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Question Number 202816 by ajfour last updated on 03/Jan/24

Commented by ajfour last updated on 03/Jan/24

$$Withwhatconstantforceshould$$$$trolleybepushedtowardsleftsuch$$$$thatinclinationofplankdontchange$$$$andsystemmovesatuniformspeed?$$

Commented by mr W last updated on 04/Jan/24

$$isthetrolleyfixedwiththeplank$$$$andthereforetheinclinationofthe$$$$plankgiven,e.g.withinclanation$$$$angle\theta ?$$

Commented by mr W last updated on 04/Jan/24

Commented by mr W last updated on 04/Jan/24

$$orthetrolleyisnotfixedwiththe$$$$plank,buttheircontactisfrictionless?$$

Commented by mr W last updated on 04/Jan/24

$$ihavesolvedbothcases.$$

Answered by a.lgnaoui last updated on 03/Jan/24

$${\mathrm{F}}_{\mathbf{x}}-{\mathbf{F}}_{\mathbf{f}}=\mathbf{constante}$$$$\mathbf{M}\frac{\mathbf{dv}}{\mathbf{dt}}-\mathit{\mu }\mathbf{m}\mathrm{v}\mathbf{x}=0$$$${\mathbf{v}}_{\mathbf{x}}=\mathbf{v}\cos \mathit{\alpha }=\mathbf{v}\sqrt{1-\frac{{\mathbf{b}}^2}{{\mathbf{L}}^2}}$$$$\Rightarrow \mathbf{M}\frac{\mathbf{dv}}{\mathbf{dt}}=\mathbf{mv}\frac{\sqrt{{\mathrm{L}}^2-{\mathbf{b}}^2}}{\mathbf{L}}$$$$\mathbf{donc}\mathbf{F}=\frac{\mathbf{mv}\sqrt{{\mathbf{L}}^2-{\mathbf{b}}^2}}{\mathbf{L}}$$$$\{\begin{array}{l}\mathbf{m};\mathbf{mssse}\mathbf{de}[\mathbf{la}\mathbf{tige}(\mathbf{L}:\mathrm{longeur})\\ \mathbf{v}=\mathbf{vitesse}[\mathbf{de}[\mathbf{deplacement}\end{array}$$

Answered by mr W last updated on 04/Jan/24

$$CASE1:$$$$trolleyandplankareconnectedwith$$$$ahingeandtheinclanationofthe$$$$plankisknown\left(\theta \right).$$

Commented by mr W last updated on 04/Jan/24

Commented by mr W last updated on 04/Jan/24

$$\tan \varphi =\mu$$$$\frac{CD}{\sin (\frac{\pi }{2}-\varphi -\theta )}=\frac{AC}{\sin \varphi }$$$$\Rightarrow CD=\frac{L\cos (\varphi +\theta )}{2\sin \varphi }$$$$=\frac{L}{2}(\frac{\cos \theta }{\tan \varphi }-\sin \theta )=\frac{L}{2}(\frac{\cos \theta }{\mu }-\sin \theta )$$$$\frac{CD}{\sin (\frac{\pi }{2}-\phi +\theta )}=\frac{CB}{\sin \phi }$$$$CD=\frac{(\frac{b}{\sin \theta }-\frac{L}{2})\cos (\phi -\theta )}{\sin \phi }$$$$=(\frac{b}{\sin \theta }-\frac{L}{2})(\frac{\cos \theta }{\tan \phi }+\sin \theta )$$$$\frac{L}{2}(\frac{\cos \theta }{\mu }-\sin \theta )=(\frac{b}{\sin \theta }-\frac{L}{2})(\frac{\cos \theta }{\tan \phi }+\sin \theta )$$$$\frac{1}{\mu }-\tan \theta =(\frac{2b}{L\sin \theta }-1)(\frac{1}{\tan \phi }+\tan \theta )$$$$\Rightarrow \frac{1}{\tan \phi }=\frac{\frac{1}{\mu }-\tan \theta }{\frac{2b}{L\sin \theta }-1}-\tan \theta$$$$\frac{R}{\sin \phi }=\frac{mg}{\sin (\varphi +\phi )}$$$$R=\frac{mg\sin \phi }{\sin (\varphi +\phi )}=\frac{mg}{\frac{\sin \varphi }{\tan \phi }+\cos \varphi }$$$$F=R\sin \varphi =\frac{mg}{\frac{1}{\tan \phi }+\frac{1}{\tan \varphi }}$$$$=\frac{mg}{\frac{\frac{1}{\mu }-\tan \theta }{\frac{2b}{L\sin \theta }-1}-\tan \theta +\frac{1}{\mu }}$$$$\Rightarrow \frac{F}{mg}=\frac{1}{(\frac{1}{\mu }-\tan \theta )(\frac{1}{\frac{2b}{L\sin \theta }-1}+1)}$$

Answered by mr W last updated on 04/Jan/24

$$CASE2:$$$$trolleyandplankarenotfixed$$$$connected,buttheircontactis$$$$frictionless.$$

Commented by mr W last updated on 04/Jan/24

Commented by mr W last updated on 04/Jan/24

$$\tan \varphi =\mu$$$$\tan {\varphi }_1={\mu }_1=0$$$$\Rightarrow {\varphi }_1=0\Rightarrow \phi =\theta$$$$\theta ⩾{\sin }^{-1}\frac{b}{L}$$$$\frac{CD}{\sin (\frac{\pi }{2}-\varphi -\theta )}=\frac{AC}{\sin \varphi }$$$$\Rightarrow CD=\frac{L\cos (\varphi +\theta )}{2\sin \varphi }$$$$=\frac{L}{2}(\frac{\cos \theta }{\tan \varphi }-\sin \theta )=\frac{L}{2}(\frac{\cos \theta }{\mu }-\sin \theta )$$$$\frac{CD}{\sin (\frac{\pi }{2}-\phi +\theta )}=\frac{CB}{\sin \phi }$$$$CD=\frac{(\frac{b}{\sin \theta }-\frac{L}{2})\cos (\phi -\theta )}{\sin \phi }$$$$=(\frac{b}{\sin \theta }-\frac{L}{2})(\frac{\cos \theta }{\tan \phi }+\sin \theta )$$$$\frac{L}{2}(\frac{\cos \theta }{\mu }-\sin \theta )=(\frac{b}{\sin \theta }-\frac{L}{2})(\frac{\cos \theta }{\tan \phi }+\sin \theta )$$$$\frac{1}{\mu }-\tan \theta =(\frac{2b}{L\sin \theta }-1)(\frac{1}{\tan \phi }+\tan \theta )$$$$\frac{1}{\mu }-\tan \theta =(\frac{2b}{L\sin \theta }-1)(\frac{1}{\tan \theta }+\tan \theta )$$$$\Rightarrow \frac{2b}{L\sin \theta }(\frac{1}{\tan \theta }+\tan \theta )-\frac{1}{\tan \theta }=\frac{1}{\mu }$$$$\Rightarrow \sin \theta \cos \theta (\frac{\sin \theta }{\mu }+\cos \theta )=\frac{2b}{L}$$$$\Rightarrow \theta =....$$$$$$$$\frac{R}{\sin \phi }=\frac{mg}{\sin (\varphi +\phi )}$$$$R=\frac{mg\sin \phi }{\sin (\varphi +\phi )}=\frac{mg}{\frac{\sin \varphi }{\tan \phi }+\cos \varphi }$$$$F=R\sin \varphi =\frac{mg}{\frac{1}{\tan \phi }+\frac{1}{\tan \varphi }}=\frac{mg}{\frac{1}{\tan \theta }+\frac{1}{\mu }}$$$$\Rightarrow \frac{F}{mg}=\frac{1}{\frac{1}{\tan \theta }+\frac{1}{\mu }}$$$$$$$$example:$$$$L=5,b=2,\mu =0.5$$$$\Rightarrow \theta \approx 0.5182(29.69°)\Rightarrow \frac{F}{mg}\approx 0.2664$$$$\Rightarrow \theta \approx 1.1710(67.09°)\Rightarrow \frac{F}{mg}\approx 0.4128$$

Answered by mr W last updated on 04/Jan/24

$$CASE3:$$$$ascase2,butwithfriction$$$$coefficient{\mu }_1\ne 0betweentheplank$$$$andtrolley.$$$$$$$$\tan \varphi =\mu$$$$\tan {\varphi }_1={\mu }_1$$$$\phi =\theta +{\varphi }_1$$$$\theta ⩾{\sin }^{-1}\frac{b}{L}$$$$\frac{CD}{\sin (\frac{\pi }{2}-\varphi -\theta )}=\frac{AC}{\sin \varphi }$$$$\Rightarrow CD=\frac{L\cos (\varphi +\theta )}{2\sin \varphi }$$$$=\frac{L}{2}(\frac{\cos \theta }{\tan \varphi }-\sin \theta )=\frac{L}{2}(\frac{\cos \theta }{\mu }-\sin \theta )$$$$\frac{CD}{\sin (\frac{\pi }{2}-\phi +\theta )}=\frac{CB}{\sin \phi }$$$$CD=\frac{(\frac{b}{\sin \theta }-\frac{L}{2})\cos (\phi -\theta )}{\sin \phi }$$$$=(\frac{b}{\sin \theta }-\frac{L}{2})(\frac{\cos \theta }{\tan \phi }+\sin \theta )$$$$\frac{L}{2}(\frac{\cos \theta }{\mu }-\sin \theta )=(\frac{b}{\sin \theta }-\frac{L}{2})(\frac{\cos \theta }{\tan \phi }+\sin \theta )$$$$\frac{1}{\mu }-\tan \theta =(\frac{2b}{L\sin \theta }-1)(\frac{1}{\tan \phi }+\tan \theta )$$$$\frac{1}{\mu }-\tan \theta =(\frac{2b}{L\sin \theta }-1)(\frac{1-{\mu }_1\tan \theta }{\tan \theta +{\mu }_1}+\tan \theta )$$$$\Rightarrow (\frac{2b}{L\sin \theta }-1)(\frac{1-{\mu }_1\tan \theta }{\tan \theta +{\mu }_1}+\tan \theta )+\tan \theta =\frac{1}{\mu }$$$$\Rightarrow \theta =....$$$$$$$$\frac{R}{\sin \phi }=\frac{mg}{\sin (\varphi +\phi )}$$$$R=\frac{mg\sin \phi }{\sin (\varphi +\phi )}=\frac{mg}{\frac{\sin \varphi }{\tan \phi }+\cos \varphi }$$$$F=R\sin \varphi =\frac{mg}{\frac{1}{\tan \phi }+\frac{1}{\tan \varphi }}=\frac{mg}{\frac{1-{\mu }_1\tan \theta }{{\mu }_1+\tan \theta }+\frac{1}{\mu }}$$$$\Rightarrow \frac{F}{mg}=\frac{1}{\frac{1-{\mu }_1\tan \theta }{{\mu }_1+\tan \theta }+\frac{1}{\mu }}$$

Answered by ajfour last updated on 04/Jan/24

Commented by ajfour last updated on 04/Jan/24

$$Case:freerollervontactbetween$$$$plankandtrolley.$$$$\mu (mg-N\cos \theta )=N\sin \theta$$$$\Rightarrow N=\frac{\mu mg}{\sin \theta +\mu \cos \theta }$$$$=\frac{\mu mg(1+t^2)}{2t+\mu (1-t^2)}$$$$\mathrm{\&}N\left(\frac{b}{\sin \theta }\right)=\frac{mgL\cos \theta }{2}$$$$\Rightarrow N=\frac{mgL}{4b}\sin 2\theta =\frac{mgL(}{4b}\frac{4t)(1-t^2)}{{(1+t^2)}^2}$$$$Ift=\tan \frac{\theta }{2}$$$$\Rightarrow \mu =(\sin \theta +\mu \cos \theta )\left(\frac{L\sin \theta \cos \theta }{2b}\right)$$$$or$$$$\frac{2t+\mu (1-t^2)}{(1+t^2)}=\frac{\mu \left(b/L\right){(1+t^2)}^2}{t(1-t^2)}$$$$t(1-t^2)\{2t+\mu (1-t^2)\}=k\mu {(1+t^2)}^3$$$$matterofdegreesix..$$$$a=(\sin \theta +c\cos \theta )\sin \theta \cos$$

Commented by mr W last updated on 04/Jan/24

$$\mu (mg-N\cos \theta )=\cancel{F+}N\sin \theta$$$$F=N\sin \theta$$

Commented by mr W last updated on 04/Jan/24

$$\theta mustfulfill$$$$\sin \theta \cos \theta (\frac{\sin \theta }{\mu }+\cos \theta )=\frac{2b}{L}$$

Commented by mr W last updated on 04/Jan/24

$$wegetthesamefor\frac{F}{mg}:$$$$N=\frac{\mu mg}{\sin \theta +\mu \cos \theta }$$$$F=N\sin \theta =\frac{\mu mg\sin \theta }{\sin \theta +\mu \cos \theta }=\frac{mg}{\frac{1}{\tan \theta }+\frac{1}{\mu }}$$$$\Rightarrow \frac{F}{mg}=\frac{1}{\frac{1}{\mu }+\frac{1}{\tan \theta }}$$

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