Question-224489

Question Number 224489 by ajfour last updated on 14/Sep/25

Commented by ajfour last updated on 14/Sep/25

Find θ. string length is s. No friction at G. (sorry for my absence) Take edge lengths of cubes=2a.

$${Find}\:\theta.\:{string}\:{length}\:{is}\:{s}.\: \\ $$$${No}\:{friction}\:{at}\:{G}. \\ $$$$\left({sorry}\:{for}\:{my}\:{absence}\right) \\ $$$${Take}\:{edge}\:{lengths}\:{of}\:{cubes}=\mathrm{2}{a}. \\ $$

Commented by fantastic last updated on 15/Sep/25

$${Sirs}\:{what}\:{is}\:{the}\:{Question}? \\ $$$${i}\:{dont}\:{understand}\:{it} \\ $$

Commented by mr W last updated on 15/Sep/25

cube ABCD is connected with a string of length s on an other cube of same size and it′s in equilibrium as shown. find the angle θ.

$${cube}\:{ABCD}\:{is}\:{connected}\:{with}\:{a} \\ $$$${string}\:{of}\:{length}\:{s}\:{on}\:{an}\:{other}\:{cube} \\ $$$${of}\:{same}\:{size}\:{and}\:{it}'{s}\:{in}\:{equilibrium} \\ $$$${as}\:{shown}.\:{find}\:{the}\:{angle}\:\theta. \\ $$

Answered by mr W last updated on 16/Sep/25

Commented by mr W last updated on 17/Sep/25

((2 sin 2θ)/( (√((1/λ^2 )−4 cos^2 θ))))−tan θ (√((1/λ^2 )−4 cos^2 θ))=1+tan θ+((2 cos 2θ)/(cos θ)) ((2 sin 2θ)/( Φ))−tan θ Φ=1+tan θ+((2 cos 2θ)/(cos θ)) tan θ Φ^2 +(1+tan θ+((2 cos 2θ)/(cos θ)))Φ−2 sin 2θ=0 Φ=(√((1/λ^2 )−4 cos^2 θ))=((−(1+tan θ+((2 cos 2θ)/(cos θ)))+(√((1+tan θ+((2 cos 2θ)/(cos θ)))^2 +16 sin^2 θ)))/(2 tan θ)) ⇒(1/λ)=(s/a)=(√(4 cos^2 θ+[((−(1+tan θ+((2 cos 2θ)/(cos θ)))+(√((1+tan θ+((2 cos 2θ)/(cos θ)))^2 +16 sin^2 θ)))/(2 tan θ))]^2 ))

$$\:\frac{\mathrm{2}\:\mathrm{sin}\:\mathrm{2}\theta}{\:\sqrt{\frac{\mathrm{1}}{\lambda^{\mathrm{2}} }−\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta}}−\mathrm{tan}\:\theta\:\sqrt{\frac{\mathrm{1}}{\lambda^{\mathrm{2}} }−\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta}=\mathrm{1}+\mathrm{tan}\:\theta+\frac{\mathrm{2}\:\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{cos}\:\theta} \\ $$$$\:\frac{\mathrm{2}\:\mathrm{sin}\:\mathrm{2}\theta}{\:\Phi}−\mathrm{tan}\:\theta\:\Phi=\mathrm{1}+\mathrm{tan}\:\theta+\frac{\mathrm{2}\:\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{cos}\:\theta} \\ $$$$\:\mathrm{tan}\:\theta\:\Phi^{\mathrm{2}} +\left(\mathrm{1}+\mathrm{tan}\:\theta+\frac{\mathrm{2}\:\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{cos}\:\theta}\right)\Phi−\mathrm{2}\:\mathrm{sin}\:\mathrm{2}\theta=\mathrm{0} \\ $$$$\Phi=\sqrt{\frac{\mathrm{1}}{\lambda^{\mathrm{2}} }−\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta}=\frac{−\left(\mathrm{1}+\mathrm{tan}\:\theta+\frac{\mathrm{2}\:\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{cos}\:\theta}\right)+\sqrt{\left(\mathrm{1}+\mathrm{tan}\:\theta+\frac{\mathrm{2}\:\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{cos}\:\theta}\right)^{\mathrm{2}} +\mathrm{16}\:\mathrm{sin}^{\mathrm{2}} \:\theta}}{\mathrm{2}\:\mathrm{tan}\:\theta} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\lambda}=\frac{{s}}{{a}}=\sqrt{\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta+\left[\frac{−\left(\mathrm{1}+\mathrm{tan}\:\theta+\frac{\mathrm{2}\:\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{cos}\:\theta}\right)+\sqrt{\left(\mathrm{1}+\mathrm{tan}\:\theta+\frac{\mathrm{2}\:\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{cos}\:\theta}\right)^{\mathrm{2}} +\mathrm{16}\:\mathrm{sin}^{\mathrm{2}} \:\theta}}{\mathrm{2}\:\mathrm{tan}\:\theta}\right]^{\mathrm{2}} } \\ $$

Commented by ajfour last updated on 16/Sep/25

thanks sir. Understood. Elegant diagrams make it look gorgeous.

Commented by mr W last updated on 16/Sep/25

let λ=(a/s) ((sin β)/(2a))=((sin ((π/2)−θ))/s)=((sin α)/(AG)) ⇒sin β=((2a cos θ)/s)=2λ cos θ ⇒cos β=(√(1−4λ^2 cos^2 θ)) α=β−((π/2)−θ)=β+θ−(π/2) AG=((s sin α)/(cos θ))=((−s cos (β+θ))/(cos θ))=((s(sin β sin θ−cos β cos θ))/(cos θ)) AG=s(tan θ sin β−cos β) SG=AG tan β=s(tan θ sin β−cos β) tan β (SG−a)=(a−AG) tan θ [s(tan θ sin β−cos β) tan β−a]=[a−s(tan θ sin β−cos β)] tan θ ((2λ^2 sin 2θ)/( (√(1−4λ^2 cos^2 θ))))−tan θ (√(1−4λ^2 cos^2 θ))=λ(1+tan θ)+((2λ cos 2θ)/(cos θ)) example: a=2, s=3 ⇒θ≈49.8620141887° (T/(sin ((π/2)−θ)))=((mg)/(sin ((π/2)−β))) ⇒(T/(mg))=((cos θ)/(cos β))=((cos θ)/( (√(1−4λ^2 cos^2 θ))))

$${let}\:\lambda=\frac{{a}}{{s}} \\ $$$$\frac{\mathrm{sin}\:\beta}{\mathrm{2}{a}}=\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\theta\right)}{{s}}=\frac{\mathrm{sin}\:\alpha}{{AG}} \\ $$$$\Rightarrow\mathrm{sin}\:\beta=\frac{\mathrm{2}{a}\:\mathrm{cos}\:\theta}{{s}}=\mathrm{2}\lambda\:\mathrm{cos}\:\theta\:\Rightarrow\mathrm{cos}\:\beta=\sqrt{\mathrm{1}−\mathrm{4}\lambda^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$$$\alpha=\beta−\left(\frac{\pi}{\mathrm{2}}−\theta\right)=\beta+\theta−\frac{\pi}{\mathrm{2}} \\ $$$${AG}=\frac{{s}\:\mathrm{sin}\:\alpha}{\mathrm{cos}\:\theta}=\frac{−{s}\:\mathrm{cos}\:\left(\beta+\theta\right)}{\mathrm{cos}\:\theta}=\frac{{s}\left(\mathrm{sin}\:\beta\:\mathrm{sin}\:\theta−\mathrm{cos}\:\beta\:\mathrm{cos}\:\theta\right)}{\mathrm{cos}\:\theta} \\ $$$${AG}={s}\left(\mathrm{tan}\:\theta\:\mathrm{sin}\:\beta−\mathrm{cos}\:\beta\right) \\ $$$${SG}={AG}\:\mathrm{tan}\:\beta={s}\left(\mathrm{tan}\:\theta\:\mathrm{sin}\:\beta−\mathrm{cos}\:\beta\right)\:\mathrm{tan}\:\beta \\ $$$$\left({SG}−{a}\right)=\left({a}−{AG}\right)\:\mathrm{tan}\:\theta \\ $$$$\left[{s}\left(\mathrm{tan}\:\theta\:\mathrm{sin}\:\beta−\mathrm{cos}\:\beta\right)\:\mathrm{tan}\:\beta−{a}\right]=\left[{a}−{s}\left(\mathrm{tan}\:\theta\:\mathrm{sin}\:\beta−\mathrm{cos}\:\beta\right)\right]\:\mathrm{tan}\:\theta \\ $$$$\:\frac{\mathrm{2}\lambda^{\mathrm{2}} \:\mathrm{sin}\:\mathrm{2}\theta}{\:\sqrt{\mathrm{1}−\mathrm{4}\lambda^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta}}−\mathrm{tan}\:\theta\:\sqrt{\mathrm{1}−\mathrm{4}\lambda^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta}=\lambda\left(\mathrm{1}+\mathrm{tan}\:\theta\right)+\frac{\mathrm{2}\lambda\:\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{cos}\:\theta} \\ $$$${example}:\:{a}=\mathrm{2},\:{s}=\mathrm{3} \\ $$$$\Rightarrow\theta\approx\mathrm{49}.\mathrm{8620141887}° \\ $$$$\frac{{T}}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\theta\right)}=\frac{{mg}}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\beta\right)} \\ $$$$\Rightarrow\frac{{T}}{{mg}}=\frac{\mathrm{cos}\:\theta}{\mathrm{cos}\:\beta}=\frac{\mathrm{cos}\:\theta}{\:\sqrt{\mathrm{1}−\mathrm{4}\lambda^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta}} \\ $$

Commented by mr W last updated on 15/Sep/25