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Question Number 169658 by Tawa11 last updated on 05/May/22

Commented by mr W last updated on 05/May/22

frictional force=active force applied on carton ⇒frictional force =150 N Answer A is correct.

$${frictional}\:{force}={active}\:{force}\:{applied} \\ $$$${on}\:{carton} \\ $$$$\Rightarrow{frictional}\:{force}\:=\mathrm{150}\:{N} \\ $$$${Answer}\:{A}\:{is}\:{correct}. \\ $$

Commented by Tawa11 last updated on 05/May/22

Ohh, God bless you sir.

$$\mathrm{Ohh},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Answered by kapoorshah last updated on 06/May/22

frictional force = coefficient of friction × normal force = η_k × (m × g) = 0.4 × 51 × 9.8 = 199.92 N since frictional force is greater than the sliding force the carton is not moving. so the frictional force = η_s × m × g = 0.5 × 51 × 9.8 = 249.9 N ≈ 250 N answer : C

$${frictional}\:{force}\:=\:{coefficient}\:{of}\:{friction}\:×\:{normal}\:{force} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\eta_{{k}} \:×\:\left({m}\:×\:{g}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{0}.\mathrm{4}\:×\:\mathrm{51}\:×\:\mathrm{9}.\mathrm{8} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{199}.\mathrm{92}\:{N} \\ $$$${since}\:{frictional}\:{force}\:{is}\:{greater}\:{than}\:{the} \\ $$$${sliding}\:{force}\:{the}\:{carton}\:{is}\:{not}\:{moving}.\: \\ $$$${so}\:{the}\:{frictional}\:{force}\:=\:\eta_{{s}} \:×\:{m}\:×\:{g} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{0}.\mathrm{5}\:×\:\mathrm{51}\:×\:\mathrm{9}.\mathrm{8} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{249}.\mathrm{9}\:{N} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\approx\:\mathrm{250}\:{N} \\ $$$${answer}\::\:{C} \\ $$

Commented by mr W last updated on 06/May/22

as far as the carton doesn′t move, there act two forces on the carton in horizontal direction: the active force F=150 N and the reactive force which is the friction force f . since both forces are in equilibrium, they must be of same size, but opposite in the direction, i.e. f=F=150 N. ⇒Answer A. certainly f must be smaller than μ_s N=μ_s mg=0.5×51×9.81=250 N, otherwise the carton begins to move. since f=150<250, it means that the carton really doesn′t move.

$${as}\:{far}\:{as}\:{the}\:{carton}\:{doesn}'{t}\:{move}, \\ $$$${there}\:{act}\:{two}\:{forces}\:{on}\:{the}\:{carton} \\ $$$${in}\:{horizontal}\:{direction}: \\ $$$${the}\:{active}\:{force}\:{F}=\mathrm{150}\:{N}\:{and} \\ $$$${the}\:{reactive}\:{force}\:{which}\:{is}\:{the} \\ $$$${friction}\:{force}\:{f}\:.\: \\ $$$${since}\:{both}\:{forces}\:{are}\:{in}\:{equilibrium},\: \\ $$$${they}\:{must}\:{be}\:{of}\:{same}\:{size},\:{but}\:{opposite}\: \\ $$$${in}\:{the}\:{direction},\:{i}.{e}. \\ $$$${f}={F}=\mathrm{150}\:{N}.\:\Rightarrow{Answer}\:{A}. \\ $$$${certainly}\:{f}\:{must}\:{be}\:{smaller}\:{than} \\ $$$$\mu_{{s}} {N}=\mu_{{s}} {mg}=\mathrm{0}.\mathrm{5}×\mathrm{51}×\mathrm{9}.\mathrm{81}=\mathrm{250}\:{N}, \\ $$$${otherwise}\:{the}\:{carton}\:{begins}\:{to}\:{move}. \\ $$$${since}\:{f}=\mathrm{150}<\mathrm{250},\:{it}\:{means}\:{that}\:{the} \\ $$$${carton}\:{really}\:{doesn}'{t}\:{move}. \\ $$

Commented by kapoorshah last updated on 06/May/22

the answer is C

$${the}\:{answer}\:{is}\:{C} \\ $$

Commented by mr W last updated on 06/May/22

then you should explain how the following is true:

$${then}\:{you}\:{should}\:{explain}\:{how}\:{the} \\ $$$${following}\:{is}\:{true}: \\ $$

Commented by mr W last updated on 06/May/22

Commented by Tawa11 last updated on 06/May/22

Sir mrW. I understand your solution since the time you solved it. The carton does not move, then f = f_r = 150N. God bless you sir.

$$\mathrm{Sir}\:\mathrm{mrW}.\:\mathrm{I}\:\mathrm{understand}\:\mathrm{your}\:\mathrm{solution}\:\mathrm{since}\:\mathrm{the}\:\mathrm{time}\:\mathrm{you}\:\mathrm{solved}\:\mathrm{it}. \\ $$$$\mathrm{The}\:\mathrm{carton}\:\mathrm{does}\:\mathrm{not}\:\mathrm{move},\:\mathrm{then}\:\:\:\mathrm{f}\:\:=\:\:\mathrm{f}_{\mathrm{r}} \:\:\:=\:\:\:\mathrm{150N}. \\ $$$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by Tawa11 last updated on 06/May/22

at kapoorshah sir. Thanks for your time. I reason like you before, but am sure 250N is not correct. That was why I posted the question to the forum. And sir mrW has cleared by doubt. [Ans = 150N]

$$\mathrm{at}\:\mathrm{kapoorshah}\:\mathrm{sir}.\:\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}. \\ $$$$\mathrm{I}\:\mathrm{reason}\:\mathrm{like}\:\mathrm{you}\:\mathrm{before},\:\mathrm{but}\:\mathrm{am}\:\mathrm{sure}\:\:\mathrm{250N}\:\mathrm{is}\:\mathrm{not}\:\mathrm{correct}. \\ $$$$\mathrm{That}\:\mathrm{was}\:\mathrm{why}\:\mathrm{I}\:\mathrm{posted}\:\mathrm{the}\:\mathrm{question}\:\mathrm{to}\:\mathrm{the}\:\mathrm{forum}. \\ $$$$\mathrm{And}\:\mathrm{sir}\:\:\mathrm{mrW}\:\:\mathrm{has}\:\mathrm{cleared}\:\mathrm{by}\:\mathrm{doubt}.\:\:\left[\mathrm{Ans}\:\:=\:\:\mathrm{150N}\right] \\ $$

Commented by mr W last updated on 06/May/22

something elementary about friction: before an object moves, the friction is called static friction. it is NOT a constant force! it is just a reaction force. if you don′t apply a force an the object to attempt to move it, the friction is zero. as far as the force you are applying on it is less than the maximal possible static friction, which is calculated by f_(max) =μ_(static) N, the object doesn′t move and the friction force f which is actually acting on the object is always equal to the active force F you are applying on the object, i.e. f=F. when you increase the active force F und reach the maximal possible static friction force f_(max) =μ_(static) N, the object begins to move. once an object moves, there acts always a friction force on the object, this friction is called kinetic friction, which is calculated by f_(kinetic) =μ_(kinetic) N. it is always there and is constant, as far as the object moves. the maximal static friction which must be overcome to make an object move is larger than the kinetic friction. we can experience this in our daily life: to make an object to start to move is more hard than just to keep it move. when the object begins to move, you can reduce the force you apply on it to keep it moving constantly. when you don′t reduce the force, the object will get an acceleration from your force and gets faster and faster, because the friction (kinetic) force acting on it is smaller than the force you are applying.

$${something}\:{elementary}\:{about}\:{friction}: \\ $$$${before}\:{an}\:{object}\:{moves},\:{the} \\ $$$${friction}\:{is}\:{called}\:{static}\:{friction}.\:{it} \\ $$$${is}\:{NOT}\:{a}\:{constant}\:{force}!\:{it}\:{is}\:{just}\:{a} \\ $$$${reaction}\:{force}.\:{if}\:{you}\:{don}'{t}\:{apply} \\ $$$${a}\:{force}\:{an}\:{the}\:{object}\:{to}\:{attempt}\:{to} \\ $$$${move}\:{it},\:{the}\:{friction}\:{is}\:{zero}.\:{as}\:{far} \\ $$$${as}\:{the}\:{force}\:{you}\:{are}\:{applying}\:{on}\:{it} \\ $$$${is}\:{less}\:{than}\:{the}\:{maximal}\:{possible} \\ $$$${static}\:{friction},\:{which}\:{is}\:{calculated} \\ $$$${by}\:{f}_{{max}} =\mu_{{static}} {N},\:{the}\:{object}\:{doesn}'{t} \\ $$$${move}\:{and}\:{the}\:{friction}\:{force}\:{f}\:{which} \\ $$$${is}\:{actually}\:{acting}\:{on}\:{the}\:{object}\:{is}\: \\ $$$${always}\:{equal}\:{to}\:{the}\:{active}\:{force}\:{F}\:{you} \\ $$$${are}\:{applying}\:{on}\:{the}\:{object},\:{i}.{e}.\:{f}={F}. \\ $$$$ \\ $$$${when}\:{you}\:{increase}\:{the}\:{active}\:{force}\:{F} \\ $$$${und}\:{reach}\:{the}\:{maximal}\:{possible} \\ $$$${static}\:{friction}\:{force}\:{f}_{{max}} =\mu_{{static}} {N}, \\ $$$${the}\:{object}\:{begins}\:{to}\:{move}.\:{once}\:{an} \\ $$$${object}\:{moves},\:{there}\:{acts}\:{always}\:{a} \\ $$$${friction}\:{force}\:{on}\:{the}\:{object},\:{this} \\ $$$${friction}\:{is}\:{called}\:{kinetic}\:{friction}, \\ $$$${which}\:{is}\:{calculated}\:{by}\:{f}_{{kinetic}} =\mu_{{kinetic}} {N}. \\ $$$${it}\:{is}\:{always}\:{there}\:{and}\:{is}\:{constant},\:{as} \\ $$$${far}\:{as}\:{the}\:{object}\:{moves}.\:{the}\: \\ $$$${maximal}\:{static}\:{friction}\:{which}\:{must} \\ $$$${be}\:{overcome}\:{to}\:{make}\:{an}\:{object}\:{move} \\ $$$${is}\:\:{larger}\:{than}\:{the}\:{kinetic}\:{friction}.\: \\ $$$${we}\:{can}\:{experience}\:{this}\:{in}\:{our} \\ $$$${daily}\:{life}:\:{to}\:{make}\:{an}\:{object}\:{to}\:{start} \\ $$$${to}\:{move}\:{is}\:{more}\:{hard}\:{than}\:{just}\:{to} \\ $$$${keep}\:{it}\:{move}. \\ $$$$ \\ $$$${when}\:{the}\:{object}\:{begins}\:{to}\:{move},\:{you} \\ $$$${can}\:{reduce}\:{the}\:{force}\:{you}\:{apply}\:{on}\:{it}\:{to} \\ $$$${keep}\:{it}\:{moving}\:{constantly}.\:{when} \\ $$$${you}\:{don}'{t}\:{reduce}\:{the}\:{force},\:{the}\:{object} \\ $$$${will}\:{get}\:{an}\:{acceleration}\:{from}\:{your} \\ $$$${force}\:{and}\:{gets}\:{faster}\:{and}\:{faster}, \\ $$$${because}\:{the}\:{friction}\:\left({kinetic}\right)\:{force} \\ $$$${acting}\:{on}\:{it}\:{is}\:{smaller}\:{than}\:{the}\:{force} \\ $$$${you}\:{are}\:{applying}. \\ $$

Commented by Tawa11 last updated on 06/May/22

God bless you sir. I really appreciate your time.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time}. \\ $$

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