Question-186046

Question Number 186046 by ajfour last updated on 31/Jan/23

Commented by ajfour last updated on 31/Jan/23

A fixed spherical crucible holds two solid balls of radii a and b while they have the same density ρ. Find theNormal Reaction between the balls.

$${A}\:{fixed}\:{spherical}\:{crucible}\:{holds} \\ $$$${two}\:{solid}\:{balls}\:{of}\:{radii}\:{a}\:{and}\:{b} \\ $$$${while}\:{they}\:{have}\:{the}\:{same}\:{density}\:\rho. \\ $$$${Find}\:{theNormal}\:{Reaction}\:{between} \\ $$$${the}\:{balls}. \\ $$

Answered by mr W last updated on 31/Jan/23

Commented by mr W last updated on 31/Jan/23

M_A =((4πa^3 ρ)/3) M_B =((4πb^3 ρ)/3) OA=r−a OB=r−b AB=a+b ((AC)/(CB))=(M_B /M_A )=(b^3 /a^3 ) cos (α+β)=(((r−a)^2 +(r−b)^2 −(a+b)^2 )/(2(r−a)(r−b))) =1−((2ab)/((r−a)(r−b)))=ξ, say ⇒sin (α+β)=(√(1−ξ^2 ))=((2(√(abr(r−a−b))))/((r−a)(r−b))) ((sin γ)/(sin (α+β)))=((OB)/(AB))=((r−b)/(a+b)) ⇒sin γ=((r−b)/(a+b))×((2(√(abr(r−a−b))))/((r−a)(r−b))) =((2(√(abr(r−a−b))))/((r−a)(a+b))) similarly, sin δ=((2(√(abr(r−a−b))))/((r−b)(a+b))) ((OC)/(sin γ))=((AC)/(sin α)) ((sin δ)/(OC))=((sin β)/(CB)) ((sin δ)/(sin γ))=((AC)/(CB))×((sin β)/(sin α)) ((r−a)/(r−b))=(b^3 /a^3 )×((sin β)/(sin α)) ⇒((sin β)/(sin α))=((a^3 (r−a))/(b^3 (r−b)))=λ, say cos α cos β−sin α sin β=ξ (√((1−sin^2 α)(1−sin^2 β)))−sin α sin β=ξ (√((1−sin^2 α)(1−λ^2 sin^2 α)))=ξ+λ sin^2 α (1−sin^2 α)(1−λ^2 sin^2 α)=(ξ+λ sin^2 α)^2 1−ξ^2 =(1+λ^2 +2ξλ) sin^2 α ⇒sin α=(√((1−ξ^2 )/(1−ξ^2 +(λ+ξ)^2 ))) (N/(M_A g))=((sin α)/(sin γ))=(((r−a)(a+b))/(2(√(abr(r−a−b)))))(√((1−ξ^2 )/(1−ξ^2 +(λ+ξ)^2 ))) ⇒N=(((r−a)(a+b))/(2(√(abr(r−a−b)))))(√((1−ξ^2 )/(1−ξ^2 +(λ+ξ)^2 )))×M_A g example: a=3, b=2, r=10, M_A =27 kg, M_B =8 kg ξ=1−((2ab)/((r−a)(r−b)))=1−((2×3×2)/(7×8))=((11)/(14)) λ=((a^3 (r−a))/(b^3 (r−b)))=((3^3 ×7)/(2^3 ×8))=((189)/(64)) N=((7×5)/(2(√(3×2×10×5))))(√((1−((11^2 )/(14^2 )))/(1−((11^2 )/(14^2 ))+(((189)/(64))+((11)/(14)))^2 )))×27×10 =((2160)/( (√(2353))))≈44.529 N or N=((8×5)/(2(√(3×2×10×5))))(√((1−((11^2 )/(14^2 )))/(1−((11^2 )/(14^2 ))+(((64)/(189))+((11)/(14)))^2 )))×8×10 =((2160)/( (√(2353))))≈44.529 N

$${M}_{{A}} =\frac{\mathrm{4}\pi{a}^{\mathrm{3}} \rho}{\mathrm{3}} \\ $$$${M}_{{B}} =\frac{\mathrm{4}\pi{b}^{\mathrm{3}} \rho}{\mathrm{3}} \\ $$$${OA}={r}−{a} \\ $$$${OB}={r}−{b} \\ $$$${AB}={a}+{b} \\ $$$$\frac{{AC}}{{CB}}=\frac{{M}_{{B}} }{{M}_{{A}} }=\frac{{b}^{\mathrm{3}} }{{a}^{\mathrm{3}} } \\ $$$$\mathrm{cos}\:\left(\alpha+\beta\right)=\frac{\left({r}−{a}\right)^{\mathrm{2}} +\left({r}−{b}\right)^{\mathrm{2}} −\left({a}+{b}\right)^{\mathrm{2}} }{\mathrm{2}\left({r}−{a}\right)\left({r}−{b}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1}−\frac{\mathrm{2}{ab}}{\left({r}−{a}\right)\left({r}−{b}\right)}=\xi,\:{say} \\ $$$$\Rightarrow\mathrm{sin}\:\left(\alpha+\beta\right)=\sqrt{\mathrm{1}−\xi^{\mathrm{2}} }=\frac{\mathrm{2}\sqrt{{abr}\left({r}−{a}−{b}\right)}}{\left({r}−{a}\right)\left({r}−{b}\right)} \\ $$$$\frac{\mathrm{sin}\:\gamma}{\mathrm{sin}\:\left(\alpha+\beta\right)}=\frac{{OB}}{{AB}}=\frac{{r}−{b}}{{a}+{b}} \\ $$$$\Rightarrow\mathrm{sin}\:\gamma=\frac{{r}−{b}}{{a}+{b}}×\frac{\mathrm{2}\sqrt{{abr}\left({r}−{a}−{b}\right)}}{\left({r}−{a}\right)\left({r}−{b}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}\sqrt{{abr}\left({r}−{a}−{b}\right)}}{\left({r}−{a}\right)\left({a}+{b}\right)} \\ $$$${similarly}, \\ $$$$\mathrm{sin}\:\delta=\frac{\mathrm{2}\sqrt{{abr}\left({r}−{a}−{b}\right)}}{\left({r}−{b}\right)\left({a}+{b}\right)} \\ $$$$\frac{{OC}}{\mathrm{sin}\:\gamma}=\frac{{AC}}{\mathrm{sin}\:\alpha} \\ $$$$\frac{\mathrm{sin}\:\delta}{{OC}}=\frac{\mathrm{sin}\:\beta}{{CB}} \\ $$$$\frac{\mathrm{sin}\:\delta}{\mathrm{sin}\:\gamma}=\frac{{AC}}{{CB}}×\frac{\mathrm{sin}\:\beta}{\mathrm{sin}\:\alpha} \\ $$$$\frac{{r}−{a}}{{r}−{b}}=\frac{{b}^{\mathrm{3}} }{{a}^{\mathrm{3}} }×\frac{\mathrm{sin}\:\beta}{\mathrm{sin}\:\alpha} \\ $$$$\Rightarrow\frac{\mathrm{sin}\:\beta}{\mathrm{sin}\:\alpha}=\frac{{a}^{\mathrm{3}} \left({r}−{a}\right)}{{b}^{\mathrm{3}} \left({r}−{b}\right)}=\lambda,\:{say} \\ $$$$\mathrm{cos}\:\alpha\:\mathrm{cos}\:\beta−\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta=\xi \\ $$$$\sqrt{\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\alpha\right)\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\beta\right)}−\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta=\xi \\ $$$$\sqrt{\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\alpha\right)\left(\mathrm{1}−\lambda^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\alpha\right)}=\xi+\lambda\:\mathrm{sin}^{\mathrm{2}} \:\alpha \\ $$$$\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\alpha\right)\left(\mathrm{1}−\lambda^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\alpha\right)=\left(\xi+\lambda\:\mathrm{sin}^{\mathrm{2}} \:\alpha\right)^{\mathrm{2}} \\ $$$$\mathrm{1}−\xi^{\mathrm{2}} =\left(\mathrm{1}+\lambda^{\mathrm{2}} +\mathrm{2}\xi\lambda\right)\:\mathrm{sin}^{\mathrm{2}} \:\alpha \\ $$$$\Rightarrow\mathrm{sin}\:\alpha=\sqrt{\frac{\mathrm{1}−\xi^{\mathrm{2}} }{\mathrm{1}−\xi^{\mathrm{2}} +\left(\lambda+\xi\right)^{\mathrm{2}} }} \\ $$$$\frac{{N}}{{M}_{{A}} {g}}=\frac{\mathrm{sin}\:\alpha}{\mathrm{sin}\:\gamma}=\frac{\left({r}−{a}\right)\left({a}+{b}\right)}{\mathrm{2}\sqrt{{abr}\left({r}−{a}−{b}\right)}}\sqrt{\frac{\mathrm{1}−\xi^{\mathrm{2}} }{\mathrm{1}−\xi^{\mathrm{2}} +\left(\lambda+\xi\right)^{\mathrm{2}} }} \\ $$$$\Rightarrow{N}=\frac{\left({r}−{a}\right)\left({a}+{b}\right)}{\mathrm{2}\sqrt{{abr}\left({r}−{a}−{b}\right)}}\sqrt{\frac{\mathrm{1}−\xi^{\mathrm{2}} }{\mathrm{1}−\xi^{\mathrm{2}} +\left(\lambda+\xi\right)^{\mathrm{2}} }}×{M}_{{A}} {g} \\ $$$$ \\ $$$${example}: \\ $$$${a}=\mathrm{3},\:{b}=\mathrm{2},\:{r}=\mathrm{10},\:{M}_{{A}} =\mathrm{27}\:{kg},\:{M}_{{B}} =\mathrm{8}\:{kg} \\ $$$$\xi=\mathrm{1}−\frac{\mathrm{2}{ab}}{\left({r}−{a}\right)\left({r}−{b}\right)}=\mathrm{1}−\frac{\mathrm{2}×\mathrm{3}×\mathrm{2}}{\mathrm{7}×\mathrm{8}}=\frac{\mathrm{11}}{\mathrm{14}} \\ $$$$\lambda=\frac{{a}^{\mathrm{3}} \left({r}−{a}\right)}{{b}^{\mathrm{3}} \left({r}−{b}\right)}=\frac{\mathrm{3}^{\mathrm{3}} ×\mathrm{7}}{\mathrm{2}^{\mathrm{3}} ×\mathrm{8}}=\frac{\mathrm{189}}{\mathrm{64}} \\ $$$${N}=\frac{\mathrm{7}×\mathrm{5}}{\mathrm{2}\sqrt{\mathrm{3}×\mathrm{2}×\mathrm{10}×\mathrm{5}}}\sqrt{\frac{\mathrm{1}−\frac{\mathrm{11}^{\mathrm{2}} }{\mathrm{14}^{\mathrm{2}} }}{\mathrm{1}−\frac{\mathrm{11}^{\mathrm{2}} }{\mathrm{14}^{\mathrm{2}} }+\left(\frac{\mathrm{189}}{\mathrm{64}}+\frac{\mathrm{11}}{\mathrm{14}}\right)^{\mathrm{2}} }}×\mathrm{27}×\mathrm{10} \\ $$$$\:\:\:\:=\frac{\mathrm{2160}}{\:\sqrt{\mathrm{2353}}}\approx\mathrm{44}.\mathrm{529}\:{N} \\ $$$${or} \\ $$$${N}=\frac{\mathrm{8}×\mathrm{5}}{\mathrm{2}\sqrt{\mathrm{3}×\mathrm{2}×\mathrm{10}×\mathrm{5}}}\sqrt{\frac{\mathrm{1}−\frac{\mathrm{11}^{\mathrm{2}} }{\mathrm{14}^{\mathrm{2}} }}{\mathrm{1}−\frac{\mathrm{11}^{\mathrm{2}} }{\mathrm{14}^{\mathrm{2}} }+\left(\frac{\mathrm{64}}{\mathrm{189}}+\frac{\mathrm{11}}{\mathrm{14}}\right)^{\mathrm{2}} }}×\mathrm{8}×\mathrm{10} \\ $$$$\:\:\:\:=\frac{\mathrm{2160}}{\:\sqrt{\mathrm{2353}}}\approx\mathrm{44}.\mathrm{529}\:{N} \\ $$

Commented by ajfour last updated on 01/Feb/23

$${Thank}\:{you}\:{sir}! \\ $$

Answered by ajfour last updated on 01/Feb/23

Commented by ajfour last updated on 01/Feb/23

(4πa^3 )ρg(r−a)sin α=(4πb^3 )(r−b)sin β ⇒ t=a^3 (r−a)sin α=b^3 (r−b)sin β ((sin (90°−α−θ))/(r−b))=((sin (90°−β+θ))/(r−a)) =((sin (α+β))/(a+b)) Ncos (α+θ)=Mgsin α Ncos (β−θ)=mgsin β cos (α+β)=k , 4πρg(((a+b)/( (√(1−k^2 )))))=R ((cos (α+θ))/(r−b))=((cos (β−θ))/(r−a))=((√(1−k^2 ))/(a+b)) N=((Mg(a+b)sin α)/( (r−b)(√(1−k^2 ))))=((a^3 Rsin α)/(r−b)) N=((mg(a+b)sin β)/((r−a)(√(1−k^2 ))))=((b^3 Rsin β)/(r−a)) sin α=(((r−b)/a^3 ))(N/R)=λ(r−b)/a^3 sin β=(((r−a)/b^3 ))(N/R)=λ(r−a)/b^3 cos (α+β)=k ⇒ (√(1−λ^2 (((r−b)/a^3 ))^2 ))(√(1−λ^2 (((r−a)/b^3 ))^2 )) −((λ^2 (r−a)(r−b))/(a^3 b^3 ))=k λ=(N/R) =((√(1−k^2 ))/( (√((((r−a)/b^3 ))^2 +(((r−b)/a^3 ))^2 +2k(((r−a)/b^3 ))(((r−b)/a^3 )))))) R=4πρg(((a+b)/( (√(1−k^2 ))))) N=((4πρg(a+b)ab(√(ab)))/( (√(((a/b))^3 (r−a)^2 +((b/a))^3 (r−b)^2 +2k(r−a)(r−b))))) k=cos (α+β)=(((r−a)^2 +(r−b)^2 −(a+b)^2 )/(2(r−a)(r−b))) N=(((4πρab(√(ab)))g(a+b))/( (√((1+(a^3 /b^3 ))(r−a)^2 +(1+(b^3 /a^3 ))(r−b)^2 −(a+b)^2 ))))

$$\left(\mathrm{4}\pi{a}^{\mathrm{3}} \right)\rho{g}\left({r}−{a}\right)\mathrm{sin}\:\alpha=\left(\mathrm{4}\pi{b}^{\mathrm{3}} \right)\left({r}−{b}\right)\mathrm{sin}\:\beta \\ $$$$\Rightarrow\:\:{t}={a}^{\mathrm{3}} \left({r}−{a}\right)\mathrm{sin}\:\alpha={b}^{\mathrm{3}} \left({r}−{b}\right)\mathrm{sin}\:\beta \\ $$$$\frac{\mathrm{sin}\:\left(\mathrm{90}°−\alpha−\theta\right)}{{r}−{b}}=\frac{\mathrm{sin}\:\left(\mathrm{90}°−\beta+\theta\right)}{{r}−{a}} \\ $$$$\:\:\:=\frac{\mathrm{sin}\:\left(\alpha+\beta\right)}{{a}+{b}} \\ $$$${N}\mathrm{cos}\:\left(\alpha+\theta\right)={Mg}\mathrm{sin}\:\alpha \\ $$$${N}\mathrm{cos}\:\left(\beta−\theta\right)={mg}\mathrm{sin}\:\beta \\ $$$$\mathrm{cos}\:\left(\alpha+\beta\right)={k}\:\:,\:\mathrm{4}\pi\rho{g}\left(\frac{{a}+{b}}{\:\sqrt{\mathrm{1}−{k}^{\mathrm{2}} }}\right)={R} \\ $$$$\frac{\mathrm{cos}\:\left(\alpha+\theta\right)}{{r}−{b}}=\frac{\mathrm{cos}\:\left(\beta−\theta\right)}{{r}−{a}}=\frac{\sqrt{\mathrm{1}−{k}^{\mathrm{2}} }}{{a}+{b}} \\ $$$${N}=\frac{{Mg}\left({a}+{b}\right)\mathrm{sin}\:\alpha}{\:\left({r}−{b}\right)\sqrt{\mathrm{1}−{k}^{\mathrm{2}} }}=\frac{{a}^{\mathrm{3}} {R}\mathrm{sin}\:\alpha}{{r}−{b}} \\ $$$${N}=\frac{{mg}\left({a}+{b}\right)\mathrm{sin}\:\beta}{\left({r}−{a}\right)\sqrt{\mathrm{1}−{k}^{\mathrm{2}} }}=\frac{{b}^{\mathrm{3}} {R}\mathrm{sin}\:\beta}{{r}−{a}} \\ $$$$\mathrm{sin}\:\alpha=\left(\frac{{r}−{b}}{{a}^{\mathrm{3}} }\right)\frac{{N}}{{R}}=\lambda\left({r}−{b}\right)/{a}^{\mathrm{3}} \\ $$$$\mathrm{sin}\:\beta=\left(\frac{{r}−{a}}{{b}^{\mathrm{3}} }\right)\frac{{N}}{{R}}=\lambda\left({r}−{a}\right)/{b}^{\mathrm{3}} \\ $$$$\mathrm{cos}\:\left(\alpha+\beta\right)={k} \\ $$$$\Rightarrow\:\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} \left(\frac{{r}−{b}}{{a}^{\mathrm{3}} }\right)^{\mathrm{2}} }\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} \left(\frac{{r}−{a}}{{b}^{\mathrm{3}} }\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:−\frac{\lambda^{\mathrm{2}} \left({r}−{a}\right)\left({r}−{b}\right)}{{a}^{\mathrm{3}} {b}^{\mathrm{3}} }={k} \\ $$$$\:\lambda=\frac{{N}}{{R}} \\ $$$$=\frac{\sqrt{\mathrm{1}−{k}^{\mathrm{2}} }}{\:\sqrt{\left(\frac{{r}−{a}}{{b}^{\mathrm{3}} }\right)^{\mathrm{2}} +\left(\frac{{r}−{b}}{{a}^{\mathrm{3}} }\right)^{\mathrm{2}} +\mathrm{2}{k}\left(\frac{{r}−{a}}{{b}^{\mathrm{3}} }\right)\left(\frac{{r}−{b}}{{a}^{\mathrm{3}} }\right)}} \\ $$$${R}=\mathrm{4}\pi\rho{g}\left(\frac{{a}+{b}}{\:\sqrt{\mathrm{1}−{k}^{\mathrm{2}} }}\right) \\ $$$${N}=\frac{\mathrm{4}\pi\rho{g}\left({a}+{b}\right){ab}\sqrt{{ab}}}{\:\sqrt{\left(\frac{{a}}{{b}}\right)^{\mathrm{3}} \left({r}−{a}\right)^{\mathrm{2}} +\left(\frac{{b}}{{a}}\right)^{\mathrm{3}} \left({r}−{b}\right)^{\mathrm{2}} +\mathrm{2}{k}\left({r}−{a}\right)\left({r}−{b}\right)}} \\ $$$${k}=\mathrm{cos}\:\left(\alpha+\beta\right)=\frac{\left({r}−{a}\right)^{\mathrm{2}} +\left({r}−{b}\right)^{\mathrm{2}} −\left({a}+{b}\right)^{\mathrm{2}} }{\mathrm{2}\left({r}−{a}\right)\left({r}−{b}\right)} \\ $$$${N}=\frac{\left(\mathrm{4}\pi\rho{ab}\sqrt{{ab}}\right){g}\left({a}+{b}\right)}{\:\sqrt{\left(\mathrm{1}+\frac{{a}^{\mathrm{3}} }{{b}^{\mathrm{3}} }\right)\left({r}−{a}\right)^{\mathrm{2}} +\left(\mathrm{1}+\frac{{b}^{\mathrm{3}} }{{a}^{\mathrm{3}} }\right)\left({r}−{b}\right)^{\mathrm{2}} −\left({a}+{b}\right)^{\mathrm{2}} }} \\ $$$$ \\ $$

Commented by mr W last updated on 01/Feb/23