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C

HAPT

Moment of Analytic Statics 1

1.1 Definition-I

Moment of a Force

Let Fbe a force and pbe the⊥distance of its line of action from the fixed pointO, then moment of FaboutO= (F)(p).

Figure 1.1

Note that:

The moment vanishes if either F =0 or p =0 i.e. if the point about which the 1

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Sign of Moment

The moment of a forceFabout a pointOis said to be+veor−veaccording as the force rotates the body in ananticlockwise orclockwisedirection respectively.

1.2 Definition-II

Moment of a Force

Let Abe any point on the line of action ofF.

Figure 1.2

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AO =γand∠OAN =θthen the moment ofFaboutO

= (F)(p)

= (F)(ON)

= (F)(γsinθ)

= (γ)(Fsinθ)

=γFsinθ

= AO(Resolved part of F perpendicular to AO) (MF)O =γ×F.

Varignon’s Theorem of Moments

The algebraic sum of moments of two forces about any point in their plane is equal to the moment of their resultant about the same point, i.e

(MP)0+ (MQ)0= (MR)0.

Note. Resultant of TwokForcesThe resultant of two like parallel forces P and Q acting

Figure 1.3

at A&B is equivalent to a force P+Q acting in the same direction at C such that AC×P =CB×Q

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Figure 1.4

by two coplanar forcesPandQi.e. contained in the same plane. Such forces are said to be concurrent.

Letγbe the position vector of Arelative toO. Then we may write γ×(P+Q) = γ×P+γ×Q

γ×R=γ×P+γ×Q (MR)O = (MP)O + (MQ)O

Note. Coplanar f orces Forces all of whose lines of action lie in one and the same plane.

Case II: When the Forces arek

LetPandQbe two like kforces acting at AandB. LetRbe the resultant of Pand Qacting atC. Then

AC×P=CB×Q. (1.2.1)

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Figure 1.5

The sum of the moments ofPandQaboutO

=OA×P+OB×Q

= (OC+CA)×P+ (OC+CB)×Q

=OC×P+CA×P+OC×Q+CB×Q

=OC×(P+Q) +CA×P+CB×Q

=OC×R−AC×P+CB×Q (∵ CA =−AC) (1.2.2) Now

|AC×P|= ACPsinθ

|CB×Q|=CBQsinθ From equation (1.2.1) we have

|AC×P| =|CB×Q|

=⇒ −AC×P+CB×Q =0 (1.2.3)

Now equation (1.2.2) becomes

=OC×R+0=OC×R

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= (MR)O Case III: WhenPandQare unlikekforces.

Figure 1.6

(MP)O+ (MQ)O =OA×P+OB×Q

= (OC+CA)×P+ (OC+CB)×Q

=OC×(P+Q) +CA×P+CB×Q Put |AC×P| =|CB×Q|

∴−AC×P+CB×Q=0

=OC×R−AC×P+CB×Q

=OC×R Hence (MP)O+ (MQ)O = (MR)O

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1.3 Generalized form of Varignon’s Theorem

The algebraic sum of moments of any number of coplanar forces about any point in their plane is equal to the moment of their resultant about the same point.

Example 1.3.1. A force F = Fxiˆ+Fyj acts at a point Aˆ (x1,y1). Find the moment of F about a second point B(x2,y2).

Solution.

Figure 1.7

γ= BA

= (x1x2)iˆ+ (y1y2)jˆ

M=γ×F

=

(x1−x2)iˆ+ (y1−y2)jˆ

×

Fxiˆ+Fy

=

(x1−x2)Fy−(y1−y2)Fx

kˆ (∵iˆ×ˆj=k andˆ jˆ×iˆ=−kˆ) M= (x1−x2)Fy−(y1−y2)Fx.

Example 1.3.2. A force F =Fxiˆ+Fyj acts at a point of coordinates x and y. Find theˆ ⊥ distance from the line of action of F to the origin O of the system of coordinates.

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Fy(x−0)−Fx(y−0) = M (1.3.1) Also,

M =|F|OL OL = M

|F|

= Fyx−Fxy qFx2+Fy2

Example 1.3.3. A10kg force acts at an ∠of60oon the corner C of a12 by1–cm plate ABCD. Determine the moment of force about A.

Figure 1.8

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Solution. Resolving the given force at C acting along AD and⊥of AD at A we get F =10cos60oiˆ+10sin60o

=10×1

2iˆ+10×

√3 2 jˆ

=5ˆi+5√ 3 ˆj

=Fxiˆ+Fy

Now (MF)A = (x1−x2)Fy+ (y1−y2)Fx

=−(1−0)5+ (12−0)5√ 3

=−5×1+5√ 3×12

=5(12√ 3−1)

Example 1.3.4. The point(5, 3)is on the line of action of the force P making an∠tan1(34) with the horizontal. Find the moment of force about the origin of coordinates.

Figure 1.9

Solution. Resolving the given forces at A acting along AB and AC we get F =P cosθiˆ+P sinθjˆ

(MF)O = (5−0)P sinθ−(3−0)P cosθ

=5P sinθ−3P cosθ

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5 5

=3P−12P 5

= 3 5P

The moment of the force about the origin of coordinates(MF)O = 35P

Example 1.3.5. Three forces P, Q and R act along the sides BC, CA, AB of a triangle ABC. Show that if their resultant passes through the ortho-centre (Intersection of altitudes), then

PsecA+QsecB+RsecC=0

Figure 1.10

Solution. Taking moment of force about O we have

POD+QOE+ROF =0 (1.3.2)

Now, in4ODC

OD

OC =cosB (1.3.3)

in4OEC

OE

OC =cosA (1.3.4)

from equation (1.3.3) and (1.3.4) we have

=⇒ OE

OD = cosA cosB

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OE=ODcosA

cosB (1.3.5)

Again in4ODB

OD

OB =cosC (1.3.6)

in4OFB

OF

OB =cosA (1.3.7)

from equation (1.3.6) and (1.3.7) we have OF

OD = cosA cosC OF =ODcosA

cosC (1.3.8)

From equations(1.3.3),(1.3.5)and(1.3.8)we get POD+QODcosA

cosB +RODcosA cosC =0 P+QcosA

cosB +RcosA

cosC =0 (∵ OD6=0)

=⇒ P

cosA+ Q

cosB + R cosC =0

=⇒PsecA+QsecB+RsecC =0

Example 1.3.6. Three forces P, Q and R act along the sides BC, CA and AB of a triangle ABC. Show that if their resultant passes through the incentre (Intersection of the internal bisectors of the angles of the triangle) and the circumcentre( Intersection of the bisectors of the sides of the triangle), then

P

cosB−cosC = Q

cosC−cosA = R cosA−cosB

Solution. Let O be the incentre of4ABC. Let OD, OE, OF be the⊥to the sides BC, CA and AB respectively. If r is the inradius then

OD=OE=OF =r Taking moments about O we have

P·OD+Q·OE+R·OF=0 (1.3.9)

⇒P·r+Q·r+R·r=0 (1.3.10)

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Figure 1.11

⇒ P+Q+R=0 (∵ r 6=0) (1.3.11)

⇒ R=−(P+Q) (1.3.12)

If O be the circumcentre

Figure 1.12

∴ ∠BOC =2A

∠AOC =2B

∠AOB=2C

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Taking moment of force about O

POD+QOE+ROF =0 OD

R1

=cosA OE

R1 =cosB OF

R1 =cosC PR1cosA+QR1cosB+RR1cosC =O

PcosA+QcosB+RcosC=O (1.3.13)

from equation (1.3.12) we have

PcosA+QcosB−(P+Q)cosC=0 PcosA+QcosB−PcosC−QcosC =0 Q(cosB−cosC)−P(cosC−cosA) = 0

Q

cosC−cosA = P

cosB−cosC. (1.3.14)

Also from equation (1.3.12)

P =−(Q+R) (1.3.15)

−(Q+R)cosA+QcosB+RcosC=0 Q(cosB−cosA) +R(cosC−cosA) =0

Q

cosC−cosA = R

cosA−cosB. (1.3.16)

From equation (1.3.14),(1.3.16) we get P

cosB−cosC = Q

cosC−cosA = R cosA−cosB.

Example 1.3.7. Forces of magnitudes1, 3, 2, 5act along the sides of a square taken in order.

Find the magnitude and line of action of their resultant.

Solution. Resolving the forces along AB and AD

Rx =R cosθ = (12) = −1 (1.3.17)

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Figure 1.13

Ry =R sinθ = (3−5) =−2 (1.3.18) Squaring and adding (1.3.17),(1.3.18) we get

R2 =1+4=5 =⇒ R =√ 5 Let the resultant meet AB in E.

Taking moment of the system at the forces about E we get

−3BE+2BC+5AE =0

−3(AE−AB) +2AB+5AE=0 (∵ ABCD is square ∴ AB= BC) 2AE+5AB=0

AE= −5 2 AB

|AE|= 5 2|AB|.

Example 1.3.8. Three forces P, 2P and 3P act along the sides AB,BC and CA of an equilateral triangle ABC. Find the magnitude and direction of their resultant and find also the point in which its line of action meet the side BC.

Solution. Let R be the resultant that meets BC in D. Resolving the forces along BX

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Figure 1.14

R cosθ =2P−3P cos60o−P cos60o

=2P−3P 2 − P

2

= 4P−4P

2 =0

=⇒ R cosθ =0 (1.3.19)

Resolving along BY

R sinθ =3P sin60o−P sin60o

=3P

√3 2 − P

√3 2

(1.3.20)

R sinθ =√

3P (1.3.21)

Squaring and adding (1.3.19) and (1.3.21) we get

R2cos2θ+R2sin2θ=3P2 =⇒ R=√ 3P.

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R sinθ R cosθ =

√3P 0 =

tanθ=tan90o =⇒ θ=90o Taking moment about B

2P×0+3P×BE+P×0=R×BL 3PBC sin60o =√

3PBD sin90o (∵θ =90o) BD= 3

2BC

Example 1.3.9. Forces P, 3P, 2P, 5P act along the sides AB,BC,CD and DA of a square ABCD. Find the magnitude and direction of their resultant and prove that if it meets AD produced at a point E such that AE : ED=5 : 4.

Figure 1.15

Solution.

R cosθ =−5P+3P =−2P (1.3.22)

R sinθ = P−2P =−P (1.3.23)

Squaring and adding (1.3.22) and (1.3.23) we get

R2cos2θ+R2sin2θ=5P2 =⇒ R=√ 5P.

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From equation (1.3.22)

R cosθ=−2P

=⇒√

5Pcosθ=−2P

=⇒cosθ= −2

√5 From equation (1.3.23)

R sinθ =−P

=⇒sinθ = −P

√5P

=⇒sinθ = −1

√5

=⇒sinθand cosθboth are negative ∴θlies in the third quadrant.

θ =π+tan1(1 2) Let a be the side of the square. Taking moments about A

RAE sinθ =−2Pa−3Pa=−5Pa AE= −5Pa

R sinθ AE= −5Pa

−P AE=5a

AE=5(AE−DE) AE=5AE−5DE

=⇒5DE=4AE

∴ AE: ED =5 : 4

Example 1.3.10. Forces3, 4, 5, 6kg act along the sides of a square taken in order. Find the magnitude, direction and line of action of resultant the said of square being a.

Solution.

R cosθ =5−3 =2 (1.3.24)

R sinθ =6−4=2 (1.3.25)

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Figure 1.16

Squaring and adding (1.3.24) and (1.3.25) we get

R2cos2θ+R2sin2θ =4+4=⇒ R=2√ 2kg tanθ =1 =⇒θ =45o

From equation (1.3.24)

R cosθ =2 cosθ = 2

2√ 2 From equation (1.3.25)

R sinθ =2 sinθ = 2

2√ 2

= √1 2

=⇒sinθand cosθboth are positive ∴θ lies in first quadrant. Taking moments about C R.CK sinθ =6×CD+3×CB

2√

2CK sin45o =6a+3a 2√

2CK

√2 =9a CK= 9a 2

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or CD+DK= 9a 2 a+DK= 9a

2 DK= 9a

2 −a

=⇒ DK= 7a 2 .

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1.4 Couple

Figure 1.17

Two forces Fand−Fhaving the same magnitude, parallel lines of action and opposite sense are said to form couple.

Moment of the Couple

Let Fand−Fform couple. LetObe the center of moment.

Figure 1.18

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Now

Moment of the couple =γA×F+γB×(−F)

= (γAγB)×F

=γ×F wereγ= (γAγB) M=|γ×F|

=γF sinθ

= Fγsinθ

= Fd

wheredis the⊥distance between lines of action ofFand −F(called arm of the couple).

Notation

CoupleFand −Fwith the⊥distance between their lines of action will be denoted by(F,d).

Resultant of a Couple

Theorem 1.4.1. The resultant of any number of couples acting in the same plane on a rigid body is a couple.

Note. If the resultant of the forces is equal to zero then system reduces to a couple R= RXi+RYj

RX =0 RY =0

=⇒forms a couple.

Example 1.4.1. ABCD is a rectangle with AB = 4cm, BC = 3cm. Forces 2, 7, 6, 10 and 5 kg weight act along AB,BC,CD,AD, and AC respectively. Show that they are equivalent to a couple whose moment is equal to46kg.cm.

Solution. Let∠BAC=θ then sinθ = 35 and cosθ = 45.

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Figure 1.19

Resolving the forces along AB and5AD cosθwe get RX =2−6+5cosθ

=−4+5×4 5 =0 RY =7−10+5sinθ

=−3+5×3 5 =0

=⇒RX =0and RY =0. The system reduces to a couple.

Now taking the moment of the forces about A we get

M =7×4+6×3 =46kg.cm.

Example 1.4.2. Forces 1, 2, 3, 5,P,Q act along AB,BC,CD,DA,AC and BD respectively and ABCD is a square of side a. Find the values of P and Q for the system to reduce to a couple. Find also the moment of the couple.

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Figure 1.20

Solution. Resolving the forces along AB and AD we get

RX =13+P cos45o−Q cos45o

=−2+√P

2 −√Q 2

=−2+ (P−Q)√1 2

RY =2−5+P sin45o+Q sin45o

=−3+√P

2 +√Q 2

=−3+ (P+Q)√1 2. If the system reduces to a couple =⇒ RX =0and RY =0

=⇒ −2+ (P−Q)√1 2 =0

=⇒(P−Q) = 2√ 2

−3+ (P+Q)√1 2 =0

=⇒(P+Q) = 3√ 2

=⇒ P= √5 2 and Q= √1

2

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2

The moment of couple

M=1a 2+2a

2 +3a 2+5a

2 = 11a 2 .

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Example 1.4.3. In a regular hexagon, forces P, 2P, 3P, 5P and 6P act along AB,BC,DC,EF and AF respectively. Show that a force can be determined to act along ED, so that the six forces are equivalent to a couple and find the moment of the couple.

Figure 1.21

Solution. Resolving the forces along AB and⊥to AB we get

RX = P+2P cos60o+3P cos60o+Q−5P cos60o−6P cos60o The systm redused to a couple if RX =0 and RY =0

=⇒ RX = P+2P 2 +3P

2 +Q−5P 2 −6P

2 =0

=−P−P+Q=0

=⇒ Q=2P.

Now let O be the center of hexagon. Let distance of O from each side=

3 2 a.

Moment of couple about O.

M =

√3 2 aP+

√3

2 a(2P)−

√3

2 a(3P)−

√3

2 a(2P) +

√3

2 a(5P)−

√3 2 a(6P)

=

√3

2 aP(1+2−3−2+5−6)

=⇒M =−3

√3 2 aP.

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on CD so as to lie on the side of CD away from AB. Forces P each act along the lines AB,BC,DA,DF,FE, and EC respectively and a force2P acts along CD. Show that the system reduces to a couple, the magnitude of whose moment is2`P(1−sinα).

Figure 1.22

Solution. The⊥distance between the sides AB and DC of the rhombus= AD sinα =

`sinα.

Similarly, the⊥distance between the sides AD and BC of the rhombus =`sinα.

The force2P along CD be replaced by two equal forces P and P along CD.

The force P along FE and a force P along CD form a couple of moment= (−P`). (clockwise)

Similarly, the force P along DF and force P along EC from a couple of moment=−P`. The force P along AB and force P along CD from a couple of moment= P` sinα.

(anticlockwise)

Then, the force P along DA and force P along BC from a couple of moment= P`sinα.

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The four couples are equivalent to a single couple of moment

=−P`−P`+P`sinα+P`sinα

=−2P`+2P`sinα+P`

=−2P`(1−sinα)

∴The magnitude of the moment of the couple =2P`(1−sinα).

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1.5 Equilibrium of a rigid body

Theorem 1.5.1. A coplanar force system acting on a rigid body is in equilibrium. Then necessary and sufficient conditions are

F =0 and M =0

where F is the resultant of the forces and M is the moment of the system about any arbitrary point in the plane.

Ist Equivalent Condition

A Coplanar forces system acting an a rigid blog is in equilibrium iff

X =0,

Y =0, M =0

where∑Xand ∑Yare the algebraic sums of the components of the forces along and⊥axes respectively.

IInd Equivalent Condition

A Coplanar forces system acting an a rigid blog is in equilibrium iff X =0, MA =0, MB =0

whereAandBbe any two arbitrary points on the plane andXis the sum of resolved parts along any line which is not⊥toAB.

IIIrd Equivalent Condition

A Coplanar force system acting an a rigid body is in equilibrium iff MA = MB =MC =0

where A,BandCare not collinear ( or non collinear).

Example 1.5.1. A ladder of weight W rests at an angleφto the horizontal with its ends resting on a smooth floor and against a smooth vertical wall the lower end being jointed by a string to the junction of the wall and the floor. Show that the tension of the string is

1

2Wcotφ. Also show that the tension of the string is 54Wcotφwhen a man whose weight is equal to that of ladder has ascended the ladder three quarters of its length.

Solution. Let AB be the ladder and C the junction of the wall of the floor

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Figure 1.23

Case I:Resolving the forces along AC and BC we get

X=O=⇒R2−T =0=⇒R2 =T Y =O =⇒ R1W =0=⇒R1 =W Taking moment about A we get MA =0

W(AF) +R2(BC) = 0 R2(BC) =W(AF)

R2(AB) sinφ=W(AE cosφ) R2(AB) sinφ=W(AB

2 )cosφ R2 = W

2 cotφ but R2= T

=⇒T = W 2 cotφ

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4

Let T1be the tension in this case.

Resolving the forces along AC and CB we get

Figure 1.24

X =O =⇒ R1 =W+W =2W R2 =T1

Taking moment about A we get

MA =R2(BC)−W(AF) +W(AM) = 0

=R2(AB sinφ)−W(AE cosφ)−W(ADcosφ) = 0

=R2(AB sinφ)−W(AB

2 ) cosφ−W(3

4ABcosφ) = 0 R2 = (W

2 +3W 4 )cosφ

sinφ

= 5w 4 cotφ but R2 =T1

=⇒T1 = 5w 4 cotφ

Example 1.5.2. A door of weight W height2a and width2b is hinged at the top and bottom.

If the reaction at the upper hinge has no vertical component. Find the components of reaction at both hinges. (Assume that the weight of the door acts at its centre)

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Solution. Let R be the horizontal component of the reaction at B and R2and R3 be the vertical and horizontal component of the reaction at A respectively. Resolving the system

Figure 1.25

along AD and⊥to AD we get

RX =R1+R3 =0 (1.5.1)

RY =R2−W =0 (1.5.2)

Taking moment about A we get MA =−W(AF) +R1(AB) =0

=−W(b) +R1(2a) =0 R1 = Wb

2a (1.5.3)

from equation (1.5.1) we get

R3=−R1=−Wb 2a

Example 1.5.3. A reactangular gate of l00kg,10m wide and6m high is placed on two symmetrically placed hinges4m apart. The lower hinge can exert only a horizontal reaction.

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from its outer end.

Figure 1.26

Solution. Let E and F be the hinges having R1,R2 and R3 as the components of the reactions along AD and⊥to AD respectively.

RX =−R1−R2 =0=⇒ R1+R2 =0 (1.5.4) RY = R3−50−100=0=⇒R3 =150 (1.5.5) Taking the moment about E we get

ME =−100(AH)−50(AK) +R2(EF) = 0

=−100(5)−50(2) +R2(4) =0

=−500−100+R2(4) =0 4R2 =600

R2 =150 from equation (1.5.4) we got

R1 =−R2 =−150

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Example 1.5.4. A uniform beam AB of weight W rests on a smooth floor at A and on a smooth inclined plane of inclinationα at B. The beam is maintained in a position inclined atθto the horizontal by a string fastened to B which passes over smooth pulling at the top of the plane and supports a weight P hanging freely. Show that the beam will rest in all positions if2P =Wsinα.

Figure 1.27

Solution. Resolving the forces along AC and⊥to AC we get R2cos(90oα)−P sin(90oα) =0 R2sinα =P cosα

(1.5.6)

R1+R2sin(90oα) +P cos(90oα)−W =0 R1+R2cosα+P sinα =W

(1.5.7)

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MB =−R1(AC) +W(EC) =0

=−R1(AB cosθ) +W(AE) =0

=−R1(AB cosθ) +W(ADcosθ) =0

=−R1(AB cosθ) +W(AB

2 )cosθ =0 R1 = W

2 (1.5.8)

From eqution (1.5.6) we get

R2= Pcosα sinα Now putting the value in R2in equation (1.5.8) we get

R1+P cosα

sinα cosα+Psinα =W R1+Pcos

2α+Psin2α

sinα =W

R1+ P

sinα =W using equation (1.5.8)

W 2 + P

sinα =W P

sinα = W 2 Wsinα =2P

Example 1.5.5. A beam of weight W is divided by its centre of gravity G into two portions AG and GB whose lengths are a and b respectively. The beam rests in a vertical plane on a smooth floor AC against a smooth wall CB. A string is attached to a hook at C and to the beam at a point D. If T be the tension of the string andαand βbe the inclinations of the beam and string respectively to the horizontal. Show that T = (a+Wa cosαb)sin(

αβ). Solution. Now resolving the forces horizontally, we have

R2Tcosβ=0=⇒R2=Tcosβ (1.5.9) Taking moments about A we get

−W(AE)−(Tsinβ)(AC) +R2(CB) = 0

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Figure 1.28

−W(AGcosα)−(Tsinβ)(ABcosα) +R2(ABsinα) =0

−W(acosα)−(Tsinβ)(a+b)cosα) +R2(a+b)sinα =0 R2(a+b)sinα =W(acosα) +Tsinβ(a+b)cosα

using equation(1.5.9) we have

R2= Tcosβ

Tcosβ(a+b)sinα =W(acosα)−Tsinβ(a+b)cosα T(a+b)[sinαcosβ−sinβcosα] =Wa cosα

T(a+b)sin(αβ) = Wa cosα

T = Wa cosα

(a+b)sin(αβ)

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1.6 Equation of The Line of Action of The Resultant

LetP1,P2,P3, . . . be forces acting at pointsA1(x1,y1),A2(x2,y2). . .

Figure 1.29

LetX1,Y1;X2,Y2;X3,Y3; . . . be their resolved parts along the axes. LetRbe their resultant acting at A(x,y), then

R cosθ= X= X1+X2+X3+. . . (1.6.1) R sinθ =Y =Y1+Y2+Y3+. . . (1.6.2) Squaring and adding equation (1.6.1),(1.6.2) we get

R2=X2+Y2 R=pX2+Y2 and

tanθ = Y X

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Moment ofX1aboutO =−Y1X1 Moment ofY1aboutO =X1Y1

∴ Moment of P1 about O = algebraic sums of the moments of X1,Y1 about O

=X1Y1−Y1X1

Similarly moment ofP2,P3aboutOare(X2Y2−Y2X2),(X3Y3−Y3X3). . . XY−YX =G

which is the equation of line of actions of the single resultant. When G is the algebraic sum of the moment of forces about origin.

Example 1.6.1. Forces3, 4and7kg act along the sides AB,CB and CA of an equilateral 4. Taking A as origin and CA as x−axis. Show that the resultant acts along the line 7x−5√

3y+4a=0a being the length of the side of the4.

Figure 1.30

Solution. Resolving the forces along CA and ⊥ to CA we get X =Rx =7−3cos60o+4cos60o

=7−3 2+4

2

= 15 2

Y =Ry=4sin60o+3cos60o

=4

√3 2 +3

√3 2

=7

√3 2

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M =−4AD =−4

√3

2 a (clockwise)

=−2

3a (∴ AD=

√3 2 a) The equation of the resultant

xY−yX =G 7√

3

2 x−15

2 y =−2√ 3a 7

2x−5

√3

2 y =−2a 7x−5√

3y+4a =0

Example 1.6.2. Find the magnitude of the resultant and line of action of a single force of8 units acting along the Y−axis (vertically upwards) and a couple in xy−plane of moment 24units.

Solution. Couple in xy plane of moment=24units=8×3.

The equation of the line of action of the resultant X=3

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C

HAPT

Catenary 1

Cable

A rope or cable is defined as material lines which can be bent arbitrarily without changing its length.

The Catenary

When a cable hangs freely, under the action of gravity between two points which are not in the same vertical line, then the curve formed by the cable is said to be a

catenary .

Figure 1.1 Span

1

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in the same horizontal line. Therefore

Span =AB =2x.

Sag

The vertical depth of the vertex below the span of the cable is called sag of the cable. Therefore

Sag =CD =y−c.

Parameter

The distanceCO=c, i.e. the depth of theX−axisbelow the vertex of the catenary is called parameter of the catenary.

1.1 Intrinsic Equation of a Catenary

If a uniform cable hangs freely under its own weight per unit length, then find the Intrinsic equation of the curve.

Proof. Consider a uniform cable ACBhanging freely under its own weight per unit length. LetCPbe a piece of cable of lengths.Therefore

Total weight ofCP=Ws,

which is acting vertically downwards. CPis in equilibrium under the following three forces:

(1) The horizontal tension H atC.

(2) The tensionTatP.

(3) Ws, weight ofCP, acting through theC.Gof the arcCP.

(SinceCPis in equilibrium , the lines of action ofWs,Tand Hmeet in one point).

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Figure 1.2

Resolving the forces horizontally and vertically we have

H =Tcosψ (1.1.1)

Ws =Tsinψ (1.1.2)

whereψis the angle which the tangent makes with theX−axis. Dividing (1.1.2) by (1.1.1) we have

tanψ= Ws

H (1.1.3)

It is convenient to write

tanψ = s

H/W or s =ctanψ, (1.1.4)

wherec = WH is the parameter of the catenary.

Eqn (1.1.4) is called the Intrinsic equation of the Catenary.

Cartesian Equation of the Catenary

If a uniform cable hangs freely under its own weight per until length, then find

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Solution. We know that Intrinsic equation of catenary is s =ctanψ.

Again tanψ = dy dx.

∴ s =cdy dx. Now putting p = dy

dx, we have

s =cp.

Differentiating above eqn with respect to x we have ds

dx =cdp dx,

∴cdp dx =

r

1+ (dy

dx)2 (∵ ds dx =

r

1+ (dy dx)2) cdp

dx = q

1+p2

∴ pdp

1+p2 = dx c . On Integrating we get

sinh1(p) = x

c +A, (1.1.5)

where A is the constant of integration.

At the lowest point C,x =0andψ=0. Therefore p = dy

dx =tan0=0.

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On putting these values in equation (1.1.5) we get 0 =0+A

=⇒ A =0.

sinh1(p) = x c p =sinh(x

c) dy

dx =sinh(x c) On Integrating we get

y =c coch(x c) +B At the lowest point C,x =0and y=c=⇒B =0.

∴ y=c coshx c. Which is called the cartesian equation of the catenary.

1.2 Relation between x, y, T, s and ψ

Parametric Equation of Catenary:

We know that, Intrinsic equation of Catenary is

s=c tanψ. (1.2.1)

Differentiating equation (1.2.1) with respect toxwe get, ds

dx =c sec2ψ dψ dx r

1+ (dy

dx)2 =c sec2ψdψ dx q

1+tan2ψ=c sec2ψdψ dx

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dx dx

dψ =c secψ dx =c secψdψ Integrating above equation, we get,

Z

dx = Z

c secψdψ

x =c log{secψ+tanψ}+C1

At the lowest point c,x =0, ψ=0

0=c log{sec0+tan0}+C1

=⇒ C1 =0.

Therefore, x =c log{secψ+tanψ}. Again differentiating equation (1.2.1) with respect to y, we get

ds

dy =c sec2ψdψ dy ds

dx dx

dy =c sec2ψdψ dy q

1+tan2ψ( 1

tanψ) = c sec2ψdψ dy secψ( 1

tanψ) =c sec2ψdψ dy dy

dψ =c secψtanψ dy =c secψtanψdψ.

Integrating we get Z

dy = Z

c secψtanψdψ y=c secψ+C2. At the lowest pointC,y =c,ψ=o

∴ c =c sec0+C2

=⇒ C2 =0

∴ y=c secψ.

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Hence, Parametric equations of Catenary are

x=c log(secψ+tanψ), y=c secψ.

1.3 Relation between T, y and s

(i)y2 =c2+s2

Since, y=c secψ. Squaring on both the sides, we get y2 =c2 sec2ψ

=c2(1+tan2ψ) y2 =c2+c2tan2ψ.

Buts=c tanψ

=⇒ y2 =c2+s2. (ii)T =Wy

Since Parameter of the Catenary is given by, c = H

W (1.3.1)

and

H =T cosψ. (1.3.2)

From equation (1.3.1) and (1.3.2),we have T cosψ=Wc

T = Wc

cosψ =⇒T =Wc secψ T =Wy (∵ y =c secψ).

Prove that the span of the cable of length2`and saghhis2` 1-2h3`22 :

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s =c sinh(x c) If 2a is the span.Then, x =a and s=`

∴ `=c sinh(a c) a

c =sinh1 (` c)

=log `

c + r

1+ (` c)2

a =c log `

c + r

1+`2 c2

But c = `2−h2 2h

= `2−h2 2h log

`

`2h2 2`

+ s

1+ `2 (`22hh2)2

= `2−h2 2h log

2h`

`2−h2 + s

1+ 4h

2`2 (`2−h2)

= `2−h2 2h log

2h`

`2−h2 + s

(`2−h2)2+4h2`2

`2−h2

= `2−h2 2h log

2h`

`2−h2 +

√`4+h4−2`2h2+4h2`2

`2−h2

= `2−h2 2h log

2h`

`2−h2 +

√`4+h4+2`2h2

`2−h2

= `2−h2 2h log

2h`

`2−h2 +

p(`2+h2)2

`2−h2

= `2−h2 2h log

2h`+`2+h2

`2−h2

= `2−h2 2h log

(h+`)2 (`+h)(`−h)

= `2−h2 2h log

h+`

`−h

= `2−h2 2h log

`+h

`−h

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= `2−h2

2h log 1+ ` 1−h` We know that, log

1+x 1−x

=2

x+ x3 3 + x

5

5 +. . .

=

1−h`22

2h

`2

2

h

` +1 3(h

`)3+1 5(h

`)5+. . .

=

1−h`22

2h

`2

2h

`

1+h

2

`2(1 3) +h

4

`4(1

5) +. . .

=`

1−h

2

`2 1+h

2

`2(1 3) + h

4

`4(1

5) +. . .

=`

1+h

2

`2(1 3)− h

2

`2

(Neglecting other higher powers) a=`

1−2

3 h2

`2

Thus, Span=2a

=2`

1−2 3

h2

`2

.

Example 1.3.1. A uniform chain of length`, which can just bear a tension of n times its weight is stretched between two points in the same horizontal line. Show that the least possible sag in the middle is`

n−qn214

Solution. Let A be(x,y). The tension at A,

T=n(weight of the chain) that is T =wy

wy=n(w`)

y =n`. (1.3.3)

For any point of the catenary, we have

y2 =c2+s2

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Figure 1.3

Therefore for the point A

n2`2 =c2+ (` 2

2

) (∵s = `

2 and y=n`) or c2 =`2

n21 4

c =` r

n21 4. Hence, the sag in the middle is

CD =y−c =n`−` r

n21 4

=`

n− r

n21 4

∴ CD=`

n− r

n21 4

.

Example 1.3.2. Ifα, βare the inclinations to the horizon of the tangents at the extremities of a catenary and`the length of the portion, show that the height of one extremities over the

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other is

`sin α+β2

cos

α−β 2

,the two extremities being on one side of the vertex of the catenary.

Solution. Consider the portion PQ of the catenary. let P= (x2,y2)and Q= (x1,y1) Tangents at P and Q make anglesα andβrespectively to the horizontal. Also arc PQ=

`(given). Suppose that arc CQ =s1and arc CP =s2

=⇒s2−s1 =`. (1.3.4)

Figure 1.4

Using the formula

s=c tanψ we have s1 =c tanβ and s2 =c tanα

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c tanα−c tanβ=`

or c= `

tanα−tanβ.

(1.3.5)

Vertical height of one extremity P above Q

=y2−y1

=c secα−c secβ

=c[secα− secβ]

= `

tanα−tanβ[ 1

cosα − 1 cosβ]

= `

sinα

cosαsinβcosβ[ 1

cosα − 1 cosβ]

=` cosβcosα sinαcosβ−cosαsinβ

=`2sin(α+2β)sin(α2β) sin(αβ)

=`2sin(α+2β)sin(α2β) sin(2α2β)

=`2sin(α+2β)sin(α2β) 2sin(α2β)cos(α2β)

=`sin(α+2β) cos(α2β).

Example 1.3.3. A given length2s of uniform chain has to be hung between two points at the same level and the tension has not to exceed the weight of a length b of the chain. Show that the greatest span is√

b2−s2log(bb+ss).

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Solution. Let A be(x,y)

T = Tension at A

=wy =wb(Given).

∴ y=b.

Figure 1.5

For any point of the catenary we have

y2 =c2+s2

=⇒ b2 =c2+s2 c =pb2−s2.

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= AB

=2x

=2c log(secφ+tanφ) (∵ x =Clog(secφ+tanφ))

=2c log(y c + s

c) (∵ y=c secφ and s=c tanφ)

=2c log(y+s c )

=2p

b2−s2 log

b+s

√b2−s2

(∵ y=b&c =pb2−s2)

=2p

b2−s2 log √

b+s√ b+s

√b+s√ b−s

=2p

b2−s2 log √

b+s

√b−s

. The greatest span

=pb2−s2 log

b+s b−s

.

Example 1.3.4. A heavy uniform string of length`, is suspended from a fixed point A and its other end is pulled horizontally by a force equal to the weight of a length a of the string.

Show that the horizontal and vertical distance between A and B are (a sinh1(`a)) and

√`2+a2−a respectively.

Solution. Now as

H =wa (Given). (1.3.6)

Also, we have

H =wc. (1.3.7)

Form equations (1.3.6) and (1.3.7) we get

wa=wc⇒c =a.

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Figure 1.6

Now from the relation at the point A we have,

y2A=c2+s2= a2+`2 (∵ c =a and s =`) yA=pa2+l2,

and yB =c but c= a

⇒yB = a.

=⇒ yA−yB =p`2+a2−a.

Again from s

c =sinh(x

c)at the point A, we have,

`

a =sinh(xA a ) or sin1(`

a) = (xA a )

⇒xA= a sinh1(` a). Therefore xA−xB = a sinh1(`

a)−0 (∵ xB =0) xA−xB = a sinh1(`

a).

y_A-y_B

X_A-X_B

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and the lowest point of rope is b m in below the point of suspension. Show that the horizontal component of tension is W2b(`2−b2), W being the weight of the rope per meter of its length.

Solution. Now as we know that

y2 =c2+s2. At the point A(x, y), we have s =`

and y=b+c.

Therefore, we have(b+c)2 =c2+`2 b2+c2+2bc =c2+`2

c = `2−b2 2b . Hence, H =Wc=W(`2−b2

2b ).

Example 1.3.6. Prove that when c is large compared with the span the catenary approximates to a parabola.

Solution. As we know that equation of catenary is

y =c cosh(x c) y =c[1+ 1

2!(x

c)2+ 1 4!(x

c)4+. . .] (∵cosh(x) = [1+x

2

2! +x

4

4! +x

6

6! +. . .]). Since x is small, c is large ∴ x

c is small we get

y =c+(x)2 2c y−c = (x)2

2c

=⇒ x2 =2c(y−c).

Which is a parabola whose latus rectum is 2c and vertex (0,c), the lowest point of the catenary.

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1.4 The Parabolic Cable

Let the cable carry a uniform load ofW0per unit length. ForCB,

The load =V =W0x

and dy

dx =tanψ= V H. Now, we have dy

dx = W0x H Z

dy = W0 H

Z xdx y= W0x

2

2H +c1.

Figure 1.7

If we choose as x−axisthe horizontal line passing throughC(0, 0)then,c1 =0.

Thereforey = W0x

2

2H .

This is a equation of parabola with axis vertical and vertex at origin.

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T = q

H2+ (W0x)2.

Example 1.4.1. If a cable is loaded with a uniform horizontal load of W and has a span a and sag h, show that H = Wa8h2 and that

T = Wa 2

1+ a

2

16h2 12

,

where H and T are the tensions at lowest and highest point of the cable,respectively.

Solution. We know that, the equation of parabolic cable is

y = Wx

2

2H .

Here y =h and x = a

2. Now, h = W(a2)2

2H h = Wa

2

8H H = Wa

2

8h .

Figure 1.8

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The tension in the parabolic cable is given by T =

q

H2+ (Wx)2 T =

r (Wa

2

8h )2+ (Wa 2)2 T =

rW2a4 64h2 +W

2a2 4 T = Wa

2 r

1+ a

2

16h2.

Example 1.4.2. A length` of a uniform chain has one end fixed at a heights h above a rough table and rests in a vertical plane so that a portion of it lies in a straight line on the table. Prove that if the chain is on the point of slipping, the length on the table is

`+µhq[(µ2+1)h2+2`hµ],

whereµis the coefficient of friction.

Solution. Resolving forces horizontally and vertically, we have

R=Wx (1.4.1)

µR=T0 =Wc. (1.4.2)

From (1.4.1) and (1.4.2), we have

µ(Wx) =Wc,

⇒c =µx. (1.4.3)

Using the formula y2 =c2+s2 from the point A we have (∵ y for A = c+h and s for A =`−x)

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Figure 1.9

y2 =c2+s2

(C+h)2 = (µx)2+ (`−x)2 (µx+h)2 = (µx)2+ (`−x)2

(µx)2+h2 = (µx)2+`2+x2−2`x+2hµx x2−2x(`+µh) + (`2µ2) =0

Therefore, x = 2(`+µhp4(`+µh)2−4(`2−h2) 2

x= (`+µh)± q

`2+µ2h2+2µh`−`2+h2 x= (`+µhqh2(µ2+1) +2µh`

Thus, x =`+µhq(µ2+1)h2+2µh`.

The positive sign before radical is neglected for otherwise x > `, which is not possible.

1.5 Tension in a Cable

We write, ds dx =

r

1+ (dy dx)2

= r

1+sinh2(x

c) (∵ s

c =tanψ= dy

dx =sinh(x c))

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Figure 1.10

ds

dx =cosh(x

c). (1.5.1)

Also, H =T cosψ

=⇒ T = Hsecψ.

For any curve dx

ds =cosψ

=⇒ ds

dx =secψ.

=⇒ T = H cosh(x

c) (from equation 1.5.1).

1.6 Maximum Tension in the Cable

To prove that the maximum tension in the cable of weightWper unit length, span 2aand length 2`is

Waq a

6(`−a)

, approximately.

Solution. We know that the maximum tension occurs at A and B. Using the formula

s=c sinh(x

c), at the point A

(60)

Figure 1.11

s =` and x =a We have,

` =c sinh(a c) Or, ` =c

a c + (a

c)31 3!+ (a

c)51

5! +. . .] (∵ sinh x=x+ x

3

3! + x

5

5! +. . .)

` =a+ a

3

6c2 (Neglecting the higher power terms)

`−a = a

3

6c2 c2 = a

3

6(`−a) c =

s a3 6(`−a).

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We know that, Tension in the cable is given by

T =Hc osh(x

c) = H cosh (a

c) (∵ x =a). If a

c is small or c is large. Then, Tmax =H =Wc (approximitely)

=W s

a3 6(`−a) Tmax =Wa

r a 6(`−a).

Example 1.6.1. A cable200m long hangs between two points at the same height. The sag is20m and the tension at either point of suspension is120kg Wt. Find the total weight of the cable.

Solution. Given that T =120kg Wt

Figure 1.12

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2s=200m Given s=100m.

y−c =20 y=20+c.

We know that, y2 =c2+s2 (20+c)2 =c2+ (100)2 400+c2+40c =c2+10000

40c =9600 c = 9600

40 =240 y =c+20

=240+20 y =260.

As, T =Wy

120= (W)(260) W = 120

260 = 6 13 Total weight=200× 6

13 =W(2s)

= 1200 13 .

Example 1.6.2. A uniform cable hangs across two smooth pegs at the same height, the ends hanging down vertically. If the free ends are each12m long and the tangent to the catenary at each peg makes an angle of60o with the horizontal, find the total length of the cable.

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Solution. At the point B, we have

y =h+c =12 and y=c secφ.

Now, y =c sec60o y =2c 12 =2c c =6.

As, s =c tanψ

=6tan60o s =6(√

3).

Figure 1.13

Total Length of the Cable =2s+2y

=2×6√

3+2×12

=12√ 3+24

=12(√

3+2).

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to two small rings which can slide on a fixed horizontal wire. Each of these rings are acted on by a horizontal force equal to`ω. Show that the distance apart of the rings is

2`log1+√ 2 . Solution.

H =ω` But H =ωc

ω`=ωc

⇒`=c We know that S=ctanψ

`=`tanψ tanψ=1

ψ=45o

Span = AB=2`log{secψ+tanψ} Span =2`log{√

2+1}.

Example 1.6.4. If T be the tension at any point P of a catenary and T0at the lowest point C, prove that

T2−T02=W2, W being the weight of the arc CP of the catenary.

Solution. We know that

T=wy

is

where,

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and T0 = The tension at the lowest point C

=wc.

Now, T2−T02 =w2y2−w2c2

=w2(y2−c2). But y2 =c2+S2

T2−T02 =w2(y2−c2)

=w2(c2+s2−c2)

=w2s2

=W2 (∵ W =ws).

Example 1.6.5. A uniform chain is hung up from two points at the same level and distant 2a apart. If z is the sag at the middle show that

z=c(cosh(a

c)−1). If z is small compared to a show that

2zc=a2 (nearly).

Solution. We know that equation of the catenary is

y =ccosh(x c).

Since the span2x =2a, x= a for the point A Given that z is the sag in the middle.

∴z = sag =DC =y−c

=ccosh(x c)−c

=c[cosh(x

c)−1] (∵ x =a at A)

=c[cosh(a

c)−1].

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Figure 1.14

Again z is small compared with a, therefore c must be very large. Now z=c[cosh(a

c)−1]

=c[1+ 1 22(a

c)2]1 4!(a

c)4+· · · −1

= a

2

2c

⇒2zc= a2 (nearly).

Example 1.6.6. A uniform chain of length`is to be suspended from two points A and B in the same horizontal line so that terminal tension is n times that at the lowest point. Show that span AB must be

√ `

n21log{n+pn2−1}. Solution. Let the coordinate of the A be(x,y). Then we have

span = AB=2clog(secψ+tanϕ) whereψis the inclination of the tangent at A to x–axis. Given that

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Figure 1.15

Tension at A=n( Tension at the lowest point c)

=nT0

⇒T =nT0=n(ωc) (∵ T0 =ωc)

ωy =nT0 =n(ωc) (∵ T =ωy)

⇒y =nc

⇒csecψ=nc (∵y =csecψ)

⇒n =secψ (1.6.1)

∴tanψ= q

sec2ψ−1

tanψ =pn2−1. (1.6.2)

Again,

the arc CA= half length of the chain s= `

2, we have, `

2 =ctanψ (∵ s=ctanψ)

= ` 2 tanψ

= `

2p

sec2ψ−1

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c = 2√

n2−1. (1.6.3)

Hence,

AB=2clog{secψ+tanψ}

= `

√n2−1log{n+pn2−1}.

Example 1.6.7. A telegraph wire is made of a given material and such a length L is stretched between two posts, distant d apart and of the same height as will produce the least possible tension at the posts. Show that

L= d

λsinhλ whereλis given by the equation

λtanλ=1.

Solution. Let T be the tension at either post. Then the catenary has x= d2 and s= L2. We know that the tension in the cable is given by

T =Hcosh(x c)

=Hcosh( d 2c)

=Wc cosh( d 2c) For maximum or minimum of T, we have

dT dc =0

=⇒ d dc

Wcosh( d 2c)

=0

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Figure 1.16

=⇒W

cosh( d

2c) +csinh( d 2c)(−d

2c2)

=0 or cosh( d

2c)− d

2csinh( d 2c) =0

or d

2ctanh( d

2c) =1.

Butλtanhλ=1 (Given)

d

2ctanh( d

2c) =λtanhλ

λ= d 2c.

⇒c= d 2λ. As we know, s=csinh(x

c). Here, s= L

2 x= d 2

and c= d

2λ Now,(L

2) = d

2λsinh( d 2c)

⇒ L= d

λsinhλ.

Example 1.6.8. If the length of the uniform chain suspended between points at the same

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span2d. Show that the equation to determine c is coth(d

c) = d c. Solution. We know that tension cable is given by

T = Hcosh(x c)

⇒T = Hcosh(d

c) =Wccosh(d c). For maximum or minimum of T, dT

dc =0

=⇒ dT

dc =W[csinh(d c)(−d

c2 ) +cosh(d c)] =0

=⇒=W[csinh(d c)(−d

c2 ) +cosh(d c)] =0

=⇒cosh(d

c) = (d

c)csinh(d c) cot h(d

c) = (d c).

References

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