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

Commented by ajfour last updated on 07/Feb/25

$$Lengthsa,b,caregiven.Wallisfriction$$$$-less.Find\theta and\varphi andT.$$

Commented by Tawa11 last updated on 07/Feb/25

$$\mathrm{Sir},\mathrm{I}\mathrm{reposted}\mathrm{Q}111432.\mathrm{Your}\mathrm{question}.$$$$\mathrm{Sir}\mathrm{Mr}\mathrm{W}\mathrm{has}\mathrm{solved}.\mathrm{Q}216388$$$$\mathrm{Weldone}\mathrm{sirs}.$$

Commented by ajfour last updated on 07/Feb/25

$$yeahisaw.carryon!$$

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

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

Commented by ajfour last updated on 07/Feb/25

$$yeahigot$$$$(3s^2+1){(as+b)}^2=16c^2{s}^2$$$$s=\sin \theta$$

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

$$example:a=7,b=1,l=6$$

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

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

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

$$yes,igotthesame.$$

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