LM#9ParticlelA very small amount of matter which nray be assumed to occupy a single point in space is called as aparticle. The 3 requirements to model a particle as an object are as follows:l. All the forces acting on a given body or object should be applied at the same point i.e. all theforces are concurrent.2. The object under consideration is not translating or rotating i.e. the sum of all the forces acting onthe object is zero.3. The body is isolated from the system and the surroundings.Rigid Body:A rigid body is a combination of a large number of padicles occupying fixed positions with respect toeach other. A rigid body is different from a particle inthe following ways:1, For rigid bodies, forces need not be concurrenl, they can act anywhere on the body at any point.2. If a body is moving or rotating, the sum of extemal forces and moments is equal to the mass ofbody times its acceleration.3. Rigid bodies need not be isolated from the system and surroundings.#4= 775,nmAC = AB s,.r65 = 7c 331.r,aIPROBLEM 3.2A 90-N force is applied to the control rodlB as shown. Knowing that thelength of the rod is 225 mm, determine the moment of the force aboutpoint B by resolving the force into horizontal and vertical components.SOLUTIONqo N{ = (90 N)cos25o= 81.568 N{, = (90 N)sin2s"= 38.036 Ntx = (0.225 m)cos65.= 0.095089 my = (o.zzs m)sin65"= 0.20392 mM, = xF, : /F,= (0.0es0se rn)(:r.o:o N) * (0.203e2 mX81.568 N)= -13.0165 N.mMa = 13.02 N.m) {\,,\4''L-PROPRIETARY MATENAL @ 2007 The Mccmw-Hill Companies, lnc. All rights reserved. No patt of this Ma ual may be disptajed, repro&lcedol distributed in any form or by aO) mean' without the prior written permission ofrhe pubtisher, or used beyokd the timited'di$tribution to teachers andeducalofs peftitted by Mccraw-Hillfor their individuol course preparation. IJ),oi are i student using this Mamtal,lou are using it *)iihout Wmissian.164.,5io4o,5 J+Lgga srogXd: il52I lt is known that a force with a moment of 1152 N m about D is requiredj to straighten the fence post CD. If the capacity ofthe winch puller ,4,B isI ZSSO N. determine the minimum value of distance d to create the,lPROBLEM 3,11specified moment about poht D knowingthat q= 0.24 m and b = 105 m.PRO\RIETARr MATER|AT. @ 2007 The Mccraw,Hiil Companies, tnc. All rights reserved. No patt of thk Manual nlav.be disp.laved. repfodrtcedal (tistriblned in a ! Jit m or ty ury ^uons, withoi the prior witen pambsion olthe publisher, or used iqowl the timited distributiotl to reuchcrs unJetlucators pernittei "by Mccra;-Hilt for rhen inrlividuai course prepiration. If you ate a s Ae using this Manual lott are using it withokt permis\tinSOLUTIONIt.05o,2+ rnThe minimum value of dcanbe found based on the equation relating the moment ofthe force T,4B about D:whereMD - OnB,,"'))(d)Ma = 1152N m(r*^ *), = rn,ou*sino: (2880 N)sinoNowor (d + o.za)'z + (1.05)2 -- 6.8906d2or 5.8906d2 - 0.48d - 1.1601 = 0Using the quadratic equation, the minimum values ofd are 0.48639 m and -0'40490 m'Since only the positive value applies hete, d : 0.48639 mor / =486mm{1.05 m(a + o.z+.;2 +(t.os)'](a + o.z+)2 + (t.os)2 = 2.62563.98 A5-mJongbeam is subjected to a variety ofloadings. (a) Replaceeach loading with an equivalent force-couple system at end B of the beam.(b) Which o"f the Ioadings are equivalent?I2OO N)(200 N1.6 kN.rnL 6 kN.tn(c)6(X) Nt)! kN.rnr)| -.'I roo NlzooNlr))4.2 kN.rnFig. P3.986mN3.99 A S-mJong beam is loaded as shown. Determine the loading ofProb. 3.98 that is equivalent to this loading.3.100 Detennine the single equivalent force and the distance liomend A to its line of action for the beam and loading shown when (a) Fs =200 N J, n4B:100 N.m, (b) FB:100 N t, MB = 600 N m' (c) Fr:100 N J, ME : -200 N m.0.8 m")r.27 kN.rrrFig. P3.99,100 NI2(l() N.rrlFig. P3.1FB) \{.c1363.q8 f$ Ds406(lrtr(g, D,6tN'^ti)A,t Bv. a-) z:------------- >/_-)l' 9'1,'dz dLI8y"'*- . ;Ld, F-/,-d.'TiI aoo tlt5a*r,t nfnRooilaocNe--------:aLt t'd,-ez- dz-,1tuSoon .446ul, n, Fx l,Gffi)_.,3,6 '1, l, 'o"3 *N'^kN'r''#a-+d, f,-/,-J- dz[ao*l4L*'^Sr l,)hrI --:--\ l--.'r=*-_:--''lf" /l'ot p.^l'heerJ'J,'^"un,ffilr;i-:. /Lbti''(4J&---------+a-+lt f-d,-)z dz7'6 lztl't'6alao 4y by' ?4__re_>d t &d,-dz dtBy4-)1-_-1 1>d, 5'd,-ilz dtoB LN'n' J+.r ktl,^l+oot))6,8 Ltt.n7se)GLN,^6t:'*"'t +'8y &tn*y+-*-------4e,---JAt 5'd.-d,- 4>lz$at's-iIPROBLEM 3.102The masses oftwo children sitting at ends I and B of a seesaw are 38 kgmd 29 ks., respectively. Determine where a third child should sit so thatthe resuldnt oi the weights of the th€e children will pass through C ifshe has a mass of (a) 21 ke, (b) 24 ke.,ot- tegnireLSOLUTIONldAW.ta/* FirstBfue = mes = (38 kc)c/ wa = mns = Qexe)slj,Wc = mc| = Q7 ks)cFor resultant weight Io Nt at C, LM, = grhen [(38 kg)s](3 m) - i(27 kg)sl(d) - [(2e ks)s](2 m) = 0,. o ='u ut' = 0.66667 mor d=0.667ml-lryc=ntcc=Qakg)gFor resultant weight ta act at C, >Mc = 0rhen [(3s kg)s](2 m) - [(24 ks)s](d) - [(2e ks)s](2 m) = 076- 5R= 0.75 m(")Qnly loadsqf L_4re,q44C*/FtCuor d = 0.750€XvilibiunPROPRIETAR\ MATERIAI. @ 2007 The Mccraw-Hill Companies, Inc. All rights reserved. No paft oJ lhh .Manual.may,be displayed, feproduced.* ami"":ii' ,"y]), ""i ay i,ri""i iin",t rhe prior writen petmission ofrhe publbher. or used bevond the linircd dktnbuhon to teachers andeducarors permirteel'by Mccrri*-iitt ft, nlii'iriii""iaun preparation. Ifyoi are a student using thls ManuaL you lre using it witho t permission'215*---3fllliIFig. P3.114Fig. P3.148* Jo o .lIrl'+t{fPROBLEM 4.2The boom on a 95001b truck is used to unload a pallet of shingles ofweight 3500 lb, Determine the reaction at each of the two (a) rearwheels B, (D) front wheels C.PROPRIETARY MATERIAZ. O 200? The Mccraw-Hill Companies, Inc. All rights reserved. Nopafl of this Manuat may be disptayett, reproducedor distributed in ary1 Jbm or by a y means, without the prior witten permissian of the pubtisher, or u.ted bqroncl the limited distribution to teaihers .)ndeducators penhitted by Mcctuw-Hill.for their indiidual course preparution. If )tou are a student using this Manual. !o11arc uting it without permonon.SOLUTIONFree.Body Diagram:\a)*) >M, = Ot (:.s r.ipg[(r.o + 1.3 + le.5cosls')ftf- zrufg.a + 1.3 + t4)ft] + (e.5 kipsxt.6 ft) = 02& = 5.4009 kipsor FB = 2.70 kips 1 <(:.s rip$[(u.scosls' - 14)ft]- (e.s tips)[(ra + r.:)n] + zr.[(r+ + r.3 + 1.6)ft] = 02F, = 759916*, o'(b)tLMB = 0:or r;:3.S0 kips | <(PROBLEM 4.6Solve Prob. 4.5 when a = 44o.Problem 4.5: A hand truck is