Question Number 29016 by ajfour last updated on 03/Feb/18
Commented by ajfour last updated on 03/Feb/18
$${Find}\:{maximum}\:{value}\:{of}\:\boldsymbol{{m}} \\ $$$${if}\:{bigger}\:{sphere}\:{has}\:{to}\:{remain} \\ $$$${in}\:{contact}\:{with}\:{wall}.{Assume} \\ $$$${friction}\:{coefficient}\:{large}\:{enough} \\ $$$${at}\:{all}\:{surfaces}.\:\:{r},\:{R}\:{constant}\:; \\ $$$${r}\:<\:{R}\:. \\ $$
Answered by ajfour last updated on 03/Feb/18
Commented by ajfour last updated on 03/Feb/18
$$\mathrm{sin}\:\theta\:=\frac{{R}−{r}}{{R}+{r}}\:\:, \\ $$$${f}_{{g}} ={f}\:\:\:{and}\:\:\:{f}={f}_{{w}} \:\:{considering} \\ $$$${zero}\:{torque}\:{on}\:{both}\:{spheres} \\ $$$${individually}\:{about}\:{their}\:{centres}. \\ $$$$\:\:\:\:{f}_{{g}} ={N}_{{w}} \:\:\:\:{considering}\:{net}\: \\ $$$${horizontal}\:{force}\:{on}\:{system} \\ $$$${consisting}\:{of}\:{both}\:{spheres}. \\ $$$${so}\:\:\:\:\:\:\:\boldsymbol{{f}}_{\boldsymbol{{g}}} =\boldsymbol{{f}}_{\boldsymbol{{w}}} =\boldsymbol{{f}}=\boldsymbol{{N}}_{\boldsymbol{{w}}} \:\:\:\:\:\:\:{eq}.\left({A}\right) \\ $$$$\:\:\left({do}\:{keep}\:{in}\:{mind}\:\right) \\ $$$${net}\:{horizontal}\:{force}\:{on}\:{system} \\ $$$$\:\:\:{f}_{{w}} +{N}_{{g}} =\left({M}+{m}\right){g}\:\:\:\:\: \\ $$$$\Rightarrow\:\:\:\:\:{f}+{N}_{{g}} \:=\:\left({M}+{m}\right){g}\:\:\:….\left({i}\right) \\ $$$$\Sigma{F}_{{x}} =\mathrm{0}\:\:\:{for}\:{smaller}\:{sphere} \\ $$$$\:{N}_{{w}} +{f}\mathrm{cos}\:\theta={N}\mathrm{sin}\:\theta \\ $$$$\Rightarrow\:\:\:\:{f}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)={N}\mathrm{sin}\:\theta\:\:\:….\left({ii}\right) \\ $$$$\Sigma{F}_{{y}} =\mathrm{0}\:\:,\:{so} \\ $$$$\:\:\:{f}_{{w}} +{N}\mathrm{cos}\:\theta+{f}\mathrm{sin}\:\theta={mg} \\ $$$$\Rightarrow\:{f}\left(\mathrm{1}+\mathrm{sin}\:\theta\right)+{N}\mathrm{cos}\:\theta\:={mg}\:\:…\left({iii}\right) \\ $$$$\Sigma{F}_{{x}} =\mathrm{0}\:\:\:{for}\:{larger}\:{sphere} \\ $$$${f}\mathrm{cos}\:\theta+{f}_{{g}} =\:{N}\mathrm{sin}\:\theta \\ $$$$\Rightarrow\:\:\:\:\:{f}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)\:={N}\mathrm{sin}\:\theta\:\:\:\left[{same}\:{as}\:\left({i}\right)\right] \\ $$$$\Sigma{F}_{{y}} =\mathrm{0}\:,\:{so} \\ $$$${f}\mathrm{sin}\:\theta+{Mg}+{N}\mathrm{cos}\:\theta\:=\:{N}_{{g}} \\ $$$${Torque}\:{about}\:{corner}\:{of}\:{ground} \\ $$$${and}\:{wall}\:{gives}: \\ $$$${MgR}+{mgr}+{N}_{{w}} \left[{R}+\left({R}+{r}\right)\mathrm{cos}\:\theta\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{N}_{{g}} {R}\:\:\:\:\:\:\:\:\:\:\:\:\:…..\left({iv}\right) \\ $$$${we}\:{have}\:{unknowns} \\ $$$$\:\:\:{f},\:{m},\:{N},\:{N}_{{g}} \:\:\:{and}\:{eq}.{s}\:\:\left({i}\right)\:{to}\:\left({iv}\right) \\ $$$${using}\:\left({ii}\right)\:{in}\:\left({iii}\right): \\ $$$$\:\:{N}\left[\frac{\mathrm{sin}\:\theta\left(\mathrm{1}+\mathrm{sin}\:\theta\right)}{\mathrm{1}+\mathrm{cos}\:\theta}+\mathrm{cos}\:\theta\right]={mg} \\ $$$$\Rightarrow\:\:\boldsymbol{{N}}=\frac{\boldsymbol{{mg}}\left(\mathrm{1}+\mathrm{cos}\:\boldsymbol{\theta}\right)}{\mathrm{1}+\mathrm{cos}\:\boldsymbol{\theta}+\mathrm{sin}\:\boldsymbol{\theta}} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{{f}}\:=\:\frac{\boldsymbol{{mg}}\mathrm{sin}\:\boldsymbol{\theta}}{\mathrm{1}+\mathrm{cos}\:\boldsymbol{\theta}+\mathrm{sin}\:\boldsymbol{\theta}}\:\:\:=\boldsymbol{{N}}_{\boldsymbol{{w}}} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{{N}}_{\boldsymbol{{g}}} =\:\left(\boldsymbol{{M}}+\boldsymbol{{m}}\right)\boldsymbol{{g}}−\boldsymbol{{f}} \\ $$$${using}\:{these}\:{in}\:\left({iv}\right): \\ $$$${MgR}+{mgr}+{f}\left[{R}+\left({R}+{r}\right)\mathrm{cos}\:\theta\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\left({M}+{m}\right){gR}−{fR} \\ $$$$\Rightarrow\:{mgr}+{f}\left[\mathrm{2}{R}+\left({R}+{r}\right)\mathrm{cos}\:\theta\right]−{mgR} \\ $$$$\:\:\:\:\:={Mg}\left({R}−{r}\right) \\ $$$$\:\frac{{mg}\mathrm{sin}\:\theta\left[\mathrm{2}{R}+\left({R}+{r}\right)\mathrm{cos}\:\theta\right]}{\mathrm{1}+\mathrm{sin}\:\theta\mathrm{cos}\:\theta}−{mg}\left({R}−{r}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={Mg}\left({R}−{r}\right) \\ $$$${m}=\frac{{M}\left({R}−{r}\right)}{\left[\frac{\mathrm{sin}\:\theta\left[\mathrm{2}{R}+\left({R}+{r}\right)\mathrm{cos}\:\theta\right]}{\mathrm{1}+\mathrm{sin}\:\theta\mathrm{cos}\:\theta}\:−\:\left({R}−{r}\right)\right]}\:\:. \\ $$
Commented by mrW2 last updated on 03/Feb/18
$${The}\:{forces}\:{and}\:{also}\:{their}\:{directions} \\ $$$${as}\:{displayed}\:{in}\:{the}\:{diagram}\:{are}\: \\ $$$${correct}.\:{I}\:{can}\:{follow}\:{your}\:{steps}\:{and} \\ $$$${they}\:{are}\:{correct}. \\ $$
Commented by ajfour last updated on 04/Feb/18
$${thank}\:\:{you}\:{Sir}. \\ $$