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ME 201Engineering Mechanics: Statics
Unit 3.2
Co-Planar Force Systems continued
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Steps to Problem Solving1. Draw a Free Body Diagram (FBD), label known
and unknowns Simple sketch of isolated particle
Sketch all forces that act on the particle Active forces
Reactive forces
Label known forces with proper magnitude & direction
Watch signs – negative means opposite sense from as drawn in FBD
2. Apply equations of equilibrium Any suitable direction
2 equations, 2 unknowns (n equations, n unknowns)
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Solving Spring Problems
Typically 3 components (in addition to FBD):
Equations of Equilibrium
Spring Equation
Geometry
kxF
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Example Problem Solution
θ A B
C
m
2 mGiven:
m = 8 kg
θ = 30º
kAB= 300 N/m
lAB=0.4 m (unstretched)
Find: lAC
30
W=8*9.81 N
TAB
TAC
0yF
081.9*830sin ACT
NTAC 157
0xF
030cos157 ABT
NTAB 136
1-FBD/Eqil EqnksTAB
m
mN
N
k
Ts AB 453.
300
136
2-Spring
θ A B
C2 m
.4+.453m2)453.4(.30cos ACl
mlAC 32.1
3-Geometry
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Solving Multiple Freebody Diagram Problems
Sketch possible FBDs
Select FBD with fewer unknowns to begin
Transfer results from first FBD to second
FBD
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Example Problem 3 Solution
W
θC
F
D
α
m
E
34
Given:
m = 20 lb
α = 30º
θ = 45º
Find: W
θC
TCD
34
TCE
W
3 unknowns
2 equations
Can only solve
for 2 unknowns
with a single
FBD
2 unknowns
2 equations
Create a second
FBD of point E
Solve for W by
using the FBDs
in succession
through TCE
θ
α
TCE
TEF
20
E
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Example Problem 3 Solution
FBD #1:
2045sin30cos CEEF TT
W
θC
F
D
α
m
E
34
θC
TCD
34
TCE
W
θ
α
TCE
TEF
20
0xF
045cos30sin CEEF TT
0yF
Solving simultaneously
lbTEF 6.54lbTCE 6.38
FBD #2:
W5
31.3445sin6.38
0xF
05
445cos6.38 CDT
0yF
lbW 8.47
lbTCD 1.34
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Solving Simultaneous Equations
Pick a tool:
Calculator
Excel
Mathematica
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Solving Simultaneous Equations
2x + 5y = -4
4x – 3y = 18x = 3
y = -2
2045sin30cos CEEF TT
045cos30sin CEEF TT lbTEF 6.54lbTCE 6.38
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Video 3g/3h – Solving Spring ProblemsFundamental Concepts
Typically 3 components (in addition to FBD):
Equations of Equilibrium
Spring Equation
Geometry
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Video 3i/3j – Solving Multiple Free Body Problems
Fundamental Concepts
Sketch possible FBDs
Select FBD with fewer unknowns to begin
Transfer results from first FBD to second
FBD
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Free Body Diagrams
To solve the following
problem, which free
body diagram(s) would
you use:
A. Point B
B. Point C
C. Link BC
D. Both points B & C
E. None of the above
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Key Concepts
Problem Solving Methodologies
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Steps to Problem Solving1. Draw a Free Body Diagram (FBD), label known
and unknowns Simple sketch of isolated particle
Sketch all forces that act on the particle Active forces
Reactive forces
Label known forces with proper magnitude & direction
Watch signs – negative means opposite sense from as drawn in FBD
2. Apply equations of equilibrium Any suitable direction
2 equations, 2 unknowns (n equations, n unknowns)
![Page 17: ME 201 Engineering Mechanics: Staticsemp.byui.edu/MILLERG/ME 201/Supplemental Material/In... · 2018. 10. 9. · ME 201 Engineering Mechanics: Statics Unit 3.2 Co-Planar Force Systems](https://reader036.vdocument.in/reader036/viewer/2022090810/611c54751b1ef6273f3a6342/html5/thumbnails/17.jpg)
Free Body Diagrams
To solve the following
problem, which free
body diagram(s) would
you use:
A. Point B
B. Point C
C. Link BC
D. Both points B & C
E. None of the above
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Solving Multiple Freebody Diagram Problems
Sketch possible FBDs
Select FBD with fewer unknowns to begin
Transfer results from first FBD to second
FBD
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Free Body DiagramIn Class Exercise
Determine the
force in each
cable and the
force F needed
to hold the 4 kg
lamp in the
position shown.
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Solving Spring Problems
Typically 3 components (in addition to
FBD):
Equations of Equilibrium
Spring Equation
Geometry
kxF
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Example Problems
#3-Find Wf
#2 – Find W (video)#1 – Find Lac (video)
θ A B
C
m
2 m
W
θC
F
D
α
m
E
34
#4-Find Labc
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Example Problem 2Given:
m = 8 kg
θ = 30º
kAB= 300 N/m
lAB=0.4 m (unstretched)
Find:
lAC
θ A B
C
m
2 m
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Example Problem 2 Solution
θ A B
C
m
2 mGiven:
m = 8 kg
θ = 30º
kAB= 300 N/m
lAB=0.4 m (unstretched)
Find:
lAC
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Example Problem 2 Solution
Solution:
1-FBD
2-Equilibrium
Equations
θ A B
C
m
2 m
30
W=8*9.81 N
TAB
TAC
0yF
081.9*830sin ACT
NTAC 157
0xF
030cos157 ABT
NTAB 136
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Example Problem 2 Solution
Solution:
1-FBD
2-Equilibrium
Equations
3-Spring
F=ks
4-Geometry
m
mN
N
k
Ts AB 453.
300
136
θ A B
C
m
2 m
ksTAB
θ A B
C2 m
.4+.453m
2)453.4(30cos ACl
mlAC 32.1
30
W=8*9.81 N
TAB
TAC
157 N
136 N
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Example Problem 3Given:
m = 20 lb
α = 30º
θ = 45º
Find:
W
W
θC
F
D
α
m
E
34
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Example Problem 3 Solution
W
θC
F
D
α
m
E
34
Given:
m = 20 lb
α = 30º
θ = 45º
Find: W
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Example Problem 3 Solution
Solution:
W
θC
F
D
α
m
E
34
θC
TCD
34
TCE
W
3 unknowns
2 equations
Can only solve
for 2 unknowns
with a single
FBD
2 unknowns
2 equations
Create a second
FBD of point E
Solve for W by
using the FBDs
in succession
through TCE
θ
α
TCE
TEF
20
E
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Example Problem 3 Solution
FBD #1:
2045sin30cos CEEF TT
W
θC
F
D
α
m
E
34
θC
TCD
34
TCE
W
θ
α
TCE
TEF
20
0xF
045cos30sin CEEF TT
0yF
Solving simultaneously
lbTEF 6.54lbTCE 6.38
FBD #2:
W5
31.3445sin6.38
0xF
05
445cos6.38 CDT
0yF
lbW 8.47
lbTCD 1.34
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In Class Exercise
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Solution
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Solution
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In Class Exercise
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Solution