Question-216416

Question Number 216416 by ajfour last updated on 07/Feb/25

Commented by ajfour last updated on 07/Feb/25

Lengths a, b, c are given. Wall is friction -less. Find θ and φ and T.

$${Lengths}\:{a},\:{b},\:{c}\:{are}\:{given}.\:{Wall}\:{is}\:{friction} \\ $$$$-{less}.\:{Find}\:\theta\:{and}\:\phi\:{and}\:{T}.\: \\ $$

Commented by Tawa11 last updated on 07/Feb/25

Sir, I reposted Q111432. Your question. Sir Mr W has solved. Q216388 Weldone sirs.

$$\mathrm{Sir},\:\mathrm{I}\:\mathrm{reposted}\:\mathrm{Q111432}.\:\:\mathrm{Your}\:\mathrm{question}. \\ $$$$\mathrm{Sir}\:\mathrm{Mr}\:\mathrm{W}\:\:\mathrm{has}\:\mathrm{solved}.\:\:\mathrm{Q216388} \\ $$$$\mathrm{Weldone}\:\mathrm{sirs}. \\ $$

Commented by ajfour last updated on 07/Feb/25

$${yeah}\:{i}\:{saw}.\:{carry}\:{on}! \\ $$

Commented by ajfour last updated on 09/Feb/25

https://youtu.be/IxV2mEpa12E?si=I9LbX6zO-MuCRg2J

Answered by mr W last updated on 07/Feb/25

Commented by mr W last updated on 07/Feb/25

l=2c FC=c sin φ ((FC)/(sin ((π/2)−θ−φ)))=(c/(sin θ)) ((sin φ)/(cos (θ+φ)))=(1/(sin θ)) tan θ tan φ=(1/2) ⇒tan φ=(1/(2 tan θ)) ...(i) l cos φ=b+a sin θ ⇒cos φ=((b+a sin θ)/l) ...(ii) ((2 tan θ)/( (√(1+4 tan^2 θ))))=((b+a sin θ)/l) or 4l^2 sin^2 θ=(1+3 sin^2 θ)(b+a sin θ)^2 3a^2 sin^4 θ+6ab sin^3 θ+(a^2 +3b^2 −4l^2 ) sin^2 θ+2ab sin θ+b^2 =0 T=((mg)/(cos θ))

$${l}=\mathrm{2}{c} \\ $$$${FC}={c}\:\mathrm{sin}\:\phi \\ $$$$\frac{{FC}}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\theta−\phi\right)}=\frac{{c}}{\mathrm{sin}\:\theta} \\ $$$$\frac{\mathrm{sin}\:\phi}{\mathrm{cos}\:\left(\theta+\phi\right)}=\frac{\mathrm{1}}{\mathrm{sin}\:\theta} \\ $$$$\mathrm{tan}\:\theta\:\mathrm{tan}\:\phi=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{tan}\:\phi=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{tan}\:\theta}\:\:\:…\left({i}\right) \\ $$$${l}\:\mathrm{cos}\:\phi={b}+{a}\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow\mathrm{cos}\:\phi=\frac{{b}+{a}\:\mathrm{sin}\:\theta}{{l}}\:\:\:…\left({ii}\right) \\ $$$$\frac{\mathrm{2}\:\mathrm{tan}\:\theta}{\:\sqrt{\mathrm{1}+\mathrm{4}\:\mathrm{tan}^{\mathrm{2}} \:\theta}}=\frac{{b}+{a}\:\mathrm{sin}\:\theta}{{l}} \\ $$$${or} \\ $$$$\mathrm{4}{l}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta=\left(\mathrm{1}+\mathrm{3}\:\mathrm{sin}^{\mathrm{2}} \:\theta\right)\left({b}+{a}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} \\ $$$$\mathrm{3}{a}^{\mathrm{2}} \mathrm{sin}^{\mathrm{4}} \:\theta+\mathrm{6}{ab}\:\mathrm{sin}^{\mathrm{3}} \:\theta+\left({a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} −\mathrm{4}{l}^{\mathrm{2}} \right)\:\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{2}{ab}\:\mathrm{sin}\:\theta+{b}^{\mathrm{2}} =\mathrm{0} \\ $$$${T}=\frac{{mg}}{\mathrm{cos}\:\theta} \\ $$

Commented by ajfour last updated on 07/Feb/25

yeah i got (3s^2 +1)(as+b)^2 =16c^2 s^2 s=sin θ

$${yeah}\:{i}\:{got} \\ $$$$\left(\mathrm{3}{s}^{\mathrm{2}} +\mathrm{1}\right)\left({as}+{b}\right)^{\mathrm{2}} =\mathrm{16}{c}^{\mathrm{2}} {s}^{\mathrm{2}} \\ $$$${s}=\mathrm{sin}\:\theta \\ $$

Commented by mr W last updated on 07/Feb/25