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Question Number 55217 by peter frank last updated on 19/Feb/19

Answered by mr W last updated on 19/Feb/19

Commented by mr W last updated on 20/Feb/19

$$\left(i\right)isclear,sincethreadisweightless.$$$$$$$$\left(ii\right)$$$$tensioninthread=T$$$$2T\cos \frac{\theta }{2}=W$$$$2T\sin \beta +\gamma \lambda =W$$$$\Rightarrow 2T(\cos \frac{\theta }{2}-\sin \beta )=\gamma \lambda$$$$\Rightarrow T=\frac{\gamma \lambda }{2(\cos \frac{\theta }{2}-\sin \beta )}$$$$$$$$anotherway:$$$$T=\gamma R$$$$R(\cos \frac{\theta }{2}-\sin \beta )=\frac{\lambda }{2}$$$$\Rightarrow R=\frac{\lambda }{2(\cos \frac{\theta }{2}-\sin \beta )}$$$$\Rightarrow T=\frac{\gamma \lambda }{2(\cos \frac{\theta }{2}-\sin \beta )}$$

Commented by peter frank last updated on 20/Feb/19

$$\mathrm{why}\cos \frac{\theta }{2}?$$

Commented by mr W last updated on 20/Feb/19

$$youcanseeinmydiagram.if\theta is$$$$theanglebetweenthetwopartsof$$$$thethreadatmidpoint,then\frac{\theta }{2}is$$$$theanglebetweenthreadandvertical$$$$direction.$$

Commented by mr W last updated on 20/Feb/19

$$oryoudrawadiagramshowingwhich$$$$anglesaremeantwith\beta and\theta inthe$$$$question.$$

Commented by peter frank last updated on 22/Feb/19

$$\mathrm{thank}\mathrm{sir}$$

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