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Question-52468




Question Number 52468 by mr W last updated on 08/Jan/19
Commented by mr W last updated on 08/Jan/19
No friction.  Find θ=?
$${No}\:{friction}. \\ $$$${Find}\:\theta=? \\ $$
Commented by ajfour last updated on 08/Jan/19
Commented by ajfour last updated on 08/Jan/19
let mutual reaction force (not shown  in figure) be T.  Fcos α− Tsin θ= mg    ..(i)  Ncos β+Tsin θ = Mg   ...(ii)  Fsin α=Tcos θ = Nsin β   ...(iii)  hence from (i)÷(ii) & using (iii)  ((((cos θcos α)/(sin α))−sin θ)/(((cos θcos β)/(sin β))+sin θ)) = λ  (=(m/M))  ⇒  (1/(tan α))−(λ/(tan β)) = (λ+1)tan θ  ⇒  θ = tan^(−1) [((tan β−λtan α)/((λ+1)tan αtan β))].
$${let}\:{mutual}\:{reaction}\:{force}\:\left({not}\:{shown}\right. \\ $$$$\left.{in}\:{figure}\right)\:{be}\:{T}. \\ $$$${F}\mathrm{cos}\:\alpha−\:{T}\mathrm{sin}\:\theta=\:{mg}\:\:\:\:..\left({i}\right) \\ $$$${N}\mathrm{cos}\:\beta+{T}\mathrm{sin}\:\theta\:=\:{Mg}\:\:\:…\left({ii}\right) \\ $$$${F}\mathrm{sin}\:\alpha={T}\mathrm{cos}\:\theta\:=\:{N}\mathrm{sin}\:\beta\:\:\:…\left({iii}\right) \\ $$$${hence}\:{from}\:\left({i}\right)\boldsymbol{\div}\left({ii}\right)\:\&\:{using}\:\left({iii}\right) \\ $$$$\frac{\frac{\mathrm{cos}\:\theta\mathrm{cos}\:\alpha}{\mathrm{sin}\:\alpha}−\mathrm{sin}\:\theta}{\frac{\mathrm{cos}\:\theta\mathrm{cos}\:\beta}{\mathrm{sin}\:\beta}+\mathrm{sin}\:\theta}\:=\:\lambda\:\:\left(=\frac{{m}}{{M}}\right) \\ $$$$\Rightarrow\:\:\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}−\frac{\lambda}{\mathrm{tan}\:\beta}\:=\:\left(\lambda+\mathrm{1}\right)\mathrm{tan}\:\theta \\ $$$$\Rightarrow\:\:\theta\:=\:\mathrm{tan}^{−\mathrm{1}} \left[\frac{\mathrm{tan}\:\beta−\lambda\mathrm{tan}\:\alpha}{\left(\lambda+\mathrm{1}\right)\mathrm{tan}\:\alpha\mathrm{tan}\:\beta}\right]. \\ $$
Commented by mr W last updated on 08/Jan/19
thank you sir!   answer is correct.
$${thank}\:{you}\:{sir}!\: \\ $$$${answer}\:{is}\:{correct}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 08/Jan/19
Commented by tanmay.chaudhury50@gmail.com last updated on 08/Jan/19
Commented by mr W last updated on 08/Jan/19
thank you sir!
$${thank}\:{you}\:{sir}!\: \\ $$
Answered by mr W last updated on 08/Jan/19
Commented by mr W last updated on 08/Jan/19
total weight of both balls =(M+m)g  its position is in the line CD  AD=((M(R+r))/(M+m))  BD=((m(R+r))/(M+m))  in ΔABC:  ∠A=(π/2)−α−θ=(π/2)−(α+θ)  ∠B=(π/2)−β+θ=(π/2)−(β−θ)  CD=((AD×sin ∠A)/(sin α))=((M(R+r))/(M+m))×((cos (α+θ))/(sin α))  CD=((BD×sin ∠B)/(sin β))=((m(R+r))/(M+m))×((cos (β−θ))/(sin β))  ⇒((M(R+r))/(M+m))×((cos (α+θ))/(sin α))=((m(R+r))/(M+m))×((cos (β−θ))/(sin β))  ⇒((cos (α+θ))/(sin α))=(m/M)×((cos (β−θ))/(sin β))  with λ=(m/M)  ⇒((cos θ)/(tan α))−sin θ=λ(((cos θ)/(tan β))+sin θ)  ⇒tan θ=(1/(1+λ))((1/(tan α))−(λ/(tan β)))  ⇒θ=tan^(−1) [(1/(1+λ))((1/(tan α))−(λ/(tan β)))]
$${total}\:{weight}\:{of}\:{both}\:{balls}\:=\left({M}+{m}\right){g} \\ $$$${its}\:{position}\:{is}\:{in}\:{the}\:{line}\:{CD} \\ $$$${AD}=\frac{{M}\left({R}+{r}\right)}{{M}+{m}} \\ $$$${BD}=\frac{{m}\left({R}+{r}\right)}{{M}+{m}} \\ $$$${in}\:\Delta{ABC}: \\ $$$$\angle{A}=\frac{\pi}{\mathrm{2}}−\alpha−\theta=\frac{\pi}{\mathrm{2}}−\left(\alpha+\theta\right) \\ $$$$\angle{B}=\frac{\pi}{\mathrm{2}}−\beta+\theta=\frac{\pi}{\mathrm{2}}−\left(\beta−\theta\right) \\ $$$${CD}=\frac{{AD}×\mathrm{sin}\:\angle{A}}{\mathrm{sin}\:\alpha}=\frac{{M}\left({R}+{r}\right)}{{M}+{m}}×\frac{\mathrm{cos}\:\left(\alpha+\theta\right)}{\mathrm{sin}\:\alpha} \\ $$$${CD}=\frac{{BD}×\mathrm{sin}\:\angle{B}}{\mathrm{sin}\:\beta}=\frac{{m}\left({R}+{r}\right)}{{M}+{m}}×\frac{\mathrm{cos}\:\left(\beta−\theta\right)}{\mathrm{sin}\:\beta} \\ $$$$\Rightarrow\frac{{M}\left({R}+{r}\right)}{{M}+{m}}×\frac{\mathrm{cos}\:\left(\alpha+\theta\right)}{\mathrm{sin}\:\alpha}=\frac{{m}\left({R}+{r}\right)}{{M}+{m}}×\frac{\mathrm{cos}\:\left(\beta−\theta\right)}{\mathrm{sin}\:\beta} \\ $$$$\Rightarrow\frac{\mathrm{cos}\:\left(\alpha+\theta\right)}{\mathrm{sin}\:\alpha}=\frac{{m}}{{M}}×\frac{\mathrm{cos}\:\left(\beta−\theta\right)}{\mathrm{sin}\:\beta} \\ $$$${with}\:\lambda=\frac{{m}}{{M}} \\ $$$$\Rightarrow\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\alpha}−\mathrm{sin}\:\theta=\lambda\left(\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\beta}+\mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\frac{\mathrm{1}}{\mathrm{1}+\lambda}\left(\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}−\frac{\lambda}{\mathrm{tan}\:\beta}\right) \\ $$$$\Rightarrow\theta=\mathrm{tan}^{−\mathrm{1}} \left[\frac{\mathrm{1}}{\mathrm{1}+\lambda}\left(\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}−\frac{\lambda}{\mathrm{tan}\:\beta}\right)\right] \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jan/19
thank you sir...in my calculation i have put  cos(θ+β)...but should be cos(β−θ)  putting that cos(β−θ)...i could have reached  destination..
$${thank}\:{you}\:{sir}…{in}\:{my}\:{calculation}\:{i}\:{have}\:{put} \\ $$$${cos}\left(\theta+\beta\right)…{but}\:{should}\:{be}\:{cos}\left(\beta−\theta\right) \\ $$$${putting}\:{that}\:{cos}\left(\beta−\theta\right)…{i}\:{could}\:{have}\:{reached} \\ $$$${destination}.. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jan/19
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jan/19
Commented by mr W last updated on 09/Jan/19
thanks for your efforts sir!
$${thanks}\:{for}\:{your}\:{efforts}\:{sir}! \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jan/19
thank you sir...
$${thank}\:{you}\:{sir}… \\ $$

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