0, i.e. forces act in the same direction. , i.e. forces act in opposite direction. The resultant of two forces is closer to the larger force.

Transcription

1 Sttics. Introduction. Sttics is tt brnc of mecnics wic dels wit te study of te system of forces in equilibrium. Mtter : Mtter is nyting wic cn be perceived by our senses of wic cn exert, or be cted on, by forces. Force : Force is nyting wic cnges, or tends to cnge, te stte of rest, or uniform motion, of body. To specify force completely four tings re necessry tey re mgnitude, direction, sense nd point of ppliction. Force is vector quntity.. rllelogrm lw of Forces. If two forces, cting t point, be represented in mgnitude nd direction by te two sides of prllelogrm drwn from one of its ngulr points, teir resultnt is represented bot in mgnitude nd direction of te prllelogrm drwn troug tt point. If nd represent te forces nd cting t point nd inclined to ec oter t n ngle. If is te resultnt of tese forces represented by te digonl of te prllelogrm nd mkes n ngle wit i.e. nd tn sin cos, ten cos Te ngle wic te resultnt mkes wit te direction of te force is given by se (i) : If = sin tn cos cos( / ) nd tn tn( / ) or / se (ii) : If se (iii) : If mx 90, i.e. forces re perpendiculr nd tn 0, i.e. forces ct in te sme direction D se (iv) : If min 80, i.e. forces ct in opposite direction Note : Te resultnt of two forces is closer to te lrger force. Te resultnt of two equl forces of mgnitude cting t n ngle is cos nd it bisects te ngle between te forces.

2 Sttics If te resultnt of two forces nd cting t n ngle mkes n ngle wit te direction of, ten sin sin nd cos cos If te resultnt of te forces nd cting t n ngle mkes n ngle wit te direction of te force, ten sin sin nd sin cos omponent of force in two directions : Te component of force in two directions mking ngles nd wit te line of ction of on nd opposite sides of it re. sin sin F nd F sin( ) sin( ). sin. sin sin( ) sin( ) - teorem : Te resultnt of two forces cting t point in directions nd represented in mgnitudes by. nd. respectively is represented by point in suc tt. µ. F ( µ) F, were is Importnt Tips Te forces,, ct long te sides,, of. Teir resultnt psses troug. () Incentre, if 0 (b) ircumcentre, if cos cos cos 0 (c) rtocentre, if sec sec sec 0 (d) entroid, if cosec cosec cosec 0 or b c Exmple: Forces M nd N cting t point mke n ngle 50. Teir resultnt cts t s mgnitude units nd is perpendiculr to M. Ten, in te sme unit, te mgnitudes of M nd N re [IT nci 99] (), 4 (b), (c), 4 (d) 4. 5 Solution: () e ve, M N MN cos 50 4 M N MN...(i) M sin50 nd, tn M N cos50 0 M N cos50 N M N 0 M...(ii) Solving (i) nd (ii), we get M nd N 4.

3 Exmple: forces is Sttics If te resultnt of two forces of mgnitude nd is perpendiculr to, ten te ngle between te ()/ (b) /4 (c) 4/5 (d) 5/6 [oorkee 997] Solution: () Let te ngle between te forces nd be. Since te resultnt of nd is perpendiculr to. Terefore, Exmple: Solution: (d) Exmple: 4 sin tn / cos 0 cos cos If te line of ction of te resultnt of two forces nd divides te ngle between tem in te rtio :, ten te mgnitude of te resultnt is () (b) (c) Let be te ngle between te forces nd. It is given tt te resultnt of nd divides te ngle between tem in te rtio :. Tis mens tt te resultnt mkes n ngle wit te direction of nd ngle wit te direction of. sin sin Terefore, nd sin sin sin cos sin sin lso sin 4 sin 4 sin 4 cos From (i) nd (ii), we get, ( cos )...(i)...(ii) Two forces X nd Y ve resultnt F nd te resolved prt of F in te direction of X is of mgnitude Y. Ten te ngle between te forces is () sin X Y (b) sin X Y (c) 4 sin X Y (d) (d) None of tese Let nd represent two forces X nd Y respectively. Let be te ngle between tem nd, te ngle wic te resultnt F (represented by ) mkes wit. Now, resolved prt of F long. D F cos D = D cos = X Y cos ut resolved prt of F long is given to by Y. Y X Y cos or Y ( cos ) X i.e., Tus, sin X Y sin or sin X Y X Y Y. sin X, sin / Exmple: 5 Te gretest nd lest mgnitude of te resultnt of two forces of constnt mgnitude re F nd G. en te forces ct n ngle, te resultnt in mgnitudes is equl to X Y Y F X Y D () F cos G sin (b) F sin G cos (c) Solution: () Gretest resultnt F Lest resultnt G F G ( F G) n solving, we get, F G (d) F G

4 Sttics 4 were nd ct n ngle, te resultnt cos F cos G sin. Tringle lw of Forces. If tree forces, cting t point, be represented in mgnitude nd direction by te sides of tringle, tken in order, tey will be in equilibrium. Here,, In tringle, we ve 0 0 Hence te forces,, re in equilibrium. onverse : If tree forces cting t point re in equilibrium, ten tey cn be represented in mgnitude nd direction by te sides of tringle, tken in order..4 olygon lw of Forces. If ny number of forces cting on prticle be represented in mgnitude nd direction by te sides of polygon tken in order, te forces sll be in equilibrium Exmple: 6 D nd E re te mid-points of te sides nd respectively of. Te resultnt of te forces is represented by E nd D is () (b) Solution: (d) e ve, E D E D = = Exmple: 7 DE is pentgon. Forces cting on prticle re represented in mgnitude nd direction by (c),, D,DE, D, nd E. Teir resultnt is given by () E (b) (c) E (d) 4 E Solution: (c) e ve, D DE D E.5 Lmi's Teorem. = D DE D DE E = E E E (d) = E E E E. If tree forces cting t point be in equilibrium, ec force is proportionl to te sine of te ngle between te oter two. Tus if te forces re, nd ;,, be te D E

5 Sttics 5 ngles between nd, nd, nd respectively. If te forces re in equilibrium, we ve,. sin sin sin Exmple: 8 Te converse of tis teorem is lso true. Solution: (d) orizontl force F is pplied to smll object of mss m on smoot plne inclined to te orizon t n ngle. If F is just enoug to keep in equilibrium, ten F = () mg cos (b) mg sin (c) mg cos (d) mg tn y pplying Lmi's teorem t, we ve F mg sin 90 sin(80 ) sin(90 ) F mg F mg tn sin cos 90 mg F Exmple: 9 kite of weigt is flying wit its string long strigt line. If te rtios of te resultnt ir pressure to te tension T in te string nd to te weigt of te kite re nd ( ) respectively, ten [oorkee 990] () T ( 6 ) (b) ( ) (c) T ( 6 ) (d) ( ) From Lmi's teorem, sin( ) T sin( 80 o ) sin( 80 o ) T...(i) sin( ) sin sin T Given, T...(ii) nd...(iii) Dividing (iii) by (ii), we get T T T ( 6 ) T ( ) ( ) Exmple: 0 Tree forces, re cting t point in plne. Te ngles between nd nd nd re 50 nd 0 respectively, ten for equilibrium, forces,, re in te rtio [MN 99; USET 000] () : : (b) : : (c) :: (d) : : Solution: (d) lerly, te ngle between nd is 60 (50 0 ) 90. y Lmi's teorem, sin0 sin90 sin50 / /

6 .6 rllel Forces. Sttics 6 () Like prllel forces : Two prllel forces re sid to be like prllel forces wen tey ct in te sme direction. Te resultnt of two like prllel forces nd is equl in mgnitude of te sum of te mgnitude of forces nd cts in te sme direction s te forces nd nd t te point on te line segment joining te point of ction nd, wic divides it in te rtio : internlly. () Two unlike prllel forces : Two prllel forces re sid to be unlike if tey ct in opposite directions. If nd be two unlike prllel force cting t nd nd is greter in mgnitude tn. Ten teir resultnt cts in te sme direction s nd cts t point on produced. Suc tt nd.. Ten in tis cse divides externlly in te inverse rtio of te forces, Importnt Tips If tree like prllel forces,, ct t te vertices,, repectively of tringle, ten teir resultnt ct t te (i) Incentre of, if (ii) ircumcentre of, if (iii) rtocentre of, if b c sin sin tn (iv) entroid of, if = =. tn tn sin Exmple: Tree like prllel forces,, ct t te corner points of tringle. Teir resultnt psses troug te circumcentre,if [ookee 995] () (b) (c) 0 (d) None of tese b c Solution: (c) Since te resultnt psses troug te circumcentre of, terefore, te lgebric sum of te moments bout it, is zero. Exmple: Hence, + + = 0. nd re like prllel forces. If is moved prllel to itself troug distnce x, ten te resultnt of nd moves troug distnce. x () (b) x (c) x (d) None of tese Solution: () Let te prllel forces nd ct t nd respectively. Suppose te resultnt + cts t. Ten,...(i) x + +

7 .7 Moment. If is moved prllel to itself troug distnce x i.e. t '. Suppose te resultnt now cts t '. Ten, ' x ' ' ' '...(ii) Now ' ' ' ' ' x ' x x x ' x ' Te moment of force bout point is given in mgnitude by te product of te forces nd te perpendiculr distnce of from te line of ction of te force. If F be force cting point of rigid body long te line nd M (= p) be te perpendiculr distnce of te fixed point from, ten te moment of force bout F. p M ( M ) (re of ) Te S.I. unit of moment is Newton-meter (N-m). Sttics 7 () Sign of te moment : Te moment of force bout point mesures te tendency of te force to cuse rottion bout tt point. Te tendency of te force F is to F turn te lmin in te clockwise direction nd of te force F is in te nticlockwise direction. Te usul convention is to regrd te moment wic is nticlockwise direction s positive nd tt in te clockwise direction s negtive. () Vrignon's teorem : Te lgebric sum of te moments of ny two coplnr forces bout ny point in teir plne is equl to te moment of teir resultnt bout te sme point. Note :.8 ouples. Ty lgebric sum of te moments of ny two forces bout ny point on te line of ction of teir resultnt is zero. onversely, if te lgebric sum of te moments of ny two coplnr forces, wic re not in equilibrium, bout ny point in teir plne is zero, teir resultnt psses troug te point. If body, ving one point fixed, is cted upon by two forces nd is t rest. Ten te moments of te two forces bout te fixed point re equl nd opposite. Two equl unlike prllel forces wic do not ve te sme line of ction, re sid to form couple. Exmple : ouples ve to be pplied in order to wind wtc, to drive gimlet, to pus cork screw in cork or to drw circles by mens of pir of compsses. () rm of te couple : Te perpendiculr distnce between te lines of ction of te forces forming te couple is known s te rm of te couple. F M p F

8 Sttics 8 () Moment of couple : Te moment of couple is obtined in mgnitude by multiplying te mgnitude of one of te forces forming te couple nd perpendiculr distnce between te lines of ction of te force. Te perpendiculr distnce between te forces is clled te rm of te couple. Te moment of te couple is regrded s positive or negtive ccording s it s tendency to turn te body in te nticlockwise or clockwise direction. Moment of couple = Force rm of te couple =.p () Sign of te moment of couple : Te moment of couple is tken wit positive or negtive sign ccording s it s tendency to turn te body in te nticlockwise or clockwise direction. ositive Negtive Note : couple of opposite sign..9 Tringle teorem of ouples. couple cn not be blnced by single force, but cn be blnced by If tree forces cting on body be represented in mgnitude, direction nd line of ction by te sides of tringle tken in order, ten tey re equivlent to couple wose moment is represented by twice te re of tringle. onsider te force long E, long nd long. Tese forces re tree concurrent forces cting t nd represented in mgnitude nd direction by te sides, nd of. So, by te tringle D E lw of forces, tey re in equilibrium. Te remining two forces long D nd long form couple, wose moment is m. L. L Since (. L) = re of te Moment =.L = (re of ) L Exmple: Solution: (c) ligt rod of lengt 0 cm. rests on two pegs 5 cm. prt. t wt distnce from te end te pegs sould be plced so tt te rection of pegs my be equl wen weigt 5 nd re suspended from nd respectively 00] [oorkee 995, USET ().75 cm., 5.75 cm. (b).75 cm., 7.75 cm. (c).75 cm., 8.75 cm. (d) None of tese Let, be te rections t te pegs nd suc tt = x esolving ll forces verticlly, we get 8 4 Tke moment of forces bout, we get... x 5 5

9 Sttics 9 Exmple: 4 4. x 4.( x 5).0 x. 75 cm x. 75cm nd cm t wt eigt from te bse of verticl pillr, string of lengt 6 metres be tied, so tt mn sitting on te ground nd pulling te oter end of te string s to pply minimum force to overturn te pillr [oorkee 997, S 000] ().5 metres (b) metres (c) metres (d) 4 metres Exmple: 5 Let te string be tied t te point of te verticl pillr, so tt = x Now moment of F bout = F. L = F. sin = F.6 cos sin = F sin To overturn te pillr wit mximum (fixed) force F, moment is mximum if sin = (mx.) 90, i.e. 45 sin Two unlike prllel forces cting t points nd form couple of moment G. If teir lines of ction re turned troug rigt ngle, tey form couple of moment H. Sow tt wen bot ct t rigt ngles to, tey form couple of moment. 6 L F ()GH (b) G + H (c) Solution: (c) e ve, G nd b H...(i) lerly, b x G H (d) None of tese Exmple: 6 G H x [from (i)] x G H Hence, required moment = G H Te resultnt of tree forces represented in mgnitude nd direction by te sides of tringle tken in order wit = 5 cm, = 5 cm, nd = 8 cm, is couple of moment () units (b) 4 units (c) 6 units (d) 6 units esultnt of tree forces represented in mgnitude nd direction by te sides of tringle tken in order is couple of moment equl to twice te re of tringle. te resultnt is couple of moment = (re of ) Here, = 5 cm, b = 5 cm nd c = 8 cm S = S = 9. b x re = S( S )( S b)( S c) 9(9 5)(9 5)(9 8) equired moment = () = 4 units..0 Equilibrium of oplnr Forces. () If tree forces keep body in equilibrium, tey must be coplnr. () If tree forces cting in one plne upon rigid body keep it in equilibrium, tey must eiter meet in point or be prllel.

10 Sttics 0 () en more tn tree forces cting on rigid body, keep it in equilibrium, ten it is not necessry tt tey meet t point. Te system of forces will be in equilibrium if tere is neiter trnsltory motion nor rottory motion. i.e. X = 0, Y = 0, G = 0 or = 0, G = 0. (4) system of coplnr forces cting upon rigid body will be in equilibrium if te lgebric sum of teir resolved prts in ny two mutully perpendiculr directions vnis seprtely, nd if te lgebric sum of teir moments bout ny point in teir plne is zero. (5) system of coplnr forces cting upon rigid body will be in equilibrium if te lgebric sum of te moments of te forces bout ec of tree non-colliner points is zero. (6) Trigonometricl teorem : If is ny point on te bse of suc tt : = m : n. Ten, (i) (ii) ( m n)cot m cot n cot were, ( n n)cot n cot m cot m n Exmple: 7 is Solution: () Exmple: 8 Two smoot beds nd, free to move on verticl smoot circulr wire, re connected by string. eigts, nd re suspended from, nd point of te string respectively. In equilibrium, nd re in orizontl line. If tn () tn (b) tn tn nd, ten te rtio tn : tn (c) tn tn [oorkee 996, USET 00] (d) None of tese esolving forces orizontlly nd verticlly t te points, nd respectively, we get T cos sin...(i) T sin cos...(ii) T cos sin...(iii) T sin cos...(iv) T cos T cos...(v) nd T sin T sin...(vi) Using (v), from (i)nd (ii), we get, From (ii) nd (vi), we ve T sin T sin or T sin T sin...(vii) dding nd subtrcting (vi) nd (vii), we get T sin...(viii) T sin...(ix) Dividing (viii) by (ix), we get T sin. T sin or cos sin. cos sin (from (v)) or T T tn tn uniform bem of lengt rests in equilibrium ginst smoot verticl plne nd over smoot peg t distnce from te plne. If be te inclintion of te bem to te verticl, ten sin is

11 () (b) (c) (d) Sttics Solution: () Let be rod of lengt nd weigt. It rests ginst smoot verticl wll t nd over peg, t distnce from te wll. Te rod is in equilibrium under te following forces : Exmple: 9. Friction. (i) Te weigt t G (ii) Te rection t (iii) Te rection S t perpendiculr to. Since te rod is in equilibrium. So, te tree force re concurrent t. In K, we ve, sin = In, we ve, sin = In G, we ve sin = ; sin.. bem wose centre of grvity divides it into two portions nd b, is plced inside smoot orizontl spere. If be its inclintion to te orizon in te position of equilibrium nd be te ngle subtended by te bem t te centre of te spere, ten ( b ) ( b ) () tn ( b )( b ) tn (b) tn tn (c) tn tn (d) tn tn ( b ) ( b ) ( b )( b ) pplying m n teorem in, we get ( G G)cot G G cot G cot G ( b)cot(90 ) b cot cot b ( b)tn b tn tn tn tn b Friction is retrding force wic prevent one body from sliding on noter. It is, terefore rection. en two bodies re in contct wit ec oter, ten te property of rougness of te bodies by virtue of wic force is exerted between tem to resist te motion of one body upon te oter is clled friction nd te force exerted is clled force of friction. F M K N S 0 S G M N G b () Friction is self djusting force : Let orizontl force pull evy body of weigt resting on smoot orizontl tble. It will be noticed tt up to certin vlue of, te body does not move. Te rection of te tble nd te weigt of te body do not ve ny effect on te orizontl pull s tey re verticl. It is te force of friction F, cting in te orizontl direction, wic blnces nd prevents te body from moving. s is incresed, F lso increses so s to blnce. Tus F increses wit. stge comes wen just begins to move te body. t tis stge F reces its mximum vlue nd is

12 Sttics equl to te vlue of t tt instnt. fter tt, if is incresed furter, F does not increse ny more nd body begins to move. Tis sows tt friction is self djusting, i.e. mount of friction exerted is not constnt, but increses grdully from zero to certin mximum limit. () Stticl friction : en one body tends to slide over te surfce of noter body nd is not on te verge of motion ten te friction clled into ply is clled stticl friction. () Limiting friction : en one body is on te verge of sliding over te surfce of noter body ten te friction clled into ply is clled limiting friction. (4) Dynmicl friction : en one body is ctully sliding over te surfce of noter body te friction clled into ply is clled dynmicl friction. (5) Lws of limiting friction/stticl friction/dynmicl friction : (i) Limiting friction cts in te direction opposite to tt in wic te body is bout to move. (ii) Te mgnitude of te limiting friction between two bodies bers constnt rtio depends only on te nture of te mterils of wic tese bodies re mde. (iii) Limiting friction is independent of te spe nd te re of te surfces in contct, so long s te norml rection between tem is sme, if te norml rection is constnt. (iv) Limiting friction f s is directly proportionl to te norml rection, i.e. f s fs µ s.; µ s fs /, were µ s is constnt wic is clled coefficient of stticl friction. In cse of dynmic friction, µ k = f k /, were µ k is te coefficient of dynmic friction. (6) ngle of friction : Te ngle wic te resultnt force mkes wit te direction of te norml rection is clled te ngle of friction nd it is generlly S denoted by. Tus is te limiting vlue of, wen te force of friction F ttins its mximum vlue. Mximum force of friction tn Norml rection Since nd µ re te components of S, we ve, S cos =, S sin = µ. Hence by squring nd dding, we get S µ nd on dividing tem, we get tn = µ. Hence we see tt te coefficient of friction is equl to te tngent of te ngle of friction.. oefficient of Friction. en one body is in limiting equilibrium in contct wit noter body, te constnt rtio wic te limiting force of friction bers to norml rection t teir point of contct, is clled te coefficient of friction nd it is generlly denoted by µ. Tus, µ is te rtio of te limiting friction nd norml rection. Mximum force of friction Hence, µ = tn Norml rection F µ F µ, were F is te limiting friction nd is te norml rection. F =

13 Sttics Note : Te vlue of µ depends on te substnce of wic te bodies re mde nd so it differs from one body to te oter. lso, te vlue of µ lwys lies between 0 nd. Its vlue is zero for perfectly smoot body. one of friction : cone wose vertex is t te point of contct of two roug bodies nd wose xis lies long te common norml nd wose semi-verticl ngle is equl to te ngle of friction is clled cone of friction.. Limiting equilibrium on n Inclined lne Let body of weigt be on te point of sliding down plne wic is inclined t n ngle to te orizon. Let be te norml rection nd µ be te limiting friction cting up te plne. Tus, te body is in limiting equilibrium under te ction of tree forces :, µ nd. esolving te forces long nd perpendiculr to te plne, we ve µ sin nd cos µ sin µ tn tn tn cos Tus, if body be on te point of sliding down n inclined plne under its own weigt, te inclintion of te plne is equl to te ngle of te friction. () Lest force required to pull body up n inclined roug plne : Let body of weigt be t point, be te inclintion of roug inclined plne to te orizontl nd be te ngle of friction. Let be te force cting t n ngle wit te plne required just to move body up te plne. sin( ) cos( ) µ tn lerly, te force is lest wen cos( ) is mximum, i.e. wen cos( ), i.e. 0 or. Te lest vlue of is sin( ) () Lest force required to pull body down n inclined plne : Let body of weigt be t te point, be te inclintion of roug inclined plne to te orizontl nd be te ngle of friction. Let be te force cting n ngle wit te plne, required just to move te body up te plne. sin( ) [ µ tn ] cos( ) lerly, is lest wen cos( ) is mximum, i.e. wen 0 or. Te lest vlue of is sin( ). Note : If, ten te body is in limiting equilibrium nd is just on te point of moving downwrds.

14 If sin( ). Sttics 4, ten te lest force required to move te body down te plne is If, or, ten te lest force required to move te body up te plne is sin( ). If, ten te body will move down te plne under te ction of its weigt nd norml rection. Importnt Tips Lest force on te orizontl plne : Lest force required to move te body wit weigt on te roug orizontl plne is sin. Exmple: 0 force of 5 Kg is required to pull block of wood weiging 40 Kg on roug orizontl surfce. Te coefficient of friction is () (b) 0 (c) 4 (d) 4 Solution: (d) In te position of limiting equilibrium, we ve µ 5 nd 40 µ kg 40 Exmple: Exmple: uniform ldder rests in limiting equilibrium, its lower end on roug orizontl plne nd its upper end ginst smoot verticl wll. If is te ngle of inclintion of te ldder to te verticl wll nd µ is te coefficient of friction, ten tn is equl to ()µ (b) µ (c) esolving te forces orizontlly nd verticlly, we get S µ nd µ S µ...(i) Tking moments bout, we get. G sin S. cos 0. G sin S. cos. sin S. cos G. sin µ. cos [from (i)] tn µ. (d) µ + body of 6 Kg. rests in limiting equilibrium on n inclined plne wose slope is 0. If te plne is rised to slope of 60, te force in Kg. weigt long te plne required to support it is () (b) G S 0

15 Sttics 5 (c) (d) In cse (i), 6 cos 0, µ 6 sin 0. µ tn 0 In cse (ii), S 6 cos 60 µs 6 sin 60 (6 cos60) 6 sin kg S 6 kg 60 Exmple: Te coefficient of friction between te floor nd box weiging ton if minimum force of 600 Kgf is required to strt te box moving is () 4 (b) 4 (c) (d) Exmple: 4 Solution: (c) esolving orizontlly nd verticlly cos µ; sin cos µ [ sin ] or [cos µ sin ] µ or µ µ cos sin sin cos.sin cos( ) cos( ) cos Now is minimum wen cos( ) is mximum, i.e. wen cos( ) Min sin ut ton wt. 000 Kg. nd 600 kg 600 sin ; tn, µ 4 4 block of mss Kg. slides down roug inclined plne strting from rest t te top. If te 4 inclintion of te plne to te orizontl is wit tn, te coefficient of friction is 0. nd te 5 ccelertion due to grvity is g = 9.8. Te velocity of te block wen it reces te bottom is ()6. (b) 5. (c) 7 (d) 8. Let be te position of te mn t ny time. lerly, g cos Let f be ccelertion down te plne. Eqution of motion is f g sin µ f g sin µ (g cos ) f g(sin µ cos ) Here, 4 tn, 5 4 sin, cos g cos g g sin

16 Sttics 6 Now, f g g g 5 5g f 4., Let v be te velocity t. Ten, v 5g u fs 0 4 f 5g g v. 4 4 we cn tke 4 4, since tn 5 Exmple: 5 v 5g , i.e., v 7m / sec circulr cylinder of rdius r nd eigt rests on roug orizontl plne wit one of its flt ends on te plne. grdully incresing orizontl force is pplied troug te centre of te upper end. If te coefficient of friction is µ. Te cylinder will topple before sliding of () r µ (b) r µ (c) r µ (d) r µ Let bse of cylinder is. r Let force is pplied t. Let rection of plne is nd force of friction is µ. Let weigt of cylinder is. In equilibrium condition,...(i) nd µ...(ii) From (i) nd (ii), we ve µ Tking moment bout te point, e ve 0 r If µ.4 entre of Grvity or r µ r Te cylinder will be topple before sliding. Te centre of grvity of body or system of prticles rigidly connected togeter, is tt point troug wic te line of ction of te weigt of te body lwys psses in wtever position te body is plced nd tis point is clled centroid. body cn ve one nd only one centre of grvity. If w, w,....., w n re te weigts of te prticles plced t te points ( x, y), ( x, y),....., n( xn, yn) respectively, ten te centre of grvity G ( x, y) is given by wx wy x, y. w w () entre of grvity of number of bodies of different spe : E D

17 (i).g. of uniform rod : Te.G. of uniform rod lies t its mid-point. Sttics 7 (ii).g. of uniform prllelogrm : Te.G. of uniform prllelogrm is te point of inter-section of te digonls. (iii).g. of uniform tringulr lmin : Te.G. of tringle lies on medin t distnce from te bse equl to one tird of te medins. () Some Importnt points to remember : (i) Te.G. of uniform tetredron lies on te line joining vertex to te.g. of te opposite fce, dividing tis line in te rtio :. (ii) Te.G. of rigt circulr solid cone lies t distnce /4 from te bse on te xis nd divides it in te rtio :. (iii) Te.G. of te curved surfce of rigt circulr ollow cone lies t distnce / from te bse on te xis nd divides it in te rtio : (iv) Te.G. of emispericl sell t distnce / from te centre on te symmetricl rdius. (v) Te.G. of solid emispere lies on te centrl rdius t distnce /8 from te centre were is te rdius. (vi) Te.G. of circulr rc subtending n ngle t te centre is t distnce from te centre on te symmetricl rdius, being te rdius, nd in rdins. sin (vii) Te.G. of sector of circle subtending n ngle t te centre is t distnce sin from te centre on te symmetricl rdius, being te rdius nd in rdins. (viii) Te.G. of te semi circulr rc lies on te centrl rdius t distnce of te boundry dimeter, were is te rdius of te rc. from Importnt Tips Let tere be body of weigt w nd x be its.g. If portion of weigt w is removed from it nd x be te.g. of wx wx te removed portion. Ten, te.g. of te remining portion is given by x w w Let x be te.g. of body of weigt w. If x, x, x re te.g. of portions of weigts w, w, w respectively, wx wx w x w x wic re removed from te body, ten te.g. of te remining body is given by x 4 w w w w Exmple: 6 Solution: () Two uniform solid speres composed of te sme mteril nd ving teir rdii 6 cm nd cm respectively re firmly united. Te distnce of te centre of grvity of te wole body from te centre of te lrger spere is [MN 980] () cm. (b) cm. (c) cm. (d) 4 cm. eigts of te speres re proportionl to teir volumes. Let be te density of te mteril, ten 4 w = eigt of te spere of rdius 6cm (6 ) 88 4 w = eigt of te spere of rdius cm ( ) 6 x = Distnce of te.g. of te lrger spere from its centre = 0 x = Distnce of te.g. of smllr spere from = 9 cm. x = Distnce of te.g. of te wole body from 6 cm cm

18 Sttics 8 Exmple: 7 Exmple: 8 Exmple: 9 Now wx x w w x w x 4 solid rigt circulr cylinder is ttced to emispere of equl bse. It te.g. of combined solid is t te centre of te bse, ten te rtio of te rdius nd eigt of cylinder is () : (b) : (c) : (d) None of tese Let be te rdius of te bse of te cylinder nd be te eigt of te cylinder. Let w nd w be te weigt of te cylinder nd emispere respectively. Tese weigts ct t teir centres of grvity G nd G respectively. Now, w = weigt of te cylinder = w = weigt of te emispere = G nd G 8 g g Since te combined.g. is t. Terefore w G w G w w ( g) g 8 g g : : 8 4 n te sme bse nd on opposite side of it, isosceles tringles nd D re described wose ltitudes re cm nd 6 cm respectively. Te distnce of te centre of grvity of te qudrilterl D from, is ()0.5 cm (b) cm (c).5 cm (d) cm Let L be te midpoint of. Ten L nd DL. Let G nd G be te centres of grvity of tringulr lmin nd D respectively. Ten, LG L 4cm. nd LG DL cm. Te.G. of te qudrilterl D is t G, te mid point of G G. GG GG cm. GL G L GG (4 ) cm cm. is uniform tringulr lmin wit centre of grvity t G. If te portion G is removed, te centre of grvity of te remining portion is t G'. Ten GG' is equl to () G (b) G (c) G (d) G D G G c G 6cm L G D E Solution: (d) Since G nd G' re te centroids of nd GD respectively. Terefore G D, GD D nd GG" GD D D 9 Now, G D nd GD D G G re of G re of G eigt of tringulr lmin G = (weigt of tringle lmin )

19 Tus, if is te weigt of lmin G, ten te weigt of lmin is. Now, G' is te.g. of te remining portion G. Terefore, ( G) ( G") G' = ( G G" ) 8 5 = D D D GG' G G' D D = D G G G" G GG" D 9 Sttics 9 D 8 9 D