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1

“”You are asking the wrong ques3ons. If you want to make the world a be<er place, tell funnier jokes!”

-‐Woody Allen

Coplanar Force Systems

Friday, September 14, 2012 Coplanar Force Systems 2

Objec4ves

¢ Understand the use of the equilibrium constraint condi4ons

¢ Understand the development of the FBD in the solu4on of equilibrium problems

2


Tools

¢  Basic Trigonometry ¢  Pythagorean Theorem ¢  Algebra ¢  Visualiza4on ¢  Posi4on Vectors ¢ Unit Vectors


Review ¢  For a system to be in equilibrium, the three following constraint condi4ons must be sa4sfied

Fx i

∑ = 0

Fy j

∑ = 0

Fz k

∑ = 0

3


Review ¢  If we take the sign of the direc4on vector 4mes the magnitude for each of the components we can then write

0

0

0

x

y

z

F

F

F

=

=

=

∑∑∑


Review ¢  The free-‐body diagram is an isola4on of an element

and the iden4fica4on of all forces which are ac4ng on that element

¢  By an element, we mean a part of a mechanical system

¢  Forces can act on a system as either an applied force from some external source or

¢  Forces can act due to the connec4on of the system to some other system in response to the applied forces – these forces are known as reac4ons

4


Ropes and Cables

¢ A rope always pulls on whatever it is connected to

¢ The force generated by a rope always has a line of ac4on along the rope

¢ Whenever the same rope passes over, but not 4ed to, mul4ple connec4ons, the force magnitude is the same through the length of the rope


Problems

¢  In order to illustrate how the process works, we will work a couple of example problems

¢  The first will be a two-‐dimensional problem and the second will be a three-‐dimensional problem

5


If the mass of cylinder C is 40 kg, determine the mass of cylinder A in order to hold the assembly in the position shown.


Determine the tension developed in cables AB, AC, and AD.

6

Homework

¢  Problem 3-‐29 (Hint: Set the force in each cable, one at a 4me, and solve for the forces in all the other cables)

¢  Problem 3-‐46 ¢  Problem 3-‐71

¢  Read through 4.4


THE FOLLOWING SLIDES ARE EXAMPLE PROBLEMS I WORKED OUT DURING A PREVIOUS COURSE.


7


Problem Given

¢ We have the system as shown ¢  The mass of the block is 30 kg ¢  Both springs are stretched

30 kg

A

BC

kAB = 1.2 kN/mkAC = 1.5 kN/m

0.5 m

0.4 m0.6 m


Problem Required

¢  The unstretched length of both springs, l0AB and l0AC

30 kg

A

BC


0.5 m

0.4 m0.6 m

8


Problem Solu4on

¢  Since we know that both springs are stretched, we can calculate the change in length in each spring

30 kg

A

BC


0.5 m

0.4 m0.6 m

0

0

AB AB AB

AC AC AC

s l ls l l

= −= −


Problem Solu4on

¢  From the geometry of the system we can calculate the stretched lengths of the two springs, lAB and lAC

30 kg

A

BC


0.5 m

0.4 m0.6 m

0

0

AB AB AB

AC AC AC

s l ls l l

= −= −

( ) ( )

( ) ( )

2 2

2 2

0.4 0.5 0.64

0.6 0.5 0.78

AB

AC

l m m m

l m m m

= + =

= + =

9


Problem Solu4on

¢  Subs4tu4ng what we know and isola4ng what we want to know we have

30 kg

A

BC


0.5 m

0.4 m0.6 m

0

0

0.640.78

AB AB

AC AC

l m sl m s

= −= −


Problem Solu4on

¢ Now we have a new subproblem ¢ We have to find sAB and sAC

30 kg

A

BC


0.5 m

0.4 m0.6 m

0

0

0.640.78

AB AB

AC AC

l m sl m s

= −= −

10


Problem Solu4on

¢ We know that in a spring, the force is propor4onal to the elonga4on

30 kg

A

BC


0.5 m

0.4 m0.6 m

AB AB AB

AC AC AC

F k sF k s

==

0

0

0.640.78

AB AB

AC AC

l m sl m s

= −= −


Problem Solu4on

¢ We know the spring constants kAB and kAC so we can isolate what we are looking for

30 kg

A

BC


0.5 m

0.4 m0.6 m

1.2

1.5

=

=

ABAB

ACAC

F skNm

F skNm 0

0

0.640.78

AB AB

AC AC

l m sl m s

= −= −

11


Problem Solu4on

¢  Finally we are at a point where we need to solve for forces, FAB and FAC

30 kg

A

BC


0.5 m

0.4 m0.6 m

0

0

0.640.78

AB AB

AC AC

l m sl m s

= −= −

1.2

1.5

=

=

ABAB

ACAC

F skNm

F skNm


Problem Solu4on

¢ We will select a system that FAB and FAC both act on if possible

30 kg

A

BC


0.5 m

0.4 m0.6 m

0

0

0.640.78

AB AB

AC AC

l m sl m s

= −= −

1.2

1.5

=

=

ABAB

ACAC

F skNm

F skNm

12


Problem Solu4on

¢  In this case it is the point A where the two springs and the box are connected

30 kg

A

BC


0.5 m

0.4 m0.6 m

0

0

0.640.78

AB AB

AC AC

l m sl m s

= −= −

1.2

1.5

=

=

ABAB

ACAC

F skNm

F skNm


Problem Solu4on

¢ We can draw the FBD of the point

0

0

0.640.78

AB AB

AC AC

l m sl m s

= −= −

30 kg

A

BC

kAB = 1.2 kN/m

FAC

0.5 m

0.4 m0.6 m

FAB

W

1.2

1.5

=

=

ABAB

ACAC

F skNm

F skNm

13


Problem Solu4on

¢  And using the constraint equa4ons, develop two expressions

0

0x x

y y

x AB AC

y AB AC

F F F

F F F W

= − =

= + − =∑∑ 0

0

0.640.78

AB AB

AC AC

l m sl m s

= −= −

30 kg

A

BC

kAB = 1.2 kN/m

FAC

0.5m

0.4m

0.6m

FAB

W


30 kg

A

BC

kAB = 1.2 kN/m

FAC

0.5m

0.4m

0.6m

FAB

W

Problem Solu4on

¢ We have five unknowns and two equa4ons so we need to extract more informa4on from the problem to proceed

0

0x x

y y

x AB AC

y AB AC

F F F

F F F W

= − =

= + − =∑∑ 0

0

0.640.78

AB AB

AC AC

l m sl m s

= −= −

14


¢  Since we have the mass of the block, finding W is just a conversion problem

30 kg

A

BC

kAB = 1.2 kN/m

FAC

0.5m

0.4m

0.6m

FAB

W

( ) 2

0

30 9.81 0

x x

y y

x AB AC

y AB AC

F F F

mF F F kgs

= − =

⎛ ⎞= + − =⎜ ⎟⎝ ⎠

∑∑

Problem Solu4on

0

0

0.640.78

AB AB

AC AC

l m sl m s

= −= −


¢ We s4ll have four unknowns and two equa4ons, we need to look at the forces

30 kg

A

BC

kAB = 1.2 kN/m

FAC

0.5m

0.4m

0.6m

FAB

W

( ) 2

0

30 9.81 0

x x

y y

x AB AC

y AB AC

F F F

mF F F kgs

= − =

⎛ ⎞= + − =⎜ ⎟⎝ ⎠

∑∑

Problem Solu4on

0

0

0.640.78

AB AB

AC AC

l m sl m s

= −= −

15


¢ We actually know the line of ac4on of each of the two forces because we know that each force acts along the axis of the spring

30 kg

A

BC

kAB = 1.2 kN/m

FAC

0.5m

0.4m

0.6m

FAB

W

( ) 2

0

30 9.81 0

x x

y y

x AB AC

y AB AC

F F F

mF F F kgs

= − =

⎛ ⎞= + − =⎜ ⎟⎝ ⎠

∑∑

Problem Solu4on

0

0

0.640.78

AB AB

AC AC

l m sl m s

= −= −


¢  Earlier in the problem we found the lAB and lAC so we can use these distances, the hypotenuse of each triangle to get more informa4on

30 kg

A

BC

kAB = 1.2 kN/m

FAC

0.5m

0.4m

0.6m

FAB

W

( ) 2

0

30 9.81 0

x x

y y

x AB AC

y AB AC

F F F

mF F F kgs

= − =

⎛ ⎞= + − =⎜ ⎟⎝ ⎠

∑∑

Problem Solu4on

0

0

0.640.78

AB AB

AC AC

l m sl m s

= −= −

0.40.640.60.780.50.640.50.78

x

x

y

y

AB AB

AC AC

AB AB

AC AC

mF FmmF FmmF FmmF Fm

=

=

=

=

16


( ) 2

0.4 0.6 00.64 0.780.5 0.5 30 9.81 00.64 0.78

x AB AC

y AB AC

m mF F Fm mm m mF F F kgm m s

= − =

⎛ ⎞= + − =⎜ ⎟⎝ ⎠

∑

∑

¢  If we subs4tute this informa4on we have two equa4ons with two unknowns

30 kg

A

BC

kAB = 1.2 kN/m

FAC

0.5m

0.4m

0.6m

FAB

W

Problem Solu4on

0

0

0.640.78

AB AB

AC AC

l m sl m s

= −= −


( ) 2

0.4 0.78 0.8120.64 0.60.5 0.5 30 9.81 00.64 0.78

AC AB AB

y AB AC

m mF F Fm mm m mF F F kgm m s

= =

⎛ ⎞= + − =⎜ ⎟⎝ ⎠∑

¢  Solving the top equa4on for FAC

30 kg

A

BC

kAB = 1.2 kN/m

FAC

0.5m

0.4m

0.6m

FAB

W

Problem Solu4on

0

0

0.640.78

AB AB

AC AC

l m sl m s

= −= −

17


( ) ( )

( )

( )

2

2

0.5 0.5 0.812 30 9.81 00.64 0.78

30 9.81

0.5 0.5 0.8120.64 0.78

⎛ ⎞= + − =⎜ ⎟⎝ ⎠⎛ ⎞⎜ ⎟⎝ ⎠=

+

∑ y AB AB

AB

m m mF F F kgm m s

mkgsF m m

m m

¢  Subs4tu4ng the value for FAC into the second equa4on

30 kg

A

BC

kAB = 1.2 kN/m

FAC

0.5m

0.4m

0.6m

FAB

W

Problem Solu4on

0

0

0.640.78

AB AB

AC AC

l m sl m s

= −= −


2260.812 183.5

AB

AC AB

F NF F N

== =

¢  Subs4tu4ng into the top equa4on the value for FAB and solving for FAC

30 kg

A

BC

kAB = 1.2 kN/m

FAC

0.5m

0.4m

0.6m

FAB

W

Problem Solu4on

0

0

0.640.78

AB AB

AC AC

l m sl m s

= −= −

18


226 0.191200

183.5 0.121500

ABAB

AB

ACAC

AC

F Ns mNk mF Ns mNk m

= = =

= = =

¢ Using the force in each spring and the spring constant we can determine the elonga4on of each spring

30 kg

A

BC

kAB = 1.2 kN/m

FAC

0.5m

0.4m

0.6m

FAB

W

Problem Solu4on

0

0

0.640.78

AB AB

AC AC

l m sl m s

= −= −


¢  Finally, subs4tu4ng in to our length equa4ons

30 kg

A

BC

kAB = 1.2 kN/m

FAC

0.5m

0.4m

0.6m

FAB

W

Problem Solu4on

0

0

0.64 0.19 0.450.78 0.12 0.66

AB

AC

l m m ml m m m

= − == − =

19


¢  Load is 500 lb ¢  System as shown

Problem Given


¢  Force in each cable

Problem Required

20


¢  Forces are FCB, FCA, and FCD ¢  All the forces are generated by ropes

Problem Solu4on


¢  A FBD of the connec4on at C would include all the unknown forces we are looking for

Problem Solu4on

21


¢  A FBD of the connec4on at C would include all the unknown forces we are looking for

Problem Solu4on

FCDFCB

FCA W


¢ We can write our constraint for equilibrium

Problem Solu4on

FCDFCB

FCA W

0

0CA CB CD

F

F F F F W

=

= + + + =

ur

ur uuur uuur uuur uur

22


¢ Now we need to find the Cartesian vector representa4on for each force

Problem Solu4on

FCDFCB

FCA W

{ } { } { } { }( )

( )

( )

2,0,0 0,6,0

2 6

6.32

2 60.32 0.95

6.32

0.32 0.95

CA CA CA

CACA

CA

CA

CA

CA

CA

CA CA

F F u

PuP

P A C ft ft

P i j ft

P ft

i j ftu i j

ft

F F i j

=

=

= − = −

= −

=

−= = −

= −

uuur uuur

uuuruuur

uuur

uuur r r

r ruuur r r

uuur r r



Problem Solu4on

FCDFCB

FCA W

{ } { } { } { }( )

( )

( )

2,0,0 0,6,0

2 6

6.32

2 60.32 0.95

6.32

0.32 0.95

CB CB CB

CBCB

CB

CB

CB

CB

CB

CB CB

F F u

PuP

P B C ft ft

P i j ft

P ft

i j ftu i j

ft

F F i j

=

=

= − = − −

= − −

=

− −= = − −

= − −

uuur uuur

uuuruuur

uuur

uuur r r

r ruuur r r

uuur r r

23



Problem Solu4on

FCDFCB

FCA W

{ } { } { } { }( )

( )

( )

0,12,8 0,6,0

6 8

10.00

6 80.6 0.8

10.00

0.6 0.8

CD CD CD

CDCD

CD

CD

CD

CD

CD

CD CD

F F u

PuP

P D C ft ft

P j k ft

P ft

j k ftu j k

ft

F F j k

=

=

= − = −

= +

=

+= = +

= +

uuur uuur

uuuruuur

uuur

uuur r r

r ruuur r r

uuur r r



Problem Solu4on

FCDFCB

FCA W

( )500W lb k= −uur r

24


¢  Summing the Cartesian vector representa4ons for all the forces

Problem Solu4on

FCDFCB

FCA W

( ) ( ) ( ) ( )0.32 0.95 0.32 0.95 0.6 0.8 500CA CB CDF i j F i j F j k lb k− + − − + + + −r r r r r r r


¢  Collec4ng terms with like unit vectors

Problem Solu4on

FCDFCB

FCA W

( ) ( ) ( )0.32 0.32 0.95 0.95 0.6 0.8 500CA CB CA CB CD CDF F i F F F j F lb k− + − − + + −r r r

25


¢  The coefficient of each unit vector must be equal to 0

¢  If it is not then we have an accelera4on along that axis and we no longer have equilibrium

¢  This gives us three equa4ons in three unknowns

Problem Solu4on

FCDFCB

FCA W


¢  The equa4ons are

Problem Solu4on

FCDFCB

FCA W

( )( )( )

0.32 0.32 0

0.95 0.95 0.6 0

0.8 500 0

CA CB

CA CB CD

CD

F F

F F F

F lb

− =

− − + =

− =

26


¢  From the first equa4on

Problem Solu4on

FCDFCB

FCA W

( )( )0.95 0.95 0.6 0

0.8 500 0

CA CB

CA CB CD

CD

F FF F F

F lb

=− − + =

− =


¢  From the third equa4on

Problem Solu4on

FCDFCB

FCA W

( )0.95 0.95 0.6 0500 6250.8

CA CB

CA CB CD

CD

F FF F FlbF lb

=− − + =

= =

27


¢  Subs4tu4ng the informa4on from the first and third equa4on in the second equa4on

Problem Solu4on

FCDFCB

FCA W

( )0.95 0.95 0.6 625 0375 197.371.9197.37

CA CB

CA CA

CA

CB

F FF F lb

lbF lb

F lb

=− − + =

= =

=

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