Question-53824

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

Commented by ajfour last updated on 26/Jan/19

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

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

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

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

$${exactly}! \\ $$

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

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

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

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

Commented by ajfour last updated on 26/Jan/19