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A cylinder B, $W_B=1000N,$ dia. 40cm, hangs by a cable $AB = 40$ cm rests against a smooth wall. Find out reaction at C and $T_AB.$

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$∝ = sin^{-1} (\dfrac {20}{40})=30°$

F.B.D

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By Applying Lami’s theorem,

$∴ \dfrac {T_AB}{\sin⁡(90) } = \dfrac { R_C}{\sin⁡(150) } = \dfrac W{\sin⁡(120)} \\ ∴ R_C=(\dfrac {\sin150}{\sin 120})1000=577.35N \\ ∴ T_AB=(\sin90/(\sin 120))W=1000(1/(\sin 120))=1154.70N$

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