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Question Number 212867 by mr W last updated on 25/Oct/24
solution to Q212752
$${solution}\:{to}\:{Q}\mathrm{212752} \\ $$
Commented by mr W last updated on 26/Oct/24
Commented by mr W last updated on 28/Oct/24
assume a≥b AF^2 =l^2 +c^2 −2cl cos θ AF^2 =a^2 +b^2 −2ab cos (α+β) cos (α+β)=((a^2 +b^2 −l^2 −c^2 +2lc cos θ)/(2ab)) ⇒sin (α+β)=(√(1−(((a^2 +b^2 −l^2 −c^2 +2lc cos θ)/(2ab)))^2 )) OC=h ((sin α)/(0.5l))=((sin (90°−α−θ))/h)=((cos (α+θ))/h) ((sin β)/(0.5l))=((sin (90°−β+θ))/h)=((cos (β−θ))/h) h=((l cos (α+θ))/(2 sin α))=((l cos (β−θ))/(2 sin β)) ((sin α)/(sin β))=((cos (α+θ))/(cos (β−θ)))=((cos α cos θ−sin α sin θ)/(cos β cos θ+sin β sin θ)) ⇒(1/(tan α))−(1/(tan β))=2 tan θ ((sin ϕ)/c)=((sin θ)/(AF)) sin ϕ=((c sin θ)/( (√(l^2 +c^2 −2cl cos θ)))) ⇒ϕ=sin^(−1) ((c sin θ)/( (√(l^2 +c^2 −2cl cos θ)))) ((sin (90°−β+ϕ+θ))/a)=((sin (90°−α−ϕ−θ))/b)=((sin (α+β))/(AF)) ((cos (β−ϕ−θ))/a)=((cos (α+ϕ+θ))/b)=(1/λ) with λ=((AF)/(sin (α+β)))=(√((l^2 +c^2 −2cl cos θ)/(1−(((a^2 +b^2 −l^2 −c^2 +2lc cos θ)/(2ab)))^2 ))) cos (α+ϕ+θ)=(b/λ) ⇒α=cos^(−1) (b/λ)−ϕ−θ cos (β−ϕ−θ)=(a/λ) ⇒β=cos^(−1) (a/λ)+ϕ+θ ⇒(1/(tan (cos^(−1) (b/λ)−ϕ−θ)))−(1/(tan (cos^(−1) (a/λ)+ϕ+θ)))=2 tan θ we can solve this equation to get θ. (T_a /(sin β))=(T_b /(sin α))=((mg)/(sin (α+β))) ⇒T_a =((mg sin β)/(sin (α+β))) ⇒T_b =((mg sin α)/(sin (α+β)))
$${assume}\:{a}\geqslant{b} \\ $$$${AF}^{\mathrm{2}} ={l}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{cl}\:\mathrm{cos}\:\theta \\ $$$${AF}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{cos}\:\left(\alpha+\beta\right) \\ $$$$\mathrm{cos}\:\left(\alpha+\beta\right)=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{l}^{\mathrm{2}} −{c}^{\mathrm{2}} +\mathrm{2}{lc}\:\mathrm{cos}\:\theta}{\mathrm{2}{ab}} \\ $$$$\Rightarrow\mathrm{sin}\:\left(\alpha+\beta\right)=\sqrt{\mathrm{1}−\left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{l}^{\mathrm{2}} −{c}^{\mathrm{2}} +\mathrm{2}{lc}\:\mathrm{cos}\:\theta}{\mathrm{2}{ab}}\right)^{\mathrm{2}} }\: \\ $$$$ \\ $$$${OC}={h} \\ $$$$\frac{\mathrm{sin}\:\alpha}{\mathrm{0}.\mathrm{5}{l}}=\frac{\mathrm{sin}\:\left(\mathrm{90}°−\alpha−\theta\right)}{{h}}=\frac{\mathrm{cos}\:\left(\alpha+\theta\right)}{{h}}\:\: \\ $$$$\frac{\mathrm{sin}\:\beta}{\mathrm{0}.\mathrm{5}{l}}=\frac{\mathrm{sin}\:\left(\mathrm{90}°−\beta+\theta\right)}{{h}}=\frac{\mathrm{cos}\:\left(\beta−\theta\right)}{{h}}\:\: \\ $$$${h}=\frac{{l}\:\mathrm{cos}\:\left(\alpha+\theta\right)}{\mathrm{2}\:\mathrm{sin}\:\alpha}=\frac{{l}\:\mathrm{cos}\:\left(\beta−\theta\right)}{\mathrm{2}\:\mathrm{sin}\:\beta} \\ $$$$\frac{\mathrm{sin}\:\alpha}{\mathrm{sin}\:\beta}=\frac{\mathrm{cos}\:\left(\alpha+\theta\right)}{\mathrm{cos}\:\left(\beta−\theta\right)}=\frac{\mathrm{cos}\:\alpha\:\mathrm{cos}\:\theta−\mathrm{sin}\:\alpha\:\mathrm{sin}\:\theta}{\mathrm{cos}\:\beta\:\mathrm{cos}\:\theta+\mathrm{sin}\:\beta\:\mathrm{sin}\:\theta} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}−\frac{\mathrm{1}}{\mathrm{tan}\:\beta}=\mathrm{2}\:\mathrm{tan}\:\theta \\ $$$$ \\ $$$$\frac{\mathrm{sin}\:\varphi}{{c}}=\frac{\mathrm{sin}\:\theta}{{AF}} \\ $$$$\mathrm{sin}\:\varphi=\frac{{c}\:\mathrm{sin}\:\theta}{\:\sqrt{{l}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{cl}\:\mathrm{cos}\:\theta}} \\ $$$$\Rightarrow\varphi=\mathrm{sin}^{−\mathrm{1}} \frac{{c}\:\mathrm{sin}\:\theta}{\:\sqrt{{l}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{cl}\:\mathrm{cos}\:\theta}} \\ $$$$ \\ $$$$\frac{\mathrm{sin}\:\left(\mathrm{90}°−\beta+\varphi+\theta\right)}{{a}}=\frac{\mathrm{sin}\:\left(\mathrm{90}°−\alpha−\varphi−\theta\right)}{{b}}=\frac{\mathrm{sin}\:\left(\alpha+\beta\right)}{{AF}} \\ $$$$\frac{\mathrm{cos}\:\left(\beta−\varphi−\theta\right)}{{a}}=\frac{\mathrm{cos}\:\left(\alpha+\varphi+\theta\right)}{{b}}=\frac{\mathrm{1}}{\lambda} \\ $$$${with}\:\lambda=\frac{{AF}}{\mathrm{sin}\:\left(\alpha+\beta\right)}=\sqrt{\frac{{l}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{cl}\:\mathrm{cos}\:\theta}{\mathrm{1}−\left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{l}^{\mathrm{2}} −{c}^{\mathrm{2}} +\mathrm{2}{lc}\:\mathrm{cos}\:\theta}{\mathrm{2}{ab}}\right)^{\mathrm{2}} }} \\ $$$$\mathrm{cos}\:\left(\alpha+\varphi+\theta\right)=\frac{{b}}{\lambda}\: \\ $$$$\Rightarrow\alpha=\mathrm{cos}^{−\mathrm{1}} \frac{{b}}{\lambda}−\varphi−\theta \\ $$$$\mathrm{cos}\:\left(\beta−\varphi−\theta\right)=\frac{{a}}{\lambda}\: \\ $$$$\Rightarrow\beta=\mathrm{cos}^{−\mathrm{1}} \frac{{a}}{\lambda}+\varphi+\theta \\ $$$$ \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{tan}\:\left(\mathrm{cos}^{−\mathrm{1}} \frac{{b}}{\lambda}−\varphi−\theta\right)}−\frac{\mathrm{1}}{\mathrm{tan}\:\left(\mathrm{cos}^{−\mathrm{1}} \frac{{a}}{\lambda}+\varphi+\theta\right)}=\mathrm{2}\:\mathrm{tan}\:\theta \\ $$$${we}\:{can}\:{solve}\:{this}\:{equation}\:{to}\:{get}\:\theta. \\ $$$$ \\ $$$$\frac{{T}_{{a}} }{\mathrm{sin}\:\beta}=\frac{{T}_{{b}} }{\mathrm{sin}\:\alpha}=\frac{{mg}}{\mathrm{sin}\:\left(\alpha+\beta\right)} \\ $$$$\Rightarrow{T}_{{a}} =\frac{{mg}\:\mathrm{sin}\:\beta}{\mathrm{sin}\:\left(\alpha+\beta\right)} \\ $$$$\Rightarrow{T}_{{b}} =\frac{{mg}\:\mathrm{sin}\:\alpha}{\mathrm{sin}\:\left(\alpha+\beta\right)} \\ $$
Commented by mr W last updated on 28/Oct/24
Commented by mr W last updated on 28/Oct/24

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