.D., P.Eng.
W = mg
F
y
0
F AH W m g
ENGI 1313 Statics I – Lecture 11
Slide 9
Example 11-01 (cont.)
FBD at Point A
F
0
x
F AB F AC
F AB cos 45
cos 60
cos 45
F AC cos 60
0
0 . 7071 F AC
F AB 0 . 7071 F AC 0 . 5176 N
F
y
0 F AB sin 45
0 . 7071 F AC sin 45
F AC sin 60
F AC sin 60
0 . 5 F AC 0 . 866 F AC 1 0
F AH 0
FAC
FAB
45
© 2007 S. Kenny, Ph.D., P.Eng.
60
W 0
FAH = W = 1N
F AC 0 . 7321 N
9
A
ENGI 1313 Statics I – Lecture 11
Slide 10
Example 11-01 (cont.)
FBD at Point B
F
x
0
FBC FBA cos 45
FBC FBA cos 45 FBE cos 30
FBE cos 30
F
y
0
0 . 5176 N sin 45
FBA sin 45
FBE sin 30
0
0
FBC 0 . 732 N cos 30 0 . 5176 N cos 45
0 . 268 N
FBE sin 30 0
0
FBE
30
F BE 0 . 7320 N
B
45
FBC
FBA = FAB = 0.5176N
10
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 11
Slide 11
Example 11-01 (cont.)
FBD at Point C
F
0
x
FCD cos FCA cos 60
FCB 0
FCD cos 0 . 7321 N cos 60 0 . 268 N 0 . 6341 N
F
y
0
FCA sin 60
FCD sin 0 . 7321 N sin 60
F
F
y
x
tan
11
1
FCD sin
FCD cos
FCD sin 0
0 . 6340 N
FCD
0 . 6340 N
sin
0 . 8966 N
0 . 6340 N
0 . 6341 N
0 . 6340 N
45
0 . 6341 N
© 2007 S. Kenny, Ph.D., P.Eng.
FCB = FBC = 0.268N FCD
C
60
FCA = FAC = 0.7321N
ENGI 1313 Statics I – Lecture 11
Slide 12
Example 11-01 (cont.)
Cord Forces
Analysis summary unit force
F AH 1
F AB 0 . 5176
F
0 . 7321
AC
F
F BE 0 . 7320
F
0 . 268
BC
F 0 . 8966
CD
12
N
Maximum force
• 200 N
F AH 1
F AB 0 . 5176
F
0 . 7321
AC
F 200
F BE 0 . 7320
F 0 . 268
BC
F 0 . 8966
CD
© 2007 S. Kenny, Ph.D., P.Eng.
F AH
F AB
F
N AC
F BE
F
BC
F
CD
ENGI 1313 Statics I – Lecture 11
200
104
146
N
146
53 . 6
179
Slide 13
Example 11-01 (cont.)
Use of Vector Algebra in
Mathematical Software
to Solve Mechanics Problems
Mathcad
• www.mathcad.com
Engineering calculations
This discussion on the use of Mathcad is
just for knowledge
It is not part of any course requirement
13
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 11
Slide 14
Example 11-01 (cont.)
Mathcad Solution
14
Set-up equilibrium
equations
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 11
Slide 15
Example 11-01 (cont.)
Mathcad Solution
15
Uses a command Find to
solves a system of linear
equations
This system of linear
equations is
based on the FBD
analysis that defines the
equilibrium equations (Fx
and Fy)
The Find command
function requires an
initial guess or estimate of
the forces and angle ()
to start the mathematical
search of the solution
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 11
Slide 16
Example 11-01 (cont.)
Mathcad Solution
16
Solve system of equations
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 11
Slide 17
Particle Equilibrium in 3D
Newton’s 1st Law
F F1 F2 F3 0
Cartesian Vector
Scalar components = 0
3 Equations
Solve for at most 3 unknowns
17
© 2007 S. Kenny, Ph.D., P.Eng.
F x ˆi
F y ˆj
F
x
0
F
y
0
F
z
0
ENGI 1313 Statics I – Lecture 11
Fz kˆ 0
Slide 18
Comprehension Quiz 11-01
In 3-D, the direction of a force is known but not
the force magnitude, how many unknowns
corresponding to that force remain?
18
A) One
B) Two
C) Three
D) Four
Answer: A
H int :
F F uˆ
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 11
Slide 19
Comprehension Quiz 11-02
In 3-D, when you don’t know either the direction
or magnitude of a force, how many unknowns do
you have corresponding to that force?
19
A) One
B) Two
C) Three
D) Four
Answer: C
H int :
© 2007 S. Kenny, Ph.D., P.Eng.
F F uˆ
uˆ cos ˆi cos ˆj cos kˆ
ENGI 1313 Statics I – Lecture 11
Slide 20
Comprehension Quiz 11-03
Four forces act at point A and the system
is in equilibrium. Select the correct force
vector F4 to balance the system. z
A)
B)
C)
D)
20
F4
F4
F4
F4
20 ˆi 10 ˆj 10 kˆ N
10 ˆi 20 ˆj 10 kˆ N
20 ˆi 10 ˆj 10 kˆ N
none of the above
F3 = 10 N
A
y
F1 = 20 N
x
Answer: D
© 2007 S. Kenny, Ph.D., P.Eng.
F2 = 10 N
ENGI 1313 Statics I – Lecture 11
Slide 21
Classification of Textbook Problems
Hibbeler (2007)
21
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 11
Slide 22
References
Hibbeler (2007)
http://wps.prenhall.com/esm_hibbeler_eng
mech_1
22
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 11