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Question Number 27303 by Tinkutara last updated on 04/Jan/18

Answered by mrW1 last updated on 05/Jan/18

$$a=a_{//}=\frac{g}{2}\left(given\right)$$$$a_{\mathrm{\perp }}=a\sin \theta =\frac{g\sin \theta }{2}$$$$mg\cos \theta -N={ma}_{\mathrm{\perp }}=\frac{mg\sin \theta }{2}$$$$\Rightarrow N=mg(\cos \theta -\frac{\sin \theta }{2})=mg(\frac{4}{5}-\frac{3}{2\times 5})=\frac{mg}{2}$$

Commented by mrW1 last updated on 05/Jan/18

Commented by Tinkutara last updated on 05/Jan/18

$$Why{a}_{\mathrm{\perp }}=a\sin \theta ?$$

Commented by mrW1 last updated on 05/Jan/18

$$thecomponentofafromMindirectionnormal$$$$tothecontactsurfaceisa\sin \theta ,which$$$$mustbethesameas{a}_{\mathrm{\perp }}fromm.$$

Commented by mrW1 last updated on 05/Jan/18

Commented by mrW1 last updated on 05/Jan/18

$$letx=displacementofwedgeM$$$$accelerationofblockMisa=\frac{d^{2}x}{{dt}^2}$$$$displacementofblockmis$$$$s_{//}=x$$$$s_{\mathrm{\perp }}=x\sin \theta$$$$\Rightarrow a_{//}=\frac{d^2{s}_{//}}{{dt}^2}=a$$$$\Rightarrow a_{\mathrm{\perp }}=\frac{d^2{s}_{\mathrm{\perp }}}{{dt}^2}=a\sin \theta$$

Commented by Tinkutara last updated on 05/Jan/18

Then why the relative acceleration of block is g/2 w.r.t wedge?

Commented by mrW1 last updated on 05/Jan/18

$$itisgiventhat{a}_{//}=\frac{g}{2}.infactitis$$$$dependentonthemassofthewedge.$$$$IthinkifM=\frac{16}{25}m{we}^{\prime }llget{a}_{//}=\frac{g}{2}.$$

Commented by mrW1 last updated on 05/Jan/18

Commented by Tinkutara last updated on 05/Jan/18

Thank you very much Sir! I got the answer.

Commented by mrW1 last updated on 05/Jan/18

$$a_{//}=a$$$$a_{\mathrm{\perp }}=a\sin \theta$$$$let\lambda =\frac{M}{m}and\alpha =\frac{a}{g}$$$$$$$$mg\sin \theta -T={ma}_{//}=ma$$$$\Rightarrow T=mg\sin \theta -ma=mg(\sin \theta -\alpha )$$$$mg\cos \theta -N={ma}_{\mathrm{\perp }}=ma\sin \theta$$$$\Rightarrow N=mg\cos \theta -ma\sin \theta =mg(\cos \theta -\alpha \sin \theta )$$$$$$$$T-T\cos \theta +N\sin \theta =Ma$$$$T(1-\cos \theta )+N\sin \theta =Ma$$$$mg(\sin \theta -\alpha )(1-\cos \theta )+mg(\cos \theta -\alpha \sin \theta )\sin \theta =Ma$$$$(\sin \theta -\alpha )(1-\cos \theta )+(\cos \theta -\alpha \sin \theta )\sin \theta =\frac{Ma}{mg}=\lambda \alpha$$$$\sin \theta -\sin \theta \cos \theta -(1-\cos \theta )\alpha +\cos \theta \sin \theta -\alpha {\sin }^2\theta =\lambda \alpha$$$$\sin \theta -(1-\cos \theta )\alpha -\alpha {\sin }^2\theta =\lambda \alpha$$$$\sin \theta =(\lambda +1-\cos \theta +{\sin }^2\theta )\alpha$$$$\sin \theta =(\lambda +2-\cos \theta -{\cos }^2\theta )\alpha$$$$\Rightarrow \alpha =\frac{\sin \theta }{\lambda +2-\cos \theta -{\cos }^2\theta }$$$$\Rightarrow a=\frac{\sin \theta }{\frac{M}{m}+2-\cos \theta -{\cos }^2\theta }\times g$$$$with\sin \theta =\frac{3}{5}and\cos \theta =\frac{4}{5}weget$$$$a=\frac{0.6g}{\frac{M}{m}+0.56}$$$$fora=\frac{g}{2}$$$$\Rightarrow \frac{M}{m}=\frac{0.6}{0.5}-0.56=\frac{16}{25}$$

Commented by Tinkutara last updated on 05/Jan/18

I really appreciate this.

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