A-rope-5m-long-is-fastened-to-two-hooks-4-0m-apart-on-a-horizontal-ceiling-to-the-rope-is-attached-a-10kg-mass-so-that-the-segments-of-the-rope-are-3-0m-and-2-0m-compute-the-tensionin-each-segment-

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Question Number 117594 by mathdave last updated on 12/Oct/20

$$Arope5mlongisfastenedtotwohooks$$$$4.0mapartonahorizontal$$$$ceiling.totheropeisattacheda10kg$$$$masssothatthesegmentsoftherope$$$$are3.0mand2.0m.computethe$$$$tensionineachsegment$$

Answered by mr W last updated on 12/Oct/20

Commented by mr W last updated on 12/Oct/20

$$\cos \alpha =\frac{3^2+4^2-2^2}{2\times 3\times 4}=\frac{7}{8}$$$$\cos \beta =\frac{2^2+4^2-3^2}{2\times 2\times 4}=\frac{11}{16}$$$$\frac{T_1}{\sin (\frac{\pi }{2}-\beta )}=\frac{T_2}{\sin (\frac{\pi }{2}-\alpha )}=\frac{mg}{\sin (\alpha +\beta )}$$$$\Rightarrow T_1=\frac{mg\cos \beta }{\sin (\alpha +\beta )}=\frac{10\times 10\times \frac{11}{16}}{\frac{\sqrt{15}}{8}\times \frac{11}{16}+\frac{7}{8}\times \frac{\sqrt{135}}{16}}$$$$=\frac{55\sqrt{15}}{3}=71.00N$$$$\Rightarrow T_2=\frac{mg\cos \alpha }{\sin (\alpha +\beta )}=\frac{10\times 10\times \frac{7}{8}}{\frac{\sqrt{15}}{8}\times \frac{11}{16}+\frac{7}{8}\times \frac{\sqrt{135}}{16}}$$$$=\frac{70\sqrt{15}}{3}=90.37N$$

Commented by Tawa11 last updated on 06/Sep/21

$$\mathrm{great}\mathrm{sir}$$

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