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Question Number 169458 by ajfour last updated on 30/Apr/22

Commented by ajfour last updated on 30/Apr/22

$$Find{\theta }_{max}understaticconditions.$$$$sayforh=\frac{L}{2}.$$

Answered by mr W last updated on 01/May/22

Commented by mr W last updated on 04/May/22

$$C=commoncenterofgravityfrom$$$$Mandm$$$$AC\times (m+M)=m\times L+M\times \frac{L}{2}$$$$\Rightarrow AC=\frac{(2m+M)L}{2(m+M)}=b,say$$$$AB=\frac{h}{\sin \theta }$$$$\tan \varphi =\mu$$$$SC=\frac{BC}{\sin \theta }=\frac{AB-AC}{\sin \theta }=\frac{\frac{h}{\sin \theta }-b}{\sin \theta }$$$$\frac{AC}{\sin \varphi }=\frac{SC}{\sin (\frac{\pi }{2}-\varphi -\theta )}=\frac{SC}{\cos (\theta +\varphi )}$$$$\frac{b}{\sin \varphi }=\frac{\frac{h}{\sin \theta }-b}{\sin \theta \cos (\theta +\varphi )}$$$$\frac{1}{\sin \varphi }=\frac{\frac{h}{b\sin \theta }-1}{\sin \theta \cos (\theta +\varphi )}$$$$let\lambda =\frac{h}{b}=\frac{2(m+M)h}{(2m+M)L}$$$$\frac{1}{\sin \varphi }=\frac{\lambda -\sin \theta }{{\sin }^2\theta \cos (\theta +\varphi )}$$$$\frac{1}{\sin \varphi }=\frac{\lambda -\sin \theta }{{\sin }^2\theta (\cos \theta \cos \varphi -\sin \theta \sin \varphi )}$$$$\Rightarrow {\sin }^2\theta (\frac{\cos \theta }{\mu }-\sin \theta )+\sin \theta =\lambda$$$$example:$$$$m=\frac{M}{2},h=\frac{L}{2},\mu =0.5$$$$\lambda =\frac{2(0.5+1)\times 0.5}{(2\times 0.5+1)}=\frac{3}{4}$$$${\sin }^2\theta (2\cos \theta -\sin \theta )+\sin \theta =\frac{3}{4}$$$$\Rightarrow {\theta }_1\approx 27.8087°and{\theta }_2\approx 68.8197°$$

Commented by BHOOPENDRA last updated on 01/May/22

$$greatsir$$

Commented by ajfour last updated on 04/May/22

$$greatwaysir.$$

Commented by mr W last updated on 04/May/22

$$followingdiagramshowsthetwo$$$$states.equilibriumisif\theta ⩽{\theta }_{1}or$$$$\theta ⩾{\theta }_2.$$

Commented by mr W last updated on 04/May/22

Answered by ajfour last updated on 02/May/22

Commented by mr W last updated on 03/May/22

$$verynicesir!$$

Commented by ajfour last updated on 03/May/22

$$Mg\left(\frac{L}{2}\cos \theta \right)+mgL\cos \theta$$$$=\frac{Fh}{\sin \theta }$$$$N=(M+m)g-F\cos \theta$$$$\mu N=F\sin \theta$$$$\Rightarrow$$$$F(\sin \theta +\mu \cos \theta )=\mu (M+m)g$$$$\Rightarrow$$$$Mg\left(\frac{L}{2}\cos \theta \right)+mgL\cos \theta$$$$=\frac{\mu gh(M+m)}{\sin \theta (\sin \theta +\mu \cos \theta )}$$$$say\frac{m}{M}=\lambda ,h=\beta L\Rightarrow$$$$\frac{\cos \theta }{2}+\lambda \cos \theta =\frac{\mu \beta (\lambda +1)}{\sin \theta (\sin \theta +\mu \cos \theta )}$$$$\Rightarrow$$$$(\sin \theta +\mu \cos \theta )\sin \theta \cos \theta (\lambda +\frac{1}{2})$$$$=\beta \mu (\lambda +1)$$$$for\mu =\lambda =\beta =\frac{1}{2}$$$$(2\sin \theta +\cos \theta )\sin \theta \cos \theta =\frac{3}{4}$$$$16{(2t+1)}^2{t}^2=9{(1+t^2)}^3$$$$t\approx 0.5274,2.5808$$$$\theta ={\tan }^{-1}t\approx 27.807°,68.819°$$

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