1 3 equilibrium system isolation and equilibrium conditions

1

3 EQUILIBRIUM

System isolation and

equilibrium conditions

2

Objectives Students must be able to #1

Course Objective• State the conditions of equilibrium, draw free body

diagrams (FBDs), analyse and solve problems involving rigid bodies in equilibrium.

Chapter Objectives• Analyse objects (particles and rigid bodies) in

equilibriums• Classify problems in equilibrium into SD and SI

categories

3

Equilibrium Definition

An object is in equilibrium when it is stationary or in steady translation

Earth

Whether object in stationary (moving in steady translation) or not,

depends on “reference frame”.

really equilibrium?Newtonian Mechanics

“Inertial Reference Frame”

- Earth Frame - Central Universe Frame

stationary

with Constant velocity

moving

relative to an “inertial reference frame”.

Equilibrium

Centrifugalacceleration

4

- Equilibrium is the most important subject in statics.

- In statics, we deal primarily with bodies at rest.

(i.e. they are in the state of “equilibrium”).

If is true body in equilibrium

3/1 Introduction

0R F

If body in equilibrium is true

- More precisely, when a body is in equilibrium, the wrench resultant of all forces acting on it is zero; i.e.

- These requirements are necessary and sufficient conditions for equilibrium; i.e.

- From the Newton’s second law of motion, a body that moves with constant velocity, rotates with constant angular velocity; i.e. “zero acceleration”, can also be treated as in a state of equilibrium.

RC

1F

3F 2F

( )AM

A

A

0C

5

Equilibrium in 2D- All physical bodies are inherently 3D, but many may be treated as 2D; e.g. when all forces are on the same plane.

3/2 Mechanical system isolation- Before we apply the equilibrium conditions

we need to know what force or couple are involved.

0

0

i

i

R F

M M

- Draw Free Body Diagram (FBD)

FBD is used to isolate body (or bodies / system of bodies) so that force/couple acting on it can be identified.

F

F

F F

Isolate body

6

Free Body Diagram (FBD)

• FBD is the sketch of the body under consideration that is isolated from all other bodies or surroundings.

• The isolation of body clearly separate cause and effects of loads on the body.

• A thorough understanding of FBD is most vital for solving problems.

7

1) Pick body/combination of bodies to be isolated

y

x

F

2) Isolating the body. Draw “complete external boundary” of the isolated body

3) Add all forces and moments (including that are applied by the removed surrounding)

4) Indicate a coordinate system

mg

N

body in interest

Most important step is solving problems in mechanics.

*** If an FBD is not drawn (when it is needed), you will get no credit ( 0 point ) for the whole problem!!!! ***

Construction of FBD

y

x

F

mg

N

Free Body Diagram

5) Indicate necessary dimensions

150 cm

50 cm

150 cm

50 cm

8

Free Body Diagram• Establish the x, y, z axes in any suitable

orientation.• Label all the known and unknown force

magnitudes and directions on the diagram

• The sense of a force having an unknown magnitude can be assumed.

Use different colours in diagrams• Body outline - blue• Load (force and couple) - red• Miscellaneous (dimension, angle, etc.) - black

Note on drawing FBD

y

x

F

mg

N

150 cm

50 cm

9

Equilibrium Solving Procedure 0

0

i

i

R F

M M

• Formulate problems from physical situations. (Simplify problems by making appropriate assumptions)• Draw the free body diagram (FBD) of objects under

consideration

• State the condition of equilibrium

• Substitute variables from the FBD into the equilibrium equations

• Substitute the numbers and solve for solutions– Delay substitute numbers– Use appropriate significant figures

• Technical judgment and engineering sense– Try to predict the answers– Is the answer reasonable?

10

Equilibrium Free Body Diagram (FBD)

• FBD is the sketch of the body under consideration that is isolated from all other bodies or surroundings.

• The isolation of body clearly separate cause and effects of loads on the body.

• A thorough understanding of FBD is most vital for solving problems.

11

Equilibrium FBD ConstructionEquilibrium

No FBDs Cannot apply equilibrium conditions

Stop...........

• Select the body to be isolated• Draw boundary of isolated body, excluding supports• Indicate a coordinate system by drawing axes• Add all applied loads (forces and couples) on the isolated

body.• Add all to support reactions (forces and couples) represent

the supports that were removed.• Beware of loads or support reactions with specific directions due

to physical meanings • Add dimensions and other information that are required in the

equilibrium equation

12

Equilibrium Help on FBD

Establish the x, y axes in any suitable orientation. Label all the known and unknown applied load

and support reaction magnitudes Beware of loads or support reactions with specific directions due to physical meanings

Otherwise, directions of unknown loads and support reactions can be assumed.

13

Equilibrium On FBD Analyses

Objective: To find support reactions• Apply the equations of equilibrium

• Load components are positive if they are directed along a positive direction, and vice versa

• It is possible to assume positive directions for unknown forces and moments. If the solution yields a negative result, the actual load direction is opposite of that shown in the FBD.

, , ,

0 or 0, 0, 0

0 or 0, 0, 0

x y z

O O x O y O z

F F F F

M M M M

Equilibrium

14

Contents

• Equilibrium of Objects– Particles (2D & 3D)– 2D Rigid Bodies– 3D Rigid Bodies

the heart () of Statics

• SD and SI Problems

0

0

i

i

R F

M M

Particle Ideal particle can not rotate.

(no couple acting on it)

Rigid Bodies

0 iR F

15

Particles FBD construction

To construct a complete FBD of a particle• Select the particle to be isolated• Draw the particle as a point• Indicate a coordinate system• Add all active forces/moments (weight, etc.)• Add all support reactions (e.g. tension in the tangential

direction of a cable, tensile and compressive forces in a compressed and stretched springs)

16

Particles Equilibrium Analyses #2

Equations of Equilibrium• Apply the equations of equilibrium

• Components are positive if they are directed along a positive axis, and negative if they are directed along a negative axis.

• Assume the directions of unknown forces in the positive x, and y axes. If the solution yields a negative result, this indicates the sense of the force is the reverse of that shown on the FBD.

0, 0, 0x y zF F F

0

ˆ ˆ ˆ( ) ( ) ( ) 0x y z

F

F i F j F k

17

Particles Equilibrium in 3D

0

0

0

x

y

z

F

F

F

1F

2F

3F

xy

z

W

BF

DF

CF

Particle Equilibrium

18

H-Ex3-1#1) The sphere has a mass of 6 kg and is supported as shown.Draw a free-body diagram of the sphere, the cord CE, and the knot at C.

Action-reaction pair - use same symbol - opposite direction

y

x

0 F 58.9 0y CEF F 58.9 NCE

0 F 0y CE CEF F

CEF

CEF

58.9 CE CEF F N

0 F cos 60 0ox CBA CDF F

0 F sin 60 0oy CBA CEF F

58.9 F

sin 60 sin 60

F cos60 58.9cot 60

CECBA o o

o oCD CE

F

F

How many unknowns?

CEF58.9 N

How many Equations?

20

Determine the tension in cables AB and AD for equilibrium of the 250-kg engine shown.

FBD of A

0.25 kNg

29.807 m/sg

21

Particle is in equilibrium.

0 cos30 0 (1)

0 sin30 0.25 kN 0 (2)

Solve (2) and subst. into (1) 4.9035 kN, 4.2466 kN

4.90 kN, 4.25 kN #

x B D

y B

B D

B D

A

F T T

F T g

T T

T T

FBD of A

0.25 kNg

22

Example Hibbeler Ex 3-3 #1

If the sack at A has a weight of 20 lb, determine the weight of the sack at B and the force in each cord needed to hold the system in the equilibrium position shown.

FBD of E

Equilibrium of FBD of

0 sin30 cos 45 0 (1)

0 cos30 sin45 20 lb 0 (2)

Solve (1) & (2) 38.637 lb, 54.641 lb

38.6 lb, 54.6 lb #

x EG EC

y EG EC

EC EG

EC EG

E

F T T

F T T

T T

T T

= ?

3unknown, 2 Eqs(at this stage)

2unknown, 2 Eqs(at this stage)

23


0 cos 45 lb (4 5) 0 (3)

0 (3 5) sin45 lb 0 (4)

Solve (3) & (4) 34.151 lb, 47.811 lb

34.2 lb, 47.8 lb #

x EC CD

y CD EC B

CD B

CD B

C

F T T

F T T W

T W

T W

FBD of C

ECT

FBD of E

38.637 lb

54.641 lbEC

EG

T

T

Recommended

38.637 lb

24


A 90-N load is suspended from the hook. The load is supported by two cables and a spring having a stiffness k = 500 N/m. Determine the force in the cables and the stretch of the spring for equilibrium. Cable AD lies in the x-y plane and cable AC lies in the x-z plane.

25


0 sin30 (4 5) 0 (1)x D C

A

F F F

(no FBD, no score)

3D particle Equilibrium

How many unknowns,how many equations?

0 0 0 x y zF F F

0 cos30 0 (2)y D BF F F

0 (3 5) 90 0 (3)z CF F

Solve (3), 150 N #

Solve (1), 240 N #

Solve (2), 207.85 N 208 N #

500

207.85 / 500

0.41569 0.416 m #AB

C

D

B

s B AB AB

AB

s

F

F

F

F ks F ks s

s

perfect answer sheet

26


Determine the force developed in each cable used to support the 40-kN (4 tonne) crate shown.

0, 0, 0x y zF F F

0

ˆ ˆ ˆ( ) ( ) ( ) 0x y z

F

F i F j F k

3D particle Equilibrium

27

Hibbeler Ex 3-7 #2Particle Equilibrium

Position vectors

ˆ ˆ ˆ3 4 8 m

ˆ ˆ ˆ3 4 8 m

AB

AC

r i j k

r i j k

Force vectors

1 ˆ ˆ ˆ( 3 4 8 )89

1 ˆ ˆ ˆ( 3 4 8 )89

ˆ

ˆ40 kN

ABB B B

AB

ACC C C

AC

D D

rF F F i j k

r

rF F F i j k

r

F F i

W k

28

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ( 3 4 8 ) ( 3 4 8 ) ( ) ( 40 ) 089 89

3 4 83 4 8ˆ ˆ ˆ( ) ( ) ( 40) 089 89 89 89 89 89

CBD

C C CB B BD

FFi j k i j k F i k

F F FF F FF i j k

Equilibrium of

0 0B C D

A

F F F F W

29


Equilibrium of

330 0 (1)

89 8944

0 0 (2)89 89

880 40 0 (3)

89 89

From (2),

16From (3), 40 0 or 23.585 kN, 23.6 kN #

896

From (1), 0 or 15.0 kN,89

CBx D

CBy

CBz

B C

CC C B

CD D

A

FFF F

FFF

FFF

F F

FF F F

FF F

15.0 kN #DF

Particle Equilibrium

31

Equilibrium of 2D Rigid Bodies

• Use similar analyses as the particles• Additional consideration

– Action forces in supports/constraints– Free-body diagram (FBD) of 2D rigid bodies– Equilibrium equations (scalar form) for rigid bodies– Two-force and three-force members

2D Equilibrium

32

Force Reaction (2D) To write an FBD, first, you will need to know what kind of force we will get when eliminating the environment/surrounding.

1) Flexible cable, belt, chain, or rope

always away from the bodyT

Ttangent to the cable

33

2) Smooth surfaces

N

- Contact force at contact point normal to the surface/contact plane

N

3) Rough surfaces

N

R

F- A rough surface can

produce a tangential force (F, friction) as well as a normal force (N)

- direction of F depend on situations (chapter 6)

- always compressive

only this direction

only this direction

Not always this direction

34

N

- Roller, rocker, or ball transmits a compressive force normal to the supporting surface

5) Freely sliding guide

N

4) Roller supports

N

N

The vector N may be up or down depend on problem. If not known, you may assume any of the two. After further calculation, if N is +, correct sense was assumed. If negative, N goes the other way.M

M existence due to its bending resistance

only this direction

not alwaysthis direction

35

Pin free to turn Pin not free to turn6) Pin connection

Rx

Ry

Rx

Ry

M

7) Built-in or fixed support

WeldA A A

F

V

M

As a general rule, if a support prevents translation of a body in a given direction, then a force is developed on the body in the opposite direction. Similarly, if rotation is prevented, a couple is exerted on the body.



36

8) Gravitational attraction

m G

W=mg

Resultant of the gravitational attraction is the weight W = mg and act toward center of the earth passing through the center mass G

9) Spring action

x

F

For linear springs , F = kx

F

F x is positive

x is negative

Normaldistance

37

You may assume either case.The sign will indicate its sense of direction later.

Equilibrium

38

Equilibrium construction of a FBD

To construct a complete FBD for a 2D rigid bodies• determine which body is to be isolated• draw external boundary of isolated body• indicate a coordinate system (axes)• add all loads (forces and couples, be they applied or

support)

No FBDs Cannot apply equilibrium conditions

NO SCORES

2D Equilibrium

39

notalways “mg”

No internal force is shown in FBD

Amass m

F1 F2F3

BA

M

Pmass m

**Write FBD before allowing any movements of vectors (sliding/free)**

BA

P

m

y

x

y

x

F1 F2F3

W=mg

VF

M

P

M

W=mgV

F

P

T

VFN

y

x

NNCorrect?

40

3/A Correct the incomplete FBD

F

Ax

Ay

y

xfor all figures below

41

3/B Correct the wrong or incomplete FBD

f mg

Ox

F

y


If rough surface

42

3/B Correct the wrong or incomplete FBD

C

D

Ax

C

y


43System Isolation

& FBDEquilibrium

We will choose problem 1 and 4 as samples

44

A

mg

mgAx

Ay

Nm0g

Byf

y

x

y

x

M

|_ with the incline

45


Draw the free-body diagram of the uniform beam shown.The beam has a mass of 100 kg.

2D Equilibrium

46

Example Hibbeler Ex 5-1 #22D Equilibrium

47


Two smooth pipes, each having a mass of 300 kg, are supported by the forks of the tractor. Draw the FBDs for each pipe and both pipes together.

2D Equilibrium

48

2D Equilibrium

49


50


Draw the FBD of the unloaded platform that is suspended off the edge of the oil rig shown. The platform has a mass of 200 kg.

2D Equilibrium

51


54

3/3 Equilibrium Conditions

R

C

2F

1F

3F

(for some point P )

- Equilibrium conditions (3D) of the body is equivalents to

PP

- We can find the wrench resultant force

“equilibrium”

0R

0C

0F

0OM

(for all point O)

We can prove this!

G

0F

P

G

AllPoint0 ?M

0P

M

0F

0PM

- For co-planer forces only (2D), eq conditions is equivalents to

Fx= 0 Fy = 0 Mo = 0 (for all point O) (2D)

“equilibrium”0F

0PM

57

Equilibrium and its Independent Equations

RC

AEquilibrium

(no tendency to initiate translation, no tendency to initiate rotation)

0R F

( )AM

0C

0R F

0AM

(some point: A)

0R F

0oM

(any point: O)

A

(trivially easy)

equilibrium

RC

1F

3F 2F

A

A

0F

A

B

G

and C A

A B C

A B C

0A

M

Important Meaning

0

0

x

x

F

F

0PM 0QM Only at most 3 independent Equations

2D

any O

( ) 0i iir F B

58

0

0

0

x

y

O

F

F

M

Equilibrium Eqn. for 2D Rigid BodiesScalar Form

The sum of the moment about any point O is zero.

Particle

Rigid Body

2D Equilibrium

59

2D Equilibrium Procedure #1

Free-Body Diagram• Establish the x, y axes in any suitable orientation.• Draw an outlined shape of the body.• Show all the forces and couple moments acting on

the body.• Label all the loadings and specify their directions

relative to the x, y axes. The sense of a force or couple moment having an unknown magnitude but known line of action can be assumed.

• Indicate the dimensions of the body necessary for computing the moments of forces.

2D Equilibrium

60

Equations of Equilibrium• When applying the force equilibrium equations,

orient the x and y axes along lines that will provide the simplest resolution of the forces into their x and y components.

• If the solution of the equilibrium equations yields a negative scalar for a force or couple moment magnitude, this indicate that the sense is opposite to that which was assumed on the FBD.

0 and 0x yF F

2D Equilibrium Procedure #22D Equilibrium

61

Equations of Equilibrium• Apply the moment equation of equilibrium about a point O

that lies at the intersection of the lines of action of two unknown forces.

In this way, the moments of these unknowns are zero about O, and a direct solution for the third unknown can be determined.

0OM

2D Equilibrium Procedure #32D Equilibrium

63

H5.7) Find the tension in the cord at C and the horizontal and vertical component of reactions at pin A. The pulley is frictionless.

The body is in equilibrium.

Fx = 0 Fy = 0 MAnyPoint = 0

0xF sin 0xT A

0yF cos 500 0yT A

0AM 500(0.2) (0.2) 0T

+

500 N

sin 250 N

500 cos 933 Nx

y

T

A T

A T

If pulley rotates, T are the same in both side ?

If couple also applies at pully,T are the same in both side?

T are the same at both side?

with constant angular velocity?

64

Independent Equations

2F

1F

3F

Body in equilibrium

** However, sometimes you can not get those 3 unknowns in 2D problem.

Fx= 0 Fy = 0 Mo = 0

(for some point O)

Fx= 0 Fy = 0 Mo = 0

(for all point O)

Hey , why you cannot get anymore, you can take another points to take moment, and get new equation?

x

y

not independent

equations!

** Whatever 2D-equilibrium conditions you used, at most 3 unknowns (scalar) may be found.

65

H5.7) Find the tension in the cord at C and the horizontal and vertical component of reactions at pin A. The pulley is frictionless.

The body is in equilibrium.

Fx = 0 Fy = 0 MAnyPoint = 0

0xF sin 0xT A

0yF cos 500 0yT A

0AM 500(0.2) (0.2) 0T

+

500 N

sin 250 N

500 cos 933 Nx

y

T

A T

A T

(0.2) ( cos )(0.2 0.2sec ) 0yA T 0DM

Not independent Equations!

(0.2) 0

0

0

y

A

D

F

M

M

66


Determine the horizontal and vertical components of reaction for the beam loaded as shown. Neglect the weight of the beam in the calculation.

2D Equilibrium

67


FBD of ADB

Find: Ay, Bx, By

2D Equilibrium

68


Equilibrium of

0 600cos 45 N 0

424.26 424 N #

0 (100 N)(2 m) (600sin45 N)(5 m)

(600cos 45 N)(0.2 m) (7 m) 0

319.50 319 N #

x x

x

B

y

y

ADB

F B

B

M

A

A

Q

2D Equilibrium

69



0 600sin45 N 100 N 200 N 0

404.76 405 N #

y y y

y

ADB

F A B

B

2D Equilibrium

70


The link shown is pin-connected at A and rests against a smooth support at B. Compute the horizontal and vertical components of reactions at the pin A.

2D Equilibrium

71


FBD of link BA

2D Equilibrium

72


Equilibrium of FBD of link

0 90 N m (60 N)(1 m) (0.75 m) 0

200 N #

0 sin30 N 0, 100 N #

0 cos30 N 60 N 0

233.21 233 N #

A B

B

x x B x

y y B

y

BA

M N

N

F A N A

F A N

A

Q

2D Equilibrium

73

T=300N T

3/57 200-kg beam has a C.G. at G. The 80-kg man is exerted a 300-N force on the rope. Calculate the force reaction at A which is a weld pin.

200(9.81)

300N

T300 N

N =mg?N

mg =80(9.81)

300 N

x

y+

xA

AM

yA

T=300N

200(9.81)

xA

AM

yA80(9.81)

300 N200(9.81)

xA

AM

yA80(9.81)

C

C

= 300N?

N is internal force.

tension is internal force.

4 unknowns, 3 Eq.You can’t solve it?

You can solve it !Use FBD of rod

3 unknowns, 3 Equations. You can solve it.

mg

74

3/57 200-kg beam has a C.G. at G. The 80-kg man is exerted a 300-N force on the rope. Calculate the force reaction at A which is a weld pin.

200(9.81)

xA

AM

yA80(9.81)

Ctension is internal force.

80(9.81)

Sun’sGravitational

force

http://www.earthshope.org/Home_files/earth%20transparent.png

Draw the FBD of the foot lever shown.

why?

77

Two-Force Member In Equilibrium

P To keep the body in equillibrium,the second force must …

P - has the same manitude

- direction of opposite side.

- be co-linear

Fx= 0 Fy = 0 Mo = 0

body in equilibrium

bodies in equilibrium

Body in Equilibrium

78

Three-Force Member for Equilibrium

1F

2F

3FO

3F

1F 2F0 F

1F

2F

3F

4 forces need to be concurrent or parallel to make body in equilibrium?

To keep the body in equillibrium,the second force must …

(co-planer)

Concurrent

concurrent Parallel (with proper moment arm)

1)

2) OR Parallel (with proper moment arm)

M = 0

Fx= 0 Fy = 0 Mo = 0

85

Draw the FBD of the foot lever shown.

why?

Two force in Equilibrium

BFBF

You should write the FBD of 2-force (in equilibrium) like this.

86

The lever ABC is pin-supported at A and connected to a short link BD as shown. If the weight of the members is negligible, determine the force of the pin on the lever at A.

Hibbeler Ex 5-13

yB

xB

yA

xA

yB

xB

yD

xD

0y yA B 400 0x xA B

400(0.5) (0.1) (0.2) 0y xA A

0x xB D 0y yB D

(0.2) (0.2) 0y xD D

x y x yD D B B

4 unknowns, 3 eq.

2 unknowns, 3 eq.

2800 N

3x y x yD D B B

1600 2800 N

3 3X yA A

x

y+

Two Force Members

2 forcemembers

Hibbeler Ex 5-13

x

y+

Assume Directions

3-Unknowns, 3-Eqs

45o

0.5 m

0.5-0.1=0.4 m

O

F AF

45o

Three Forces are concurrent.

1 0.7tan 60.26

0.4

Solve by using 3-force member Concept

0.5+0.2=0.7 m

88

The lever ABC is pin-supported at A and connected to a short link BD as shown. If the weight of the members is negligible, determine the force of the pin on the lever at A.

2 forcemembers

60.26

0.5 m

O

F AF

[ 0]

sin60.26 sin45 0

y

A

F

F F

1074.9 N

1319.9 NAF

F

[ 0]

cos60.26 cos45 400 N 0 x

A

F

F F

Ans

x

y+

89

Any two-force members or three-force members here?

91

3/27 In a procedure to evaluate the strength of the triceps muscle, a person pushed down on a load cell with the palm of his hand as indicated in the figure.If the load-cell reading is 160 N, determine the vertical tensile force F generated by the triceps muscle. The mass of the lower arm is 1.5 kg with mass center at G.

92

0.250.15 0.15

x

y

160N

T

1.5(9.81)

Free Body Diagram

Ox

Oy

Because what we want to know is the force T of triceps. Taking summation of moments at point O will eliminate Ox and Oy which are unknowns out, and the calculation will be easier.

0OM

Ans

0.25 0.15(1.5)(9.81) 0.15 0.15 160 0T

1832 NT

+

3 Unknown, 3 Equations.

If what you want o know is Oy ?

You can solve it.By easier way.

If what you want o know is Ox ?

93

Concrete slab with mass of 25,000 kg is pulled by cable of tension P. Determine the tension T in the horizontal cable using only 1 equilibrium equation.

Key Here we have 3 unknowns (P,R,T). To use only 1equilibrium eq. To determine T, we need to take moment at the point where the other unknown forces (R,P) passes.

Given: = 60°

TR

W = mg60

30D

30 A

P

6 m

x

y

Technique: Finding unknown force only in 1 step

94

0MD +

0)60cos4()W()ADlength()T(

( ) (6 tan 30 ) (25000)(9.8) (4cos60 ) 0T

490500)464.3()T(

N600,141T ANS

TR

W = mg

60

D

30 A

P

6 m

4 m

x

y

30

96

Alternative Equilibrium equations- Recall: In general, you have these three equilibrium equations:

(for some point O) Fx= 0 Fy = 0 Mo = 0 (E)

A

B

A

B

C

line AB not |_ to the x direction A B C not on the same straight line

MB = 0 MA = 0

x

- Alternatively, you may use either

R=0 R=0

Fx= 0 MB = 0 MA = 0 Mc = 0

not |_

97

Calculate the magnitude of the force at pin A.

Free Body Diagram

x

y

How many unknowns ?

We have 3 unknowns of

AxAy By

Ax

Ay By 80 N-m

500 N

400 N

D

0 0 0A x yM F F

0 0 0A B xM M F

0 0 0A B DM M M

98

we can determine Ay from 0MB

080)25)(.400()125)(.500()15(.)A( y

N3.783Ay

we can determine Ax from 0Fx

+

+

0500Ax N500Ay

Thus, the resultant of force at pin A:

N3.929)3.783()500(R 22 ANS

Free Body Diagram

Ax

Ay By

80 N-m

500 N

400 NCalculate the magnitude of the force at pin A.

x

y

99

Constraints and Statical determinacy** need to know when a problem can be solved or what force/couple can be found.

1) Statically Determinate Improper Constraint

2) Statically Indeterminate

- Not enough constraint

- Not in equilibrium

- can’t get all unknowns

Unknown 3 , Independent eq 3



unknown 3 , independent 2

No solution: (cant maintain moment)

100

Independent Equations

1F

2F 3F

1) Co-linear xEquilibrium

Independent Equations on categories of 2D problem.

(Choose point O on the line of action)

Fx= 0 Fy = 0 Mo = 0

Depend on x-axis selection

2F

1F

3F

Body in equilibrium

** However, sometimes you can not get those 3 unknowns in 2D problem.

Fx= 0 Fy = 0 Mo = 0

(for some point O)

Fx= 0 Fy = 0 Mo = 0

(for all point O)

Hey , why you cannot get anymore, you can take another points to take moment, and get new equation?

x

y

not independent

equations!

** Whatever 2D-equilibrium conditions you used, at most 3 unknowns (scalar) may be found.

Choose this direction as x axis

101

Fx = 0

Fy = 0

Mo = 0

1F

2F

3F

2) Concurrent at a point

O x

yFx = 0

Fy = 0

Mo = 0

1F3) Parallel

x

y

2F 3F

4) General

x

y1F

M

2F

3F

4F

Fx = 0

Fy = 0

Mz = 0

Depend on x-axis selection

concurrent atpoint O

103

Note on Solving Problems

1. If we have more unknowns than the number of independent equations, then we have a statically indeterminate situation. We cannot solve these problems using just statics.

3. The order in which we apply equations may affect the simplicity of the solution. For example, solving FX = O first allows us to find the first unknown quickly.

4. If the answer for an unknown comes out as negative number, then the sense (direction) of the unknown force is opposite to that assumed when starting the problem.

2. The point (axis) which we check for the moment, affects the simplicity of the solution. Choosing appropriate one is the key for fast problem solving.

x

N

If you get N<0, something

wrong!

104

AOBC?

3/48 The small crane is mounted on one side of the bed of a pickup truck. For the position =45, q determine the magnitude of the force supported by the pin at O and the oil pressure P against the 50-mm-diameter piston of the hydraulic cylinder BC.

Body in interest:

AOC?

Because We also want to find force P exerted by BC to the object AOC

105

a

C

B

O

0.34

0.36

q

0.11E

Fq

G

D

C

x

yFBD

WOx

Oy

C

B

Ignore mass of link CB

2 Force Member

// with CB

which dirction?

Direction // CB

=45qFind P ,O

120(9.8)[(0.785 0.34)cos45 ]oW

M

tanCH

EO

( )FD FG OB

ED DO

(0.34sin 0.11cos ) 0.36tan

0.11sin 0.34sin

56.2o

( cos )(0.36) N-mF

M F

0MO +To find C:

0W F

M M 5060 NF

2 2

50602.58 MPa

0.052 2

FP

d

Ans

FH

CE OB

EO

( )FD FG OB

EO

Can you find ?a

106

x

yFDB

W

Ox

Oy

=45qFind P ,O

cos 0

2820

x x

x

F F O

O N

sin 0

3030 N

y y

y

F W F O

O

To find reactions at O

2 2| | 4140 N ( )x yO O O

Ans

a

C

B

O0.34

q

0.11

F

F

56.2o

0.785A

107

3/46 It is desired that a person be able to begin closing the van hatch from the open position shown with a 40-N vertical force P. As a desired exercise, determine the necessary force in each of the two gas-pressurized struts AB. The mass center of the 40-kg door is 37.5 mm directly below point A. Treat the problem as two-dimensional.

A

B

O

P= 40N

30

600 mm

550 mm

175 mm

1125 mm

Unknowns: F, Ox, Oy

Equilibrium eqs:

Ox

Oy

W=mg2F

Why direction: CB?

108

30

600 mm

550 mm

175 mm

A

B

OG x

y

2 2 2 175 550 600 2(550)(600)cos )

16.787 o

0MO +(1.125) 2 (0.55sin ) (0.55cos(30 )) 0P F W

0F696.3176115.21010645000

N803989.802F ANS

1125 mm

P = 40

2F

W=(40)(9.81)

30o

30

W=mg

O

A

B

G

2F

Ox

Oy

2: easy but ? ?

P F WM M M

Find F

30

(0.55sin )W

M W 2

(2 )(0.55sin )F

M F (40)(1.125)P

M

110

x=?

2 cos ( )T bx

mg

2 cosT

: known

Find appropriate L to make Bodies in equilibrium.

: known q =

112

3/52 The rubber-tired tractor shown has a mass of 13.5 Mg with the center of mass at G and is used for pusshing or pulling heavy loads. Determine the load P which the tractor can pull at a constant speed of 5 km/h up the 15-percent grade if the driving force exerted by the ground on each of its four wheel is 80% of the normal force under the wheel. Also find the total normal reaction N under the rear pair of wheels at B.

Normal reaction at A and B are equal?

If not, which one should be larger?

113

PWNA

0.8 NA

NB

0.8NB

FBD

xy

6001200825

0.8 sin 0.8 0x A BF N W N P

cos 0y A BF N W N cos 1200 sin 825 1800 600 0A BM W W N P

124.7 kNBN

85.1 kNP Ans

W=mg=13.5*1000*9.81=132400 N

tan 0.15 8.531o

6.3 kNAN

114

Recommended Problem3/34 3/49 3/35 3/54 3/57

1 3 equilibrium system isolation and equilibrium conditions

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