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Question Number 53824 by ajfour last updated on 26/Jan/19

Commented by ajfour last updated on 26/Jan/19

Cone is solid, wall is smooth. Find maximum length of string possible to hold the cone against the wall.

$${Cone}\:{is}\:{solid},\:{wall}\:{is}\:{smooth}.\:{Find} \\ $$$${maximum}\:{length}\:{of}\:{string}\:{possible} \\ $$$${to}\:{hold}\:{the}\:{cone}\:{against}\:{the}\:{wall}. \\ $$

Answered by mr W last updated on 26/Jan/19

Commented by mr W last updated on 26/Jan/19

if the wall is smooth, the maximum string length is when the string passes the point G. r=h tan α e=(h/4)=centroid of cone tan θ=((h−e)/r)=(3/(4 tan α)) ⇒l_(max) =(h/(sin θ))=h(√(1+(((4 tan α)/3))^2 ))

$${if}\:{the}\:{wall}\:{is}\:{smooth},\:{the}\:{maximum} \\ $$$${string}\:{length}\:{is}\:{when}\:{the}\:{string}\: \\ $$$${passes}\:{the}\:{point}\:{G}. \\ $$$${r}={h}\:\mathrm{tan}\:\alpha \\ $$$${e}=\frac{{h}}{\mathrm{4}}={centroid}\:{of}\:{cone} \\ $$$$\mathrm{tan}\:\theta=\frac{{h}−{e}}{{r}}=\frac{\mathrm{3}}{\mathrm{4}\:\mathrm{tan}\:\alpha} \\ $$$$\Rightarrow{l}_{{max}} =\frac{{h}}{\mathrm{sin}\:\theta}={h}\sqrt{\mathrm{1}+\left(\frac{\mathrm{4}\:\mathrm{tan}\:\alpha}{\mathrm{3}}\right)^{\mathrm{2}} } \\ $$

Commented by ajfour last updated on 26/Jan/19

correct answer Sir; elegant method, i am not getting this right answer.

$${correct}\:{answer}\:{Sir};\:{elegant}\:{method}, \\ $$$${i}\:{am}\:{not}\:{getting}\:{this}\:{right}\:{answer}. \\ $$

Commented by ajfour last updated on 26/Jan/19

Tcos θ=Mg Mg((h/4))=Thcos θ−Trsin θ ⇒ (h/4)=h−rtan θ = h−htan αtan θ ⇒ tan θ = (3/(4tan α)) l=h/sin θ = h(√(1+((16)/9)tan^2 α)) .

$${T}\mathrm{cos}\:\theta={Mg} \\ $$$${Mg}\left(\frac{{h}}{\mathrm{4}}\right)={Th}\mathrm{cos}\:\theta−{Tr}\mathrm{sin}\:\theta \\ $$$$\Rightarrow\:\frac{{h}}{\mathrm{4}}={h}−{r}\mathrm{tan}\:\theta\:=\:{h}−{h}\mathrm{tan}\:\alpha\mathrm{tan}\:\theta \\ $$$$\Rightarrow\:\:\mathrm{tan}\:\theta\:=\:\frac{\mathrm{3}}{\mathrm{4tan}\:\alpha} \\ $$$$\:\:{l}={h}/\mathrm{sin}\:\theta\:=\:{h}\sqrt{\mathrm{1}+\frac{\mathrm{16}}{\mathrm{9}}\mathrm{tan}\:^{\mathrm{2}} \alpha}\:\:\:. \\ $$

Commented by mr W last updated on 26/Jan/19

exactly!

$${exactly}! \\ $$

Answered by mr W last updated on 26/Jan/19

Commented by mr W last updated on 26/Jan/19

the same question but with friction on the wall: μ f=μN tan ϕ=(f/N)=μ in this case the maximum string length is when string passes point F. r=h tan α e=(h/4) tan θ=((h−e)/(r+e tan ϕ))=((3h)/(4(h tan α+((hμ)/4))))=(3/(4 tan α+μ)) sin θ=(3/(√(3^2 +(4 tan α+μ)^2 )))=(1/(√(1+(((μ+4 tan α)/3))^2 ))) ⇒l_(max) =(h/(sin θ))=h(√(1+(((μ+4 tan α)/3))^2 ))

$${the}\:{same}\:{question}\:{but}\:{with}\:{friction} \\ $$$${on}\:{the}\:{wall}:\:\mu \\ $$$${f}=\mu{N} \\ $$$$\mathrm{tan}\:\varphi=\frac{{f}}{{N}}=\mu \\ $$$${in}\:{this}\:{case}\:{the}\:{maximum}\:{string} \\ $$$${length}\:{is}\:{when}\:{string}\:{passes}\:{point}\:{F}. \\ $$$${r}={h}\:\mathrm{tan}\:\alpha \\ $$$${e}=\frac{{h}}{\mathrm{4}} \\ $$$$\mathrm{tan}\:\theta=\frac{{h}−{e}}{{r}+{e}\:\mathrm{tan}\:\varphi}=\frac{\mathrm{3}{h}}{\mathrm{4}\left({h}\:\mathrm{tan}\:\alpha+\frac{{h}\mu}{\mathrm{4}}\right)}=\frac{\mathrm{3}}{\mathrm{4}\:\mathrm{tan}\:\alpha+\mu} \\ $$$$\mathrm{sin}\:\theta=\frac{\mathrm{3}}{\sqrt{\mathrm{3}^{\mathrm{2}} +\left(\mathrm{4}\:\mathrm{tan}\:\alpha+\mu\right)^{\mathrm{2}} }}=\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\left(\frac{\mu+\mathrm{4}\:\mathrm{tan}\:\alpha}{\mathrm{3}}\right)^{\mathrm{2}} }} \\ $$$$\Rightarrow{l}_{{max}} =\frac{{h}}{\mathrm{sin}\:\theta}={h}\sqrt{\mathrm{1}+\left(\frac{\mu+\mathrm{4}\:\mathrm{tan}\:\alpha}{\mathrm{3}}\right)^{\mathrm{2}} } \\ $$

Commented by ajfour last updated on 26/Jan/19

need some time comprehending them, Sir.

$${need}\:{some}\:{time}\:{comprehending} \\ $$$${them},\:{Sir}. \\ $$

Commented by mr W last updated on 26/Jan/19

the principle is: when a body is in equilibrium under three forces, then these three forces must intersect at one point. such that the string is as long as possible, the normal force from the wall should be as high as possible, the upmost position of the normal force is the upmost point of the base of the cone.

$${the}\:{principle}\:{is}:\:{when}\:{a}\:{body}\:{is}\:{in} \\ $$$${equilibrium}\:{under}\:{three}\:{forces},\:{then} \\ $$$${these}\:{three}\:{forces}\:{must}\:{intersect}\:{at} \\ $$$${one}\:{point}. \\ $$$${such}\:{that}\:{the}\:{string}\:{is}\:{as}\:{long}\:{as} \\ $$$${possible},\:{the}\:{normal}\:{force}\:{from}\:{the} \\ $$$${wall}\:{should}\:{be}\:{as}\:{high}\:{as}\:{possible}, \\ $$$${the}\:{upmost}\:{position}\:{of}\:{the}\:{normal} \\ $$$${force}\:{is}\:{the}\:{upmost}\:{point}\:{of}\:{the} \\ $$$${base}\:{of}\:{the}\:{cone}. \\ $$

Commented by ajfour last updated on 26/Jan/19

Thanks Sir, very useful concept; understand your solution better now.

$${Thanks}\:{Sir},\:{very}\:{useful}\:{concept}; \\ $$$${understand}\:{your}\:{solution}\:{better}\:{now}. \\ $$

Commented by Otchere Abdullai last updated on 26/Jan/19

thus why i said this man W is a professor!

$${thus}\:{why}\:{i}\:{said}\:{this}\:{man}\:{W}\:{is}\:{a}\: \\ $$$${professor}! \\ $$

Commented by ajfour last updated on 27/Jan/19

more than that, an Ideal one!

$${more}\:{than}\:{that},\:{an}\:{Ideal}\:{one}! \\ $$

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