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11/25/11 1

EquilibriumofParticles

Lecture2

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➔ Apar0cleissaidtobeinequilibriumwhentheresultantofalltheforcesac0ng

onitiszero.➔ Iftwoforcesaretheonlyforcesac0ngon

abodyinequilibrium,thentheforcesareequalandopposite.

15N

15N

A

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➔ Iftherearethreeforces,whendrawing

theforcepolygon,itwillcloseifthe

par0cleisinequilibrium.

F2

AF1

F3

F2

A

F1

F3

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➔ Ingeneral,ifthereareseveralforces,when

drawingtheforcepolygon,itwillcloseifthe

par0cleisinequilibrium.

F1 A

F2

F3

F4F1

A

F2

F3

F4

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➔ Theforcepolygonisagraphical

representa0onoftheequilibriumof

forcesac0ngonapar0cle.➔ Mathema0cally,forequilibrium:

∑== 0 F  R ( ) ( ) 0=+

∑j Ri R

Y  X 

Vectorform:

Scalarform: ∑ == 0 X  X 

F  R ∑ == 0Y Y 

F  R

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Isthispar0cleinequilibrium?

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➔ Space diagram represents the sketch of the

physical problem.

➔ Free body diagram shows the particle and all

the forces acting on it

Steps:

1. Isolate / detach the body from any

contact.

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2. Indicate all  the forces that act onthe particle.

These include active forces - tendto set the particle in motion e.g.from cables and weights and

reactive forces caused byconstraints or supports that preventmotion.

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1

 3. Label known forces with their magnitudes and directions.

Use letters to representmagnitudes and directions of unknown forces.

Assume direction of force whichmay be corrected later.

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2 - 11

Space Diagram: A sketch showing

the physical conditions of the

 problem.

 Free-Body Diagram: A sketch showing

only the forces on the selected particle.

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2 - 12

In a ship-unloading operation, a

3500-lb automobile is supported bya cable. A rope is tied to the cable

and pulled to center the automobile

over its intended position. What is

the tension in the rope?

SOLUTION:

•  Construct a free-body diagram for the

 particle at the junction of the rope and

cable.

•  Apply the conditions for equilibrium by

creating a closed polygon from theforces applied to the particle.

•  Apply trigonometric relations to

determine the unknown force

magnitudes.

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2 - 13

SOLUTION:

•  Construct a free-body diagram for the

 particle at A.

•  Apply the conditions for equilibrium.

•  Solve for the unknown force magnitudes.

°=

°=

° 58sin

lb3500

2sin120sin

 AC  AB T T 

lb3570= ABT 

lb144= AC T 

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 F F 

 F  BC 

 AC 

o

o AC = =

sin

cos. .............( )

75

753 73 1  

∑ Fy = 0 i.e. FBC sin 75o

- FAC cos 75o

- 1962 = 0

 F F 

 F  BC 

 AC 

 AC =

+

= +

1962 0 26

09662031 2 0 27 2

.

.. . ......( )  

From Equations (1) and (2), 3.73 FAC = 2031.2 + 0.27 FAC 

FAC = 587 N

From (1), FBC = 3.73 x 587 = 2190 N

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➔ Apar0cleisinequilibriumifthe resultantofallthe forces ac.ng on the par.cle is zero. Thecomponent Rx, Ry, Rz of the resultantaregivenbytheprevioussec0onarezero.

0== Fx Rx 0==∑ Fy Ry 0==∑ Fz  Rz 

Note: The above equa.ons represent the necessary and sufficient condi.ons

fortheequilibriumofapar.cleinspaceTheycanbeusedtosolveproblems

dealingwiththeequilibriumofapar.cleinvolvingnomorethan3unknowns

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A2-kgcylinderishungbymeansoftwocablesABandAC,whichareaWachedtothetopofaver0calwall.AhorizontalforcePperpendiculartothewallholdsthecylinderintheposi0onshown.Determinethemagnitude

ofPandthetensionineachcable.

PA

12m

8m10m

200kg 2m

C

B1.2m

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8m

10my

x

12m

B

C

A

k

W

i

 j

O

1.2m

 ABT 

 ABλ 

 AC T 

 AC λ 

2m

P

Free-bodyDiagram.PointAwaschosenasafreebody;thispoint issubjectedto

fourforces,threeofwhichareofunknownmagnitude.

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8m

10my

x

12m

B

C

A

k

W

i

 j

O

1.2m

 ABT 

 ABλ 

 AC 

T 

 AC λ 

2m

P

P = P i W = -mg j = 200(9.81)  j 

W = -1962 N j 

(1)

(2)

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In the case of TAB and TAC, it isnecessary first to determine the

components and magnitudes of the

vectors ABandAC.Deno0ngbyλAB

theunitvectoralongAB,wewrite

8m

10my

x

12m

B

C

A

k

W

i

 j

O

1.2m

 ABT 

 ABλ 

 AC T 

 AC λ 

2m

P

AB=-(1.2m)i+(1m) j+(8m)k

|AB|=12.862m

k  ji AB

 ABˆ6220.0ˆ7775.0ˆ09330.0

862.12

++−==λ 

k T  jT iT T T   AB AB AB AB AB ABˆ6220.0ˆ7775.0ˆ09330.0 ++−== λ 

(3)

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Deno0ngbyλACtheunitvectoralongAC,

wewriteinasimilarway

AC=-(1.2m)i+(1m) j+(1m)k

|AC|=14.193m

8m

10my

x

12m

B

C

A

k

W

i

 j

O

1.2m

 ABT 

 ABλ 

 AC T 

 AC λ 

2m

P

k  ji AC 

 AC ˆ7046.0ˆ7046.0ˆ08455.0

193.14

−+−==λ 

k T  jT iT T T   AC  AC  AC  AC  AC  AC ˆ7046.0ˆ7046.0ˆ08455.0 −+−== λ 

(4)

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EquilibriumCondi0on.SinceAisinequilibrium,wemusthave:ΣF=0

TAB+TAC+P+W=0

( ) ( )

( ) 0ˆ7046.06220.0

ˆ19627046.07775.0ˆ08455.009330.0

=−

+

−+++−−

k T T 

 jT T i P T T 

 AC  AB

 AC  AB AC  AB

008455.009330.0:0 =+−−=∑ P T T  F  AC  AB X 

19627046.07775.0:0 =++=∑ AC  ABY T T  F 

07046.06220.0:0 =−+=∑ AC  AB Z T T  F 

 N T  N T  N  P  AC  AB

12381402235 ===Solvingsimultaneously,

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C(-2, 0, -2)m

D(-3, 0, 3)m

B (4, 0, 2)m

A (0, -4, 0)m

100 kg

x

y

z

The 100 kg cylinder is

suspended f rom theceiling by cables attached

at points B, C, and D.

What are the tensions in

cables AB, AC, and AD?

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DrawtheFree-BodyDiagrameisolatepartofthecablesystemnear

pointA(consideredasapar0cle)andcompletethefree-bodydiagramby

showingtheforcesexertedbythetensionsinthecables.Themagnitudes

ofthevectorsTAB,TAC,andTADarethetensionsincableAB,AC,andAD,

respec0vely.

100 kg

x

y

z A 

B

C

D

-(100)(9.81) N j

A 

TAD

TAC

TAB

0981 =−++=

∑jT T T  F   AD AC  AB

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➔ SirEli

➔ Room:MH1

➔ Consulta0onHours:

➛ TTh:2:3PM–6PM

➛ F:1–6PM

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( ) ( ) ( ) ( )mk  jik  z  z  j y yi x xr   A B A B A B A B ˆ2ˆ4ˆ4ˆˆˆ ++=−

+−

+−

=

C(-2, 0, -2)m

D(-3, 0, 3)mB (4, 0, 2)m

A (0, -4, 0)m

100 kg

x

y

z

k  jir 

r e

 A B

 A B

 A Bˆ333.0ˆ667.0ˆ667.0 ++==

k  jiT eT T  AB A B AB AB

ˆ333.0ˆ667.0ˆ667.0 ++==

k  jiT eT T   AC  AC  AC  AC ˆ408.0ˆ816.0ˆ408.0 −+−==

k  jiT eT T   AD A D AD ADˆ514.0ˆ686.0ˆ514.0 −+−==

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-(100)(9.81) N j

A 

TAD

TAC

TAB

0981 =−++=∑ jT T T  F   AD AC  AB

( ) ( ) jT T T iT T T 

 AD AC  AB AD AC  AB

ˆ981686.0816.0667.0ˆ514.0408.0667.0 −+++−−=

( ) 0ˆ514.0408.0333.0 =+−+ k T T T  AD AC  AB

0514.0408.0667.0 =−−=∑ AD AC  AB xT T T  F 

981686.0816.0667.0 =++=∑ AD AC  AB y T T T  F 

0514.0408.0333.0 =+−=∑ AD AC  AB z T T T  F 

TAB=519N TAC=636N TAD=168N