Question Number 186645 by ajfour last updated on 07/Feb/23
Commented by ajfour last updated on 07/Feb/23
$${Walls}\:{are}\:{frictionless}.\:{Balls}\:\left({solid}\right) \\ $$$${have}\:{the}\:{same}\:{density}\:\rho.\:{Find} \\ $$$${minimum}\:{value}\:{of}\:{maximum} \\ $$$${static}\:{friction}\:{coefficient}\:{at}\:{the} \\ $$$${ground}\:{for}\:{the}\:{larger}\:{ball}\:{such} \\ $$$${that}\:{static}\:{situation}\:{could}\:{prevail}. \\ $$
Answered by ajfour last updated on 07/Feb/23
$$\mathrm{sin}\:\theta=\frac{\sqrt{\mathrm{2}}\left({b}−{a}\right)}{{b}+{a}} \\ $$$${If}\:\:{J}\:{be}\:{normal}\:{reaction}\:{mutual}\:{to}\: \\ $$$${the}\:{balls}.\: \\ $$$${N}=\left({M}+{m}\right){g}=\frac{\mathrm{4}\pi\rho{g}\left({b}^{\mathrm{3}} +{a}^{\mathrm{3}} \right)}{\mathrm{3}} \\ $$$${f}={J}\mathrm{sin}\:\theta\:\:\:\:\:\&\:\:{J}\mathrm{cos}\:\theta={mg} \\ $$$$\Rightarrow\:\:\mu\left({M}+{m}\right){g}={mg}\mathrm{tan}\:\theta \\ $$$$\mu_{{s},{max}} \geqslant\left(\frac{{a}^{\mathrm{3}} }{{b}^{\mathrm{3}} +{a}^{\mathrm{3}} }\right)\frac{\sqrt{\mathrm{2}}\left({b}−{a}\right)}{\:\sqrt{\mathrm{6}{ab}−\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}} \\ $$$${If}\:\:\frac{{b}}{{a}}={p} \\ $$$$\mu_{{s},{max}} \geqslant\frac{\sqrt{\mathrm{2}}\left({p}−\mathrm{1}\right)}{\left({p}^{\mathrm{3}} +\mathrm{1}\right)\sqrt{\mathrm{6}{p}−{p}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$ \\ $$
Answered by mr W last updated on 07/Feb/23
Commented by mr W last updated on 07/Feb/23
$$\mathrm{sin}\:\theta=\frac{\sqrt{\mathrm{2}}\left({b}−{a}\right)}{{a}+{b}} \\ $$$${N}_{\mathrm{1}} ={m}_{\mathrm{1}} {g}\:\mathrm{tan}\:\theta \\ $$$${N}_{\mathrm{3}} =\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right){g} \\ $$$${f}=\mu{N}_{\mathrm{3}} ={N}_{\mathrm{1}} \\ $$$$\mu=\frac{{m}_{\mathrm{1}} {g}\:\mathrm{tan}\:\theta}{\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right){g}} \\ $$$$\:\:=\frac{{a}^{\mathrm{3}} }{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} }×\frac{\sqrt{\mathrm{2}}\left({b}−{a}\right)}{\:\sqrt{\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{2}\left({b}−{a}\right)^{\mathrm{2}} }} \\ $$