# Question-68342

Question Number 68342 by Tony Lin last updated on 09/Sep/19
Commented by Tony Lin last updated on 09/Sep/19
$${when}\:{m}\:{slide}\:{to}\:{the}\:{bottom} \\$$$${if}\:{there}\:{is}\:{no}\:{fraction} \\$$$${find}\:{v}_{{m}} \&{v}_{{M}} \\$$$${find}\:{a}_{{m}} \&{a}_{{M}} \\$$
Answered by mr W last updated on 09/Sep/19
Commented by mr W last updated on 10/Sep/19
$${METHOD}\:\mathrm{1} \\$$$${position}\:{of}\:{block}\:{M}:\:{x} \\$$$${a}_{{M}} =\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} } \\$$$${position}\:{of}\:{block}\:{m}: \\$$$${x}_{{m}} ={x}+{s}\:\mathrm{cos}\:\theta \\$$$${y}_{{m}} ={s}\:\mathrm{sin}\:\theta \\$$$${a}_{{m},{x}} ={a}_{{M}} +{a}\:\mathrm{cos}\:\theta\:{with}\:{a}=\frac{{d}^{\mathrm{2}} {s}}{{dt}^{\mathrm{2}} } \\$$$${a}_{{m},{y}} ={a}\:\mathrm{sin}\:\theta \\$$$$\\$$$${Ma}_{{M}} ={N}\:\mathrm{sin}\:\theta \\$$$$\Rightarrow{N}=\frac{{Ma}_{{M}} }{\mathrm{sin}\:\theta} \\$$$${ma}_{{m},{x}} =−{N}\:\mathrm{sin}\:\theta \\$$$${m}\left({a}_{{M}} +{a}\:\mathrm{cos}\:\theta\right)=−{N}\:\mathrm{sin}\:\theta \\$$$${Ma}_{{M}} ={N}\:\mathrm{sin}\:\theta \\$$$$\Rightarrow\left(\frac{{M}}{{m}}+\mathrm{1}\right){a}_{{M}} +{a}\:\mathrm{cos}\:\theta=\mathrm{0}\:\:\:…\left({i}\right) \\$$$$\\$$$${ma}_{{m},{y}} =−{mg}+{N}\:\mathrm{cos}\:\theta \\$$$${ma}\:\mathrm{sin}\:\theta=−{mg}+{N}\:\mathrm{cos}\:\theta \\$$$$\Rightarrow\frac{{M}}{{m}}×\frac{{a}_{{M}} }{\mathrm{tan}\:\theta}−{a}\:\mathrm{sin}\:\theta={g}\:\:\:…\left({ii}\right) \\$$$$\\$$$$\Rightarrow{a}_{{M}} =\frac{{g}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}{\frac{{M}}{{m}}+\mathrm{sin}^{\mathrm{2}} \:\theta} \\$$$$\Rightarrow{a}=−\frac{{g}\left(\frac{{M}}{{m}}+\mathrm{1}\right)\mathrm{sin}\:\theta}{\frac{{M}}{{m}}+\mathrm{sin}^{\mathrm{2}} \:\theta} \\$$$$\Rightarrow{a}_{{m},{x}} =−\frac{{g}\frac{{M}}{{m}}\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}{\frac{{M}}{{m}}+\mathrm{sin}^{\mathrm{2}} \:\theta} \\$$$$\Rightarrow{a}_{{m},{y}} =−\frac{{g}\left(\frac{{M}}{{m}}+\mathrm{1}\right)\mathrm{sin}^{\mathrm{2}} \:\theta}{\frac{{M}}{{m}}+\mathrm{sin}^{\mathrm{2}} \:\theta} \\$$$${v}_{{m},{y}} =\sqrt{\mathrm{2}{ha}_{{m},{y}} } \\$$$$\Rightarrow{v}_{{m},{y}} =\mathrm{sin}\:\theta\sqrt{\frac{\mathrm{2}{hg}\left(\frac{{M}}{{m}}+\mathrm{1}\right)}{\frac{{M}}{{m}}+\mathrm{sin}^{\mathrm{2}} \:\theta}} \\$$$$\\$$$${v}_{{m},{x}} =\sqrt{\frac{\mathrm{2}{ha}_{{m},{x}} }{\mathrm{tan}\:\theta}} \\$$$$\Rightarrow{v}_{{m},{x}} =\mathrm{cos}\:\theta\sqrt{\frac{\mathrm{2}{hg}\frac{{M}}{{m}}}{\frac{{M}}{{m}}+\mathrm{sin}^{\mathrm{2}} \:\theta}} \\$$
Commented by Tony Lin last updated on 10/Sep/19
$${thanks}\:{sir} \\$$
Answered by mr W last updated on 09/Sep/19
Commented by mr W last updated on 09/Sep/19
$${METHOD}\:\mathrm{2} \\$$$${Ma}_{{M}} ={N}\:\mathrm{sin}\:\theta \\$$$$\Rightarrow{N}=\frac{{Ma}_{{M}} }{\mathrm{sin}\:\theta} \\$$$$\\$$$${N}+{ma}_{{M}} \mathrm{sin}\:\theta={mg}\:\mathrm{cos}\:\theta \\$$$$\frac{{Ma}_{{M}} }{\mathrm{sin}\:\theta}+{ma}_{{M}} \mathrm{sin}\:\theta={mg}\:\mathrm{cos}\:\theta \\$$$$\left(\frac{{M}}{{m}}+\mathrm{sin}^{\mathrm{2}} \:\theta\right){a}_{{M}} ={g}\:\mathrm{sin}\:\theta\mathrm{cos}\:\theta \\$$$$\Rightarrow{a}_{{M}} =\frac{{g}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}{\frac{{M}}{{m}}+\mathrm{sin}^{\mathrm{2}} \:\theta} \\$$$$\\$$$${ma}_{{M}} \mathrm{cos}\:\theta+{mg}\:\mathrm{sin}\:\theta={ma} \\$$$${a}={a}_{{M}} \mathrm{cos}\:\theta+{g}\:\mathrm{sin}\:\theta \\$$$$\Rightarrow{a}=\frac{{g}\left(\frac{{M}}{{m}}+\mathrm{1}\right)\mathrm{sin}\:\theta}{\frac{{M}}{{m}}+\mathrm{sin}^{\mathrm{2}} \:\theta} \\$$$${a}_{{m},{x}} ={a}\:\mathrm{cos}\:\theta−{a}_{{M}} =\frac{{g}\left(\frac{{M}}{{m}}+\mathrm{1}\right)\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}{\frac{{M}}{{m}}+\mathrm{sin}^{\mathrm{2}} \:\theta}−\frac{{g}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}{\frac{{M}}{{m}}+\mathrm{sin}^{\mathrm{2}} \:\theta} \\$$$$\Rightarrow{a}_{{m},{x}} =\frac{\frac{{M}}{{m}}{g}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}{\frac{{M}}{{m}}+\mathrm{sin}^{\mathrm{2}} \:\theta} \\$$$${a}_{{m},{y}} ={a}\:\mathrm{sin}\:\theta \\$$$$\Rightarrow{a}_{{m},{y}} =\frac{{g}\left(\frac{{M}}{{m}}+\mathrm{1}\right)\mathrm{sin}^{\mathrm{2}} \:\theta}{\frac{{M}}{{m}}+\mathrm{sin}^{\mathrm{2}} \:\theta} \\$$
Commented by Tony Lin last updated on 10/Sep/19
$${thanks}\:{sir} \\$$
Answered by Tanmay chaudhury last updated on 09/Sep/19
Commented by Tanmay chaudhury last updated on 09/Sep/19
$${Mg}+{Ncos}\theta={N}_{{G}} \\$$$${MA}={Nsin}\theta \\$$$${for}\:{m} \\$$$${mgsin}\theta+{mAcos}\theta={ma} \\$$$${mgcos}\theta={N}+{mAsin}\theta \\$$$${MA}={sin}\theta\left({mgcos}\theta−{mAsin}\theta\right) \\$$$${A}\left({M}+{msin}^{\mathrm{2}} \theta\right)={mgsin}\theta{cos}\theta \\$$$${A}=\frac{{mgsin}\theta{cos}\theta}{{M}+{msin}^{\mathrm{2}} \theta} \\$$$${A}={acc}\:{of}\:{wedge} \\$$$${a}={gsin}\theta+\left(\frac{{mgsin}\theta{cos}\theta}{{M}+{msin}^{\mathrm{2}} \theta}\right){cos}\theta \\$$$${a}=\frac{{Mgsin}\theta+{mgsin}^{\mathrm{3}} \theta+{mgsin}\theta{cos}^{\mathrm{2}} \theta}{{M}+{msin}^{\mathrm{2}} \theta} \\$$$${a}=\frac{{Mgsin}\theta+{mgsin}\theta\left({sin}^{\mathrm{2}} \theta+{cos}^{\mathrm{2}} \theta\right)}{{M}+{msin}^{\mathrm{2}} \theta} \\$$$${a}=\frac{\left({M}+{m}\right){gsin}\theta}{{M}+{msin}^{\mathrm{2}} \theta} \\$$$$\\$$$$\\$$
Commented by Tony Lin last updated on 10/Sep/19
$${thanks}\:{sir} \\$$