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Question Number 23559 by ajfour last updated on 01/Nov/17

Commented by ajfour last updated on 01/Nov/17

For the situation shown, find the tension T_0 at midpoint P. Also find the length l of rope in terms of a. Given mass of rope is m. Angle of rope with the horizontal is 60° at the ends.

$${For}\:{the}\:{situation}\:{shown},\:{find}\: \\ $$$${the}\:\boldsymbol{{tension}}\:\boldsymbol{{T}}_{\mathrm{0}} \:{at}\:{midpoint}\:{P}. \\ $$$${Also}\:{find}\:{the}\:{length}\:\boldsymbol{{l}}\:{of}\:{rope}\:{in} \\ $$$${terms}\:{of}\:\boldsymbol{{a}}.\:{Given}\:{mass}\:{of}\:{rope} \\ $$$${is}\:\boldsymbol{{m}}.\:{Angle}\:{of}\:{rope}\:{with}\:{the} \\ $$$${horizontal}\:{is}\:\mathrm{60}°\:{at}\:{the}\:{ends}. \\ $$

Answered by mrW1 last updated on 01/Nov/17

T_0 =((mg)/2)×tan 30^° =((mg)/(2(√3)))

$$\mathrm{T}_{\mathrm{0}} =\frac{\mathrm{mg}}{\mathrm{2}}×\mathrm{tan}\:\mathrm{30}^{°} =\frac{\mathrm{mg}}{\mathrm{2}\sqrt{\mathrm{3}}} \\ $$

Commented by ajfour last updated on 02/Nov/17

....and the length of rope,sir ?

$$....{and}\:{the}\:{length}\:{of}\:{rope},{sir}\:? \\ $$

Commented by ajfour last updated on 01/Nov/17

Correct Sir .

$${Correct}\:{Sir}\:. \\ $$

Commented by mrW1 last updated on 02/Nov/17

α=(T_0 /(ρg))=((LT_0 )/(mg))=(L/(2(√3))) (=(L/(2tan θ))) tan θ=sinh ((a/α)) ⇒(a/α)=sinh^(−1) (tan θ) ⇒((2a×tan θ)/L)=sinh^(−1) (tan θ) ⇒L=2a×((tan θ)/(sinh^(−1) (tan θ))) =2a×((tan θ)/(ln (tan θ+(√(1+tan^2 θ))))) =2a×((√3)/(ln ((√3)+2))) ≈2.63a

$$\alpha=\frac{\mathrm{T}_{\mathrm{0}} }{\rho\mathrm{g}}=\frac{\mathrm{LT}_{\mathrm{0}} }{\mathrm{mg}}=\frac{\mathrm{L}}{\mathrm{2}\sqrt{\mathrm{3}}}\:\:\left(=\frac{\mathrm{L}}{\mathrm{2tan}\:\theta}\right) \\ $$$$\mathrm{tan}\:\theta=\mathrm{sinh}\:\left(\frac{\mathrm{a}}{\alpha}\right) \\ $$$$\Rightarrow\frac{\mathrm{a}}{\alpha}=\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\theta\right) \\ $$$$\Rightarrow\frac{\mathrm{2a}×\mathrm{tan}\:\theta}{\mathrm{L}}=\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\theta\right) \\ $$$$\Rightarrow\mathrm{L}=\mathrm{2a}×\frac{\mathrm{tan}\:\theta}{\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\theta\right)} \\ $$$$=\mathrm{2a}×\frac{\mathrm{tan}\:\theta}{\mathrm{ln}\:\left(\mathrm{tan}\:\theta+\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta}\right)} \\ $$$$=\mathrm{2a}×\frac{\sqrt{\mathrm{3}}}{\mathrm{ln}\:\left(\sqrt{\mathrm{3}}+\mathrm{2}\right)} \\ $$$$\approx\mathrm{2}.\mathrm{63a} \\ $$

Commented by ajfour last updated on 02/Nov/17

l=((2(√3)a)/(ln (2+(√3)))) ??

$$\:\:\boldsymbol{{l}}=\frac{\mathrm{2}\sqrt{\mathrm{3}}\boldsymbol{{a}}}{\mathrm{ln}\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)}\:\:?? \\ $$

Commented by mrW1 last updated on 02/Nov/17

yes, with θ=60°.

$$\mathrm{yes},\:\mathrm{with}\:\theta=\mathrm{60}°. \\ $$

Commented by ajfour last updated on 02/Nov/17

thank you Sir.

$${thank}\:{you}\:{Sir}. \\ $$

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