3

HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade

Engineering Mechanics

STATICS

Equilibrium of a Particle

LECTURE 3


Condition of the Equilibrium of a ParticleA particle is in equilibrium provided

• it is at rest if originally at rest or

• has a constant velocity if originally in motion

To maintain equilibrium, it is necessary to satisfy Newton’s first law of motion, which requires that the resultant force acting on a particle to be equal to zero.

This condition may be stated mathematically asnF = 0

The above equation is necessary and sufficient


Condition of the Equilibrium of a ParticleAdditionally, if the particle is moving

⇒ Newton's second Law: ΣF = ma

But to satisfy equilibrium

ΣF = 0 ⇒ ma = 0 ⇒ a = 0

⇒ particle has constant velocity


The Free-Body Diagram

To apply the equation of equilibrium;All the known and unknown forces (nF ) must be considered

The best way to do that is through drawing the particle’s

FREE BODY DIAGRAM


The Free-Body Diagram

Procedure:

1. Draw outline shape of the particle to be isolated

2. Show all forces acting on the particle

3. Identify each force


Springs• With linear elastic springs,

deformation is linearly proportional to the applied force

• The elasticity of the spring is defined by means of the spring stiffness k

F = ks

s = l – lo

If s > 0F must “pull”

If s < 0 F must “push”


Cables and Pulleys

Remarks:

• Weights of Cables are not to be considered

• Cables support only Tension in the direction of the Cable

• For any θ the cable is subjected to a constant tension T throughout its length


Example 3-1


Coplanar Force System

If a particle is subjected to a system of forces that lie in the x-y plane

For the above vector equation to be satisfied then

These scalar equations must be satisfied


Scalar Notation

• In the above example, when drawing the FBD, we assumed the senseof the unknown F is to the right.

• But the equation resulted in F = -10 N, which indicates that F must have a sense to the left to hold the particle in equilibrium


Example 3-2

FBD

Equilibrium Equations

From the above Equilibrium Equations, then


Example 3-3

FBD



Example 3-3

FBD

FBDNewton’s 3rd Law

Principle of Action and Reaction



Problem 3-16/1

The cylinder D has a mass of 20 kg. If a force of F=100N is applied horizontally to the ring at A, determine the largest dimension d so that the force in cable AC is zero.


Problem 3-16/2Free-Body diagram of ring A:• Force from cable AC (FAC=0)• Force from cable AB (FAB)• Weight of cylinder D {W = 20(9.81) = 196.2N}• Force F = 100N

F = 100N x

y

W = 20(9.81) = 196.2N

FAB

θ⇒


Problem 3-16/3

)(

)(coscos

2196.2inθ0196.2inθ

0

1100θ0100θ

0:mEquilibriu of Equations

=⇒=−⇒

=↑+

=⇒=+−⇒

=⎯→⎯

∑

∑+

sFsF

F

FF

F

AB

AB

y

AB

AB

x

F = 100Nx

y

W = 20(9.81) = 196.2N

FAB

θ


Problem 3-16/4

NF

FF

sFF

AB

AB

AB

AB

AB

2220cos62.99

100(1)in Substitute

9962θ9621θ

100196.2

θθ

12

2196.2inθ1100θ

.

..tancossin

)()(

)()(cos

=°

=⇒

°=⇒=⇒

=⇒

=

=

F = 100Nx

y

W = 20(9.81) = 196.2N

FAB

θ


Problem 3-16/5

θ

mddd

4225199262

51θ22

d1.5θ

:geometry From9962θ

..).(tan

.)(tan

tan

.

=⇒−=⇒

−=⇒

+=

°=


Problem 3-22/1

The springs on the rope assembly are originally stretched 1 ft when θ = 0°. Determine the vertical force F that must be applied so that θ = 30°.


Problem 3-22/2Free-Body diagram of A:• Tension from cable AB (Fs)• Tension from cable AD (Fs)• Vertical Force F

Fs

x

F

Fs

30ο30ο⇒


Problem 3-22/3

lbFF

FsF

F

lbFkxF

ftxllx

ft

s

y

s

s

3390303392

0inθ2

0

33931130

3111312- stretched are springs the30θwhen

31230cos2

cosθ2BA

o

.))(sin.(

)(

.).(

..

.

=⇒=−°⇒

=−⇒

=↑+

==⇒

==−=⇒

=°=

=°

==

∑Fs

x

F

Fs

30ο30ο


Three Dimensional Force System

⎪⎩

⎪⎨

⎧

===

⇒

=++⇒

=

∑∑∑

∑∑∑

∑

000

0:components , , their into resolved are forces theIf

0:require wemequilibriu particleFor

z

y

x

zyx

FFF

FFF

F

kjikji

00

00

00

321

321

321

=++−⇒=

=−+⇒=

=+−⇒=

∑∑∑

zzzz

yyyy

xxxx

FFFF

FFFF

FFFF


Problem 3-59/1

If the maximum allowable tension in cables AB and AC is 500 lb, determine the maximum height z to which the 200-lb crate can be lifted. What horizontal force F must be applied? Take y = 8 ft.


Problem 3-59/2

y = 8ft⇒

Free-Body diagram of A:• Tension from cable AB (FAB = 500 lb)• Tension from cable AC (FAC = 500 lb)• Weight of the crate (W = 200 lb) • Force F

A

FAB = 500 lb

200 lb

F

FAC = 500 lb

x

z

y

5 ft

4 ft

8 ft

5 ft

4 ft

8 ft

(4-z) ft

x’y’

z’


Problem 3-59/3

lb489

4500489

4000489

2500

500489

4489

84895

489485

485

axes as x'-y'-z'origin, asA Consider

222

222

2222

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−+

−+

−+−

−+

−=⇒

=

−+

−+

−+−

−+

−=

−+=−+−+−=

−+−−=

k)(

)(j)(

i)(

F

uF

k)(

)(j)(

i)(

u

)()()()(

k)(jir

AB

ABAB

AB

AB

zz

zz

zz

zz

zzr

z

AB

FAC = 500 lb

FAB = 500 lb

200 lb

Fx

y

z

5 ft5 ft

4 ft

4 ft

8 ft

8 ft

(4-z) ft

x’y’

z’


Problem 3-59/4

{ }

{ }jF

kW

k)(

)(j)(

i)(

F

FF

k)(

)(j)(

i)(

F

AC

ACAB

AB

F

zz

zz

zz

zz

=

−=

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−+

−+

−+−

−+=

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−+

−+

−+−

−+

−=

lb200

lb489

4500489

4000489

2500

thuslsymmetrica are and

lb489

4500489

4000489

2500

222

222

FAC = 500 lb

FAB = 500 lb

200 lb

Fx

y

z

5 ft5 ft

4 ft

4 ft

8 ft

8 ft

(4-z) ft

x’y’

z’


Problem 3-59/5

( )

( ))(

)()(

)()(

1489

8000

0489

8000

0489

4000489

4000

0

0489

2500489

2500

0

2

2

22

22

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−+=⇒

=+⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−+

−⇒

=+−+

−−+

−⇒

=

=−+

+−+

−⇒

=

∑

∑

zF

Fz

Fzz

F

zz

F

y

x

y = 8ft

FAC = 500 lb

FAB = 500 lb

200 lb

Fx

y

z

5 ft5 ft

4 ft

4 ft

8 ft

8 ft

(4-z) ft

x’y’

z’


Problem 3-59/6

( )

( )

( ))()(

)(

)()(

)()(

)(

2489

41000200

0200489

41000

0200489

4500489

4500

0

1489

8000

2

2

22

2

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−+

−=⇒

=−⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−+

−⇒

=−−+

−+

−+

−⇒

=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−+=

∑

z

z

z

z

zz

zz

F

zF

z

y = 8ft

FAC = 500 lb

FAB = 500 lb

200 lb

Fx

y

z

5 ft5 ft

4 ft

4 ft

8 ft

8 ft

(4-z) ft

x’y’

z’


Problem 3-59/7

( )

( )

Fz

F

z

z

zF

16004

8z-4200

(1)by (2) Dividing

2489

41000200

1489

8000

2

2

=−⇒

=⇒

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−+

−=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−+=

)(

)()(

)(

y = 8ft

FAC = 500 lb

FAB = 500 lb

200 lb

Fx

y

z

5 ft5 ft

4 ft

4 ft

8 ft

8 ft

(4-z) ft

x’y’

z’


Problem 3-59/8

( )

( )

22

2

2

2

1600898000

160089

18000

489

18000

489

8000(1) From

16004

⎟⎠⎞

⎜⎝⎛+=⎟

⎠⎞

⎜⎝⎛⇒

⎟⎠⎞

⎜⎝⎛+

=⇒

−+=⇒

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−+=

=−

FF

F

Fz

F

zF

Fz)(

y = 8ft

FAC = 500 lb

FAB = 500 lb

200 lb

Fx

y

z

5 ft5 ft

4 ft

4 ft

8 ft

8 ft

(4-z) ft

x’y’

z’


Problem 3-59/9

ftzF

z

lbF

FF

FF

0729314

931831

160016004

83189

10146

8910146

8916008000

1600898000

7

27

2

22

22

..

.

.

.

=−=⇒

===−

=×

=⇒

=×⇒

=−

⇒

⎟⎠⎞

⎜⎝⎛+=⎟

⎠⎞

⎜⎝⎛

y = 8ft

FAC = 500 lb

FAB = 500 lb

200 lb

Fx

y

z

5 ft5 ft

4 ft

4 ft

8 ft

8 ft

(4-z) ft

x’y’

z’

3

Documents

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