ME 201 Engineering Mechanics: Staticsemp.byui.edu/MILLERG/ME 201/Supplemental Material/In... ·...

ME 201Engineering Mechanics: Statics

Unit 3.2

Co-Planar Force Systems continued

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)

Solving Spring Problems

Typically 3 components (in addition to FBD):

Equations of Equilibrium

Spring Equation

Geometry

kxF

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

Solving Multiple Freebody Diagram Problems

Sketch possible FBDs

Select FBD with fewer unknowns to begin

Transfer results from first FBD to second

FBD

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


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

Solving Simultaneous Equations

Pick a tool:

Calculator

Excel

Mathematica

Solving Simultaneous Equations

2x + 5y = -4

4x – 3y = 18x = 3

y = -2


045cos30sin CEEF TT lbTEF 6.54lbTCE 6.38

Video 3g/3h – Solving Spring ProblemsFundamental Concepts

Typically 3 components (in addition to FBD):


Spring Equation

Geometry

Video 3i/3j – Solving Multiple Free Body Problems

Fundamental Concepts




FBD

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

Key Concepts

Problem Solving Methodologies

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)

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

Solving Multiple Freebody Diagram Problems




FBD

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.

Solving Spring Problems

Typically 3 components (in addition to

FBD):


Spring Equation

Geometry

kxF

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

Example Problem 2Given:

m = 8 kg

θ = 30º

kAB= 300 N/m


Find:

lAC

θ A B

C

m

2 m


θ A B

C

m

2 mGiven:

m = 8 kg

θ = 30º

kAB= 300 N/m


Find:

lAC


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


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

Example Problem 3Given:

m = 20 lb

α = 30º

θ = 45º

Find:

W

W

θC

F

D

α

m

E

34


W

θC

F

D

α

m

E

34

Given:

m = 20 lb

α = 30º

θ = 45º

Find: W


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


FBD #1:


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

In Class Exercise

Solution

In Class Exercise

Solution

ME 201 Engineering Mechanics: Staticsemp.byui.edu/MILLERG/ME 201/Supplemental Material/In... ·...

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