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Question-145751




Question Number 145751 by ajfour last updated on 07/Jul/21
Commented by ajfour last updated on 07/Jul/21
Find the minimum energy that  need be expended by pulling  on one of the carpets with  a force (may vary) in the  shown direction to detach  them completely from one  another. Both carpets have  mass m and friction coeff.  for every surface is uniform  and equal to μ.
$${Find}\:{the}\:{minimum}\:{energy}\:{that} \\ $$$${need}\:{be}\:{expended}\:{by}\:{pulling} \\ $$$${on}\:{one}\:{of}\:{the}\:{carpets}\:{with} \\ $$$${a}\:{force}\:\left({may}\:{vary}\right)\:{in}\:{the} \\ $$$${shown}\:{direction}\:{to}\:{detach} \\ $$$${them}\:{completely}\:{from}\:{one} \\ $$$${another}.\:{Both}\:{carpets}\:{have} \\ $$$${mass}\:{m}\:{and}\:{friction}\:{coeff}. \\ $$$${for}\:{every}\:{surface}\:{is}\:{uniform} \\ $$$${and}\:{equal}\:{to}\:\mu. \\ $$
Commented by ajfour last updated on 09/Jul/21
   ?
$$\:\:\:? \\ $$
Commented by ajfour last updated on 10/Jul/21
I dunnow, sir, i′ve put the  question asking the minimum  energy required to detach the  carpets from one another by  applying force in the shown  way: to which carpet? that  also need be found out.
$${I}\:{dunnow},\:{sir},\:{i}'{ve}\:{put}\:{the} \\ $$$${question}\:{asking}\:{the}\:{minimum} \\ $$$${energy}\:{required}\:{to}\:{detach}\:{the} \\ $$$${carpets}\:{from}\:{one}\:{another}\:{by} \\ $$$${applying}\:{force}\:{in}\:{the}\:{shown} \\ $$$${way}:\:{to}\:{which}\:{carpet}?\:{that} \\ $$$${also}\:{need}\:{be}\:{found}\:{out}. \\ $$
Commented by mr W last updated on 10/Jul/21
the force is acted on the lower carpet,  right?
$${the}\:{force}\:{is}\:{acted}\:{on}\:{the}\:{lower}\:{carpet}, \\ $$$${right}? \\ $$
Answered by mr W last updated on 10/Jul/21
assume we pull on the upper carpet.  say the overlapping length of both   carpets is x.  total friction acting on the upper  carpet is μmg, no matter how large  x is.  such that the upper carpet moves,   F≥μmg    on the lower carpet:  acting friction on the contact surface  to upper carpet is (x/b)×μmg.  friction on its contact surface to   ground is μmg+(x/b)×μmg, which is   always larger than the acting friction   on its upper surface, that means the   lower carpet doesn′t move.  total energy needed to detach both  carpets completely is then   E≥F×b=μmgb
$${assume}\:{we}\:{pull}\:{on}\:{the}\:{upper}\:{carpet}. \\ $$$${say}\:{the}\:{overlapping}\:{length}\:{of}\:{both}\: \\ $$$${carpets}\:{is}\:{x}. \\ $$$${total}\:{friction}\:{acting}\:{on}\:{the}\:{upper} \\ $$$${carpet}\:{is}\:\mu{mg},\:{no}\:{matter}\:{how}\:{large} \\ $$$${x}\:{is}. \\ $$$${such}\:{that}\:{the}\:{upper}\:{carpet}\:{moves},\: \\ $$$${F}\geqslant\mu{mg} \\ $$$$ \\ $$$${on}\:{the}\:{lower}\:{carpet}: \\ $$$${acting}\:{friction}\:{on}\:{the}\:{contact}\:{surface} \\ $$$${to}\:{upper}\:{carpet}\:{is}\:\frac{{x}}{{b}}×\mu{mg}. \\ $$$${friction}\:{on}\:{its}\:{contact}\:{surface}\:{to}\: \\ $$$${ground}\:{is}\:\mu{mg}+\frac{{x}}{{b}}×\mu{mg},\:{which}\:{is}\: \\ $$$${always}\:{larger}\:{than}\:{the}\:{acting}\:{friction}\: \\ $$$${on}\:{its}\:{upper}\:{surface},\:{that}\:{means}\:{the}\: \\ $$$${lower}\:{carpet}\:{doesn}'{t}\:{move}. \\ $$$${total}\:{energy}\:{needed}\:{to}\:{detach}\:{both} \\ $$$${carpets}\:{completely}\:{is}\:{then}\: \\ $$$${E}\geqslant{F}×{b}=\mu{mgb} \\ $$
Commented by ajfour last updated on 10/Jul/21
Thanks sir, i hadn′t thought  carefully, better question   would be if the ground were  frictionless.
$${Thanks}\:{sir},\:{i}\:{hadn}'{t}\:{thought} \\ $$$${carefully},\:{better}\:{question}\: \\ $$$${would}\:{be}\:{if}\:{the}\:{ground}\:{were} \\ $$$${frictionless}.\: \\ $$$$ \\ $$

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