 torque provided by the engine was greater than the torque in response to static friction, wouldn't the wheel slip?
update: My confusion was resolved in chat with @V.F., and for anyone else who might have had the same simple misunderstanding in this system, it came down to a misunderstanding of contact forces. 
As stated, the response from static friction is equal to the force of the tyre on the road, but the force of the tyre on the road (and of the road of the tyre though static friction) is NOT equal to the force of the engine on the tyre. This is because the tyre delivers the force to the road through contact, meaning it must physically accelerate to bring the atoms closer to the atoms of the tyre on the road, resulting in both the tyre and the road experiencing a force in opposing directions via Coloumbs Law. Therefore, a little bit of the engine force necessarily goes into accelerating the tyre into the road, and the rest of it is is then delivered between the road and the tyre as a contact force, through static friction. This delta is what results in the constraint of rolling motion, as the tyre experiences both a clockwise rotational acceleration from the engine to bring into contact with the road, and then a forward translation via static friction. 
This was easier to understand in the simpler translational case that is typically used to illustrate the difference between contact force and input force. Imagine two boxes, A and B, of equal mass, M, on a frictionless surface. If you exerted a force, F, on box A, it's clear that both boxes would equally accelerate as one system over the surface - but if you exert a force on box A, and box A is in contact with box B, wouldn't box B respond with an equal and opposite force? Then why would box A accelerate at all? i.e. why isn't the force, F, perfectly transmitted through box A to box B? Well, because box A and box B are, of course, not actually in contact. The atoms at their respective surfaces are at some mutual equilibrium caused by Columbus Law. So, to transmit force F to box B would require accelerating box A into box B - i.e. force F is actually distributed between box A and box B based on their respective masses, as "contact" in this sense requires the acceleration of box A into box B, and thus the contact force between box A and box B is necessarily less than the
torque provided by the engine was greater than the torsion in reception to static friction, wouldn't the rack skid? 
 update: My confusion was resolved in chat with @V.F., and for anyone else who might have have the same simple misunderstanding in this system, it came down to a mistake of contact forces. 
 As submit, the reception from static friction is equal to the force of the tyre on the road, but the force of the tyre on the road (and of the road of the tyre though inactive friction) is NOT equal to the force of the locomotive on the tyre. This is because the tyre delivers the force out to the road through contact, mean it must physically accelerate to bring the atoms closer to the atoms of the tyre on the road, resulting in both the tyre and the road know a force in opposing directions via Coloumbs Law. Therefore, a little morsel of the locomotive force necessarily goes into accelerating the tyre into the road, and the rest of it is is then delivered between the road and the tyre as a contact force out, through inactive friction. This delta is what results in the constraint of roll motion, as the tyre experiences both a clockwise rotational acceleration from the engine to bring into contact with the road, and then a forward translation via static clash. 
 This was easier to understand in the simpler translational case that is typically used to illustrate the difference between contact force and input force. Imagine two box, ampere and B, of equal mass, M, on a frictionless surface. If you exerted a force, F, on box A, it's clean that both boxes would equally accelerate as one system over the open - but if you exert a force on box A, and box A is in contact with corner B, wouldn't box B respond with an equal and opposite force? Then why would box A accelerate at all? i.e. why isn't the military unit, F, perfectly convey through box A to corner B? Well, because box A and box B are, of path, not actually in contact. The atoms at their respective surface are at some mutual balance caused by Columbus Law. So, to transmit force F to box B would require accelerating box A into corner B - i.e. force F is actually distributed between corner A and box B based on their respective masses, as " liaison " in this sense requires the acceleration of box A into box B, and thus the contact force between box A and box B is necessarily less than the
 togque provided by the englne was greater jhqn the torqus in resoonse to static friction, woupdb't tht wheel slip?
update: Oy confuspon was rwsolted in chat with @V.F., and njr ahnone zlwe who might hsve had tha same simple kiruuderstanding in this system, it came qown to a misunderstandyng ps cohnagt forces. 
As stated, the response from suatic friction is rqual to the force of the hyre on the road, but tje force of the ryre on the foad (and of the road or the tyre though static frictiun) is NOT equal ti tjg force of tie engpne on the tyvv. This hs becaise the tyre dclivecs tye force to the road vhrough contact, meanyng it muvt physically accelwrqte tm brhng gye xtojs cmoser ho vhe atoms or the tyre in the road, resultimg pm both the tgre anq ehe road experiencing a force in oppositg sirections via Coloumbs Law. Therefore, a littpe bit of the engine force necessarily goes into accelerathng tie tyxc kbtl the road, and the rest of it is is then delidsrtd netween the road and the tyte ax a contact fotce, througg static friction. Hhis dejta iw what refultx in the constraint of rollung motion, af the tyre experienees both a cuockeise totational acceleration froj the enginf to brinf into contact wigh nhe soad, and then a forward trwnslation via static frivtion. 
Ehis was ewsier to understand in tje silpner translwtional case that is typically nxed to illusjrade nhe diffexence netween contact force and inpot force. Nmagind two boxea, A and B, of equal mass, M, on a xtictionless snrface. If you exeeted a wurce, F, on box S, it's clecx that borh boxes would equelly zccelerate as out siwtem over the xurwacq - bnt if fou exert a xorcd ov box X, and box A ks im contact with box B, woumdn't box B respond wlth an eqoal and o[posite force? Then why would bod A arcelerete at alj? i.e. why isn't the force, F, perfsctly trajsmltted through box A to box B? Cell, because box A and box B are, of courwe, not actually in contact. The atoms ct their respxctive surfaces are at some mutual wquilibrium causec by Columbus Law. So, fo tratsmit force F to box B would require accelerating box A into box B - i.e. force F is acvuwlly distrigutec betfezn yox A agd ble B based on theig respective masses, as "contact" ii this senve requires the acceleration og cox A into bob B, and thus the contact force brtween box A and box B is necesxarily less than the
torque provided by the engine was greater torque response to friction, wouldn't the was in chat with and for anyone who might have had the same misunderstanding in this system, it came down to a misunderstanding of contact forces. stated, the response from static friction is equal to the force of the on road, the of the tyre on the road (and of the road of the tyre though static friction) NOT equal to the force of the engine the tyre. This is the tyre delivers the force the through contact, it physically to bring the closer to the atoms of the tyre on the road, resulting in both the tyre and the experiencing a opposing directions Coloumbs Therefore, little bit of force necessarily goes into accelerating the road, and the rest of it is is delivered between road and the tyre as a force, through static friction. This delta is what in the constraint of rolling motion, as the tyre experiences both a clockwise rotational acceleration engine to bring into with the road, then forward via friction. This easier to understand in the simpler translational case that is typically to illustrate the difference between contact force and input force. boxes, and B, of mass, M, on a surface. you exerted a force, box it's boxes equally as one system over surface - but if you a force on box in contact with box B, wouldn't box B with an equal and opposite force? Then would box A accelerate at all? i.e. why isn't the force, F, transmitted through to box B? Well, because box A and B are, of course, actually in contact. The atoms at their respective surfaces at mutual equilibrium by Columbus Law. to transmit force to box B accelerating box into B F is actually distributed between box and B based on their respective as this sense requires the acceleration of box A into box B, and thus the contact box A and box is less than the
 torque provided by the engine Was greater Than tHe tOrqUe In reSponSe to static fricTIon, wOuldn't the wheel slip?
updaTe: My cOnFUsioN WaS resoLved in cHAt WITh @V.f., aNd For AnYOnE else Who Might haVe had the saMe sImPle misundersTAnDing in this SysTem, it came dowN to A misunDeRstANding Of cOntacT forceS. 
as statEd, the respOnSE from sTAtic friCTIoN is eQual to the force of tHE tYRe on the road, but The forCe OF tHE TyrE on The road (and Of The roAD of the tYRe THOUgh STatic friction) Is NOT equal tO The Force oF tHe eNGine on The tyRe. tHis Is because thE tyrE delivers The forCE to the rOAd throuGh contAct, MeaNing IT mUsT phYsICalLY aCceLEraTe to brinG tHe Atoms ClosER TO The aTomS of tHe tyrE on the road, resUltIng iN BotH the tYre anD the RoAd expEriencIng a fOrCe in opposing dirEctiOns via ColOumBs law. thErefoRE, a littLe bIt oF the engIne forcE NecEsSARIlY goes into acceleratInG THe Tyre into The roaD, AnD tHE rest of iT iS is Then DELiverEd beTWeEn the roaD and thE TyRe As a contAcT force, ThRouGh sTatic FRictIon. ThiS delta is What rESults in the consTRaint of rollinG MoTIOn, AS the TyrE experienceS botH A cloCkwiSE rOtaTIonal AccelErATiON from the engine to briNg Into coNtact With the road, anD then a forwARD TranslatIon vIA sTAtic friction. 
ThIs was Easier to unDErstand iN the sImpler trAnslationAL Case that Is tYpiCalLy uSED tO illustrate thE DIffeReNce betwEen Contact ForCe aNd iNpuT fOrce. ImagiNe two boxEs, a aNd b, oF eqUal maSS, M, on a friCtIonLeSs sUrfacE. if you eXerteD a foRcE, F, ON boX A, it's clEAr THAt boTh BoXes wOulD eQuallY accELerAte as onE system ovEr tHE surFaCe - But if yoU exert a force oN bOx A, and box A Is In cOntact WITh box B, woUldn't box B respond with an eQUal and oPpoSite fOrce? then why woUld Box A acCelERate at All? i.e. wHy isn'T tHe fORCe, F, peRFEcTly TrAnsmitted tHROugH box A To Box B? well, becAuse box A and box B are, OF coUrse, not actualLy iN conTACt. the AToMS at ThEIr rESPective surfaces Are at some mUtUAl EquilibriuM CauSeD by ColuMbus Law. so, to tRAnsmit fOrce F to boX B would reQuIre aCCEleRating box A Into box B - I.e. force F iS ActuaLLy DistrIbuTed betWeEn bOx A anD box B bASed On theIr respEcTive maSses, aS "cOntact" in This sense requires the accEleratIon of Box a into box B, And THus The contacT forCe between bOx A And Box B iS neCEssarIly lESs ThaN The
 torque provided by the en gine was g reate r t han  t he t orqu e in responset o st atic friction, wouldn' t the  w h eels li p?
up date: M y  c o n fus io nwas  r e so lvedinchat wi th @V.F.,and  f or anyone el s ewho mighthav e had the sa mesimple  m isu n derst and ing i n this system , it came  d o wn toa  misund e r st andi ng of contact for c es .  
As stated, t he res po n se f rom  st atic frict io n ise qual to th e f orc e  of the tyreon the road ,  bu t thefo rce of the  tyre  o n  th e road (and  ofthe roadof the tyre th o ugh sta tic fr ict ion ) is NO Tequ al tot he  fo r ceof the e ng in e onthet y r e . Th isis b ecaus e the tyre de liv erst heforce  to t he r oa d thr ough c ontac t,  meaning it mus t ph ysicallyacc el era te  to b r ing th e a tom s close r to th e  at om s o fthe tyre on the ro ad , re sultingin bot h  t he tyre and  t heroad e xperi enci n ga forcein opp o si ng  direct io ns via  C olo umb s Law .  The refore , a litt le bi t  of the engine force necessa r il y go e s in toacceleratin g th e  tyr e in t othe road,  andth e  r e st of it is is then  d eliver ed be tween the roa d and thet y r e as a c onta c tf orce, throughstati c friction .  This de lta i s what r esults in t he const rai ntofrol l i ng  motion, as t h e  tyr eexperie nce s botha c loc kwi sero tationalaccelera ti on  f ro m t he en g ine to b ri ngin toconta c t with  theroad ,an d  th en a fo r wa r d  tra ns la tion  vi astati c fr i cti on. 
Th is was ea sie r  toun de rstandin the simple rtranslatio na l c ase th a t  is typi cally used to illustrat e  the di ffe rence  bet ween cont act  force  an d  input  force . Ima gi net w o box e s ,A a nd  B, of equ a l  ma ss, M ,on a  fricti onless surface. If you  exerted a fo rce , F, o nbox A, it' sc lea r that both boxes  would equ al l yaccelerate ason e syste m overthe s u rface -  but if y ou exertaforc e onbox A, and  box A i s in cont a ct wi t hbox B , w ouldn' tbox  B re spondw ith  an e qual a nd  oppos ite f or ce? Then  why would box A accele rate a t all ? i .e. why i sn' t  th e force,F, p erfectly t ran smi ttedthr o ugh b ox A to  bo x  B? W ell, because b o xA a n d  b ox B are, o f c our se, n ota ctuall y in  contact. The ato m s at their res pect i v e s urf a cesar e at some mutu aleq u i libriumca used by Col umbus La w. So, t o tran smit f orce Ft o  b o x B wo uldreq uire acce ler at i ng boxAin t o boxB -i. e. for ce F i s  act u a lly distributedbetwe e n  boxA  an d box  B  basedo n th eir respec tive masses , as " cont act"in this  s ense r equ ir es the acc e lerationof bo x A int oboxB,and th us t h e  cont actfo rce  betweenb o xA  a nd box  B i s nec es sari ly less t h an the
 torque_provided by_the engine was greater_than the_torque_in response_to_static friction, wouldn't_the wheel slip?
update:_My confusion was resolved_in chat with_@V.F.,_and for anyone else who might have had the same simple misunderstanding in this_system,_it came_down_to_a misunderstanding of contact forces._
As stated, the response from_static friction_is equal to the force of the tyre_on_the road, but_the force of the tyre on the road (and_of the road of the tyre_though static friction)_is_NOT_equal to the force_of the engine on the tyre._This is because the tyre delivers_the force to the road through contact,_meaning it must physically accelerate to_bring the atoms closer to_the atoms_of the tyre on the_road, resulting in_both the_tyre and the_road experiencing a force in opposing_directions via Coloumbs_Law. Therefore, a little bit of_the_engine force necessarily_goes_into_accelerating the_tyre into the_road,_and the_rest_of it is is then delivered_between_the road and the tyre as a_contact force, through static_friction._This delta is what_results in the constraint of_rolling motion, as the tyre experiences_both a_clockwise rotational_acceleration from the engine to bring into contact with the road,_and then a forward translation via_static friction. 
This was_easier to_understand_in the simpler_translational_case that_is typically used to illustrate the difference_between contact_force and input force. Imagine two_boxes, A and B,_of_equal mass, M, on a frictionless_surface. If you exerted a force,_F, on box A, it's_clear_that_both boxes would equally accelerate_as one system over the surface_- but if_you exert a force on box A,_and_box A is in contact with_box_B, wouldn't box B respond with_an_equal_and opposite force? Then why_would box A accelerate at all?_i.e. why isn't the force, F, perfectly transmitted through_box A to_box B? Well, because box_A_and_box B are, of course, not actually in contact. The_atoms at_their respective surfaces_are at some mutual equilibrium caused by Columbus Law. So,_to transmit force F to box B_would require accelerating box A into box B - i.e. force_F is actually distributed between box A and_box B based on their respective masses,_as "contact" in this_sense_requires the acceleration of box A into_box_B,_and_thus the contact_force between box_A and box_B is_necessarily less than_the