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Question Number 186645 by ajfour last updated on 07/Feb/23

Commented by ajfour last updated on 07/Feb/23

Walls are frictionless. Balls (solid) have the same density ρ. Find minimum value of maximum static friction coefficient at the ground for the larger ball such that static situation could prevail.

$${Walls}\:{are}\:{frictionless}.\:{Balls}\:\left({solid}\right) \\ $$$${have}\:{the}\:{same}\:{density}\:\rho.\:{Find} \\ $$$${minimum}\:{value}\:{of}\:{maximum} \\ $$$${static}\:{friction}\:{coefficient}\:{at}\:{the} \\ $$$${ground}\:{for}\:{the}\:{larger}\:{ball}\:{such} \\ $$$${that}\:{static}\:{situation}\:{could}\:{prevail}. \\ $$

Answered by ajfour last updated on 07/Feb/23

sin θ=(((√2)(b−a))/(b+a)) If J be normal reaction mutual to the balls. N=(M+m)g=((4πρg(b^3 +a^3 ))/3) f=Jsin θ & Jcos θ=mg ⇒ μ(M+m)g=mgtan θ μ_(s,max) ≥((a^3 /(b^3 +a^3 )))(((√2)(b−a))/( (√(6ab−(a^2 +b^2 ))))) If (b/a)=p μ_(s,max) ≥(((√2)(p−1))/((p^3 +1)(√(6p−p^2 −1))))

$$\mathrm{sin}\:\theta=\frac{\sqrt{\mathrm{2}}\left({b}−{a}\right)}{{b}+{a}} \\ $$$${If}\:\:{J}\:{be}\:{normal}\:{reaction}\:{mutual}\:{to}\: \\ $$$${the}\:{balls}.\: \\ $$$${N}=\left({M}+{m}\right){g}=\frac{\mathrm{4}\pi\rho{g}\left({b}^{\mathrm{3}} +{a}^{\mathrm{3}} \right)}{\mathrm{3}} \\ $$$${f}={J}\mathrm{sin}\:\theta\:\:\:\:\:\&\:\:{J}\mathrm{cos}\:\theta={mg} \\ $$$$\Rightarrow\:\:\mu\left({M}+{m}\right){g}={mg}\mathrm{tan}\:\theta \\ $$$$\mu_{{s},{max}} \geqslant\left(\frac{{a}^{\mathrm{3}} }{{b}^{\mathrm{3}} +{a}^{\mathrm{3}} }\right)\frac{\sqrt{\mathrm{2}}\left({b}−{a}\right)}{\:\sqrt{\mathrm{6}{ab}−\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}} \\ $$$${If}\:\:\frac{{b}}{{a}}={p} \\ $$$$\mu_{{s},{max}} \geqslant\frac{\sqrt{\mathrm{2}}\left({p}−\mathrm{1}\right)}{\left({p}^{\mathrm{3}} +\mathrm{1}\right)\sqrt{\mathrm{6}{p}−{p}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$ \\ $$

Answered by mr W last updated on 07/Feb/23

Commented by mr W last updated on 07/Feb/23

sin θ=(((√2)(b−a))/(a+b)) N_1 =m_1 g tan θ N_3 =(m_1 +m_2 )g f=μN_3 =N_1 μ=((m_1 g tan θ)/((m_1 +m_2 )g)) =(a^3 /(a^3 +b^3 ))×(((√2)(b−a))/( (√((a+b)^2 −2(b−a)^2 ))))

$$\mathrm{sin}\:\theta=\frac{\sqrt{\mathrm{2}}\left({b}−{a}\right)}{{a}+{b}} \\ $$$${N}_{\mathrm{1}} ={m}_{\mathrm{1}} {g}\:\mathrm{tan}\:\theta \\ $$$${N}_{\mathrm{3}} =\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right){g} \\ $$$${f}=\mu{N}_{\mathrm{3}} ={N}_{\mathrm{1}} \\ $$$$\mu=\frac{{m}_{\mathrm{1}} {g}\:\mathrm{tan}\:\theta}{\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right){g}} \\ $$$$\:\:=\frac{{a}^{\mathrm{3}} }{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} }×\frac{\sqrt{\mathrm{2}}\left({b}−{a}\right)}{\:\sqrt{\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{2}\left({b}−{a}\right)^{\mathrm{2}} }} \\ $$

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