Menu Close

Question-186645




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

Leave a Reply

Your email address will not be published. Required fields are marked *