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36 .A Text BooTe of E n g l n ~ e r ; n g MechaniC8SolationThe free body diagram of one of the halves of the cylinder withminimum value of P is shown inFig. 3'8. - .PWe know that t.he centre of gravity of a semicircle is at a distance of 4r/3:t from it s base measured ,tiling the radiug. Taking moments about A, and (,quating th e same. AW 41'Pxr+TX 3n w

p '2= Px2r Fig. 382WPxr - Xl'3n2W()r p ADS.= 3n

Esample 3'4. .A t J , n i f o r m ~ e l oj 600 mm diameter, t I ) ~ i g h i n , .5 IcN ~ t 8 against a rigid rectangular block of 150 mm heigld 08 81010_in Ftg. 3'9.

Fig. 39Fin.d the lead puU. through the centre of the wheel, required justto tu m 1M wheel over the corner A oj t/ae bloclc. Also find the reactionofth" block.Tak" all the s u r f a c e ~ to be 81/'1.00'11,.

(Jadavpar Unit'ersity, 1985)Solution. Given: Diallleter' of whee l = 600 rum ; Weight orwheel = 5 kN ; of the hlll(:k = 150 mm .

PJe8116 refer to Art. 6.

:17. M o m M t ~ nnd ' ' 'fir ApplicationsLeast pldl required ju .1 to turn the wheel over the corner

Le t P = Least pull requirl'l\just to turn thewheel.A little consideration will show ISOmmtha.t for t ho lea.st pull, it must be.applied normal to AO. The system.of forces is shown in Fig . 3'10.From the geometry of the figure,we find that

R "AIIsJ...

$IrNFig. 310

sin 8 = 150 300 =:s 0'5 or 8 = 30.and AB = V(3OO)2 - (150)1 = 260 mm

Now taking moments about A and equating th e same,P x300 = 5x260 = 1300

P = 1300/300 = 4'33 kN Ans.Reaction of the block

Le t R = Reaction of th e block.Resolving th e forces horizontally and equating th e same,R cos 30 = P sin 30

R = P sin 30 P ta n 30 = 4'33 X 0'577 kNcos 30= 2'5 kN Ans."3,9. Applications of MomentsThough the moments have a. number of applications, in th eneld of Engineering Mechanics, ye t t1;e following are important fromth e subject point of view:

1. Position of a result ant force. 2. L e v e ~ . .3'10. Posit ion of Resultant Force by Moments

It is also known as analytical method for t he resultant force."The position of 1\ resultant force may be found out by moments asdiscussed below :1. First of all, find ou t t he magnitude an d direction of theresultant force by the method of resolution as discussed in

Art. 2'14.2. Now equate th e moment of the resultant force with thealgebraic sum of moments of the given, system of forces.about any point . This ma.y also be found out by equatingth e sum of clockwise moments a.nd that of th e anticlockwise moments about th e point, through which th e resultantforce

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46A Text Bock of E7Iginctrillg Muha1,iC6Esample 39. Fig . 318 thows a crank ever ABC with a tension.,-ifllJ..T. The lever weighB 0'2 Nlmm.

.,.

A.y 200mm

lOON Fig. 318

Determine the tension developed in the spring when a load of 100.N '-' applied at A . (Bombay University , 1985)

SoludoD . Given W e i ~ h t of lever = 0'2 N/mm P = 100 N ;a = 200 mm ; b = 100 mm ;

Weight of llever AB = 02 X 200 =4 0 N acting at 100 mmfrom A or B. Now ta king mom ents about th e hinge B and equatingth e same. .T x l OO = (100 x 200) + (40 X 100) = 24 000 Nmm

T = 24 000/100 = 240 N ADS.Esample 3'10. Find the tension requireJ i t ~ the operating wire,to raise the 8ignal thrOttgh the system of levers as 8hown in Fig. 3'19.Al l dimensions are in mm . (Calcutta U nivers i,y, 1986)SolutioD . Given: Wei ght of signal arm, W = 150 N Le t T = Tension required in operating wire, and

P =Tension in wire AB. First of all , taking moments about the -fulcrum (PI) of the signal and equating th e :same, Px75 = 150x175

= 26250 Nmm P = 26 250/75 = 350 N

Now t aking moments about th e fulcrum ,(F s) of the operating wire and equating the same, T x, 75 = 350x 150 = 52500 N-mm

T = 52 500/75 = 700 N ADS. Fig. 3)9IT ho we ight of lever BC will have no moment abou t t he hing/! UTherefore i ta weight has been ignored.

Moments and tleir ApplicationsEsample 311. The lever ABC of a component of a mathine ishinged at B, and is subjected to a system of coplaner forces as shovmin Fig . 3'20.

'200 kgf

300kgf

Fig. 320Neglecting friction, find the magnitude of the force (P) to keep thelever in equilibrium. Also determine the magnitude a"d direction ofthe reaction at B. (Oxford Univer8ity)Sol.tioD. Given: Vertical force at 0 = 200 kgf ; Horizontalforce at C = 300 kgf

M agnitt.Ule of the force (P)Taking moments about th e hinge B and equating the same, P X 10 sin 60 _ 200 X 12 cos 30+300 X 12 cos 60 Px 10 x 0'866 = : 200 x 12 x 0'866+300 x 12xO'5

8'66 P = 2078 +1800 = 3878: . P = 3878/8'66 = 447'8 kgf Alas .

Magnitude of the reaction at BResolving the forces horizontally,

'f.H = 300+P COB 20 =- 300 +447'8 x0'9397 kgf= 720'8 kgfand now resolving th e forces vertically,

'f.V = 200 - P sin 20 = 200 - 447'8 x 0'3420 kg{= 4685 kgfMagnitude of the reaction at B,

R = V ( ~ H ) t ' f . V ) 2 = v't7208)1 + (4685) kg{=722'3 kg{ AD S .

D ~ T e c t i o " of the reaction at BLe t 8 -= Angle, which the reaction at B makes wit.h thehori zontal.We know that l:V 4685 = 0'0650 Or 0=3 45' ADS.ta n 0 = ""fJj 720'8

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III{,o II(I}'I N

I'WIJ be,lIlIS SlIppulll:d ".., ..,hown ill lig I ICi'! and sl'p'lJall'd hy a mlkr al C. IDlnlly CUllYIS shown in the rlgun!. Iklermine the reactions at A.B and C.

tll(' upper hcam E B. 1"1)1 ): F" = 0

lOOONI

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1m Ie l 5 m 2m JD

()

11} ,8)

1. 68 Module - 1 1.69

5 kN/m-----Fig.1.l47h g. 1.l45 4k N 6k N 2 kN

6kN/m I . IB4kN/m Rn

Fig.1.l48Solution.B Consider the freebody dia gram of beam shown in fig. 1.148.12kN

9kN For I Fy = 0 R - 4 - 6 - 2 + R = 0A B\ RA + RB=12 kN1111 For I M = 0, tak ing' moments abou t A,

IS__ R1.m-\_\ 4 x 1 + 40 + 6 x 5 + 2 x 5.33 - x 7 =0RB = 12.09 kN

til '.. \ ,I RA + 12.09 = 12RA = - 0.09 kNRA =0.09 kN downward

ample 1.48 A beam 6 m long is loaded as shown in fig 1.1 49. Calcula te the reactions a t A and B.

Solution. 20 kNConsider the freebody diagram of bea m shown in fig. 1.150.

2mFor I Fy = 0 10kN1mI Rj\" 10 .1() t RI I = () A t I B/\ 2m J) 2 m C 2mI{ ' I ' I{ 'm .. { ). II .

1 141'1

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'1 )I

) 1 ./ 4 + 4 - 2+2 . /2- -= - 2.01 1-\

wilh triangular or trapezoidal cross section. Theyiglll ... lor slight adjustments in the position of a body ,ill!1I1 compared to the weight lifted . Hence generally

w

1-- P

o

wh g. 1.249

The problems on wedges are generally the problems

J.lIR IIRI

Fig.l.2S0

It. !lin (t.. LJ ()r'\II 'cqulihl illll\ oCtile wedge

PR l : ~ l l h i l l g tile forces horizontally,~ l l R , + 112 R2 cos CJ.. + R2 sin CJ. - P =0 JA:#3Resolving the forces vertically, k

Fig. 1.251R3 + 112R2 sin a - R2cos CJ. =O.Example 1.102

Find the horizontal force P on the 10 wedge shown in fig. 1.252 to raise the 1500 N load.The coefficient of fr iction is 0.3 at all contact surfaces.

1S00NO.lR 1(1500 N----,

p

Fig .l .252

Solution.Consider the equilibrium ofthe load of 1500 N.

Resolving the forces horizontallyRl - 0.3 R2 cos 10 - sin 10 =0

Rl =R2 (0.3 cos 10 + sin 10 )=0.47 R2 -------(i)

Resolving the forces verticallyR2 cos 10 - 0.3 R2 sin 10 - 1500 - 0.3 Rl =0

cos 10 - 0.3 R2 sin 10 - 0.3 (0.47 R2 ) = 1500

Fig .l. 2S3

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