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Question Number 71645 by ajfour last updated on 18/Oct/19

Commented by ajfour last updated on 18/Oct/19

If the three inner solid spheres have radii p,q,r and the same density ρ, while the outer sphere has radius R, and negligible mass, find how high from ground are points A,B, C when spheres are in static equilibrium.

Ifthethreeinnersolidsphereshaveradiip,q,randthesamedensityρ,whiletheouterspherehasradiusR,andnegligiblemass,findhowhighfromgroundarepointsA,B,Cwhenspheresareinstaticequilibrium.

Answered by mr W last updated on 19/Oct/19

Commented by ajfour last updated on 19/Oct/19

very meticulous Sir, I am trying to follow, the outer sphere is massless, have you used to fact to simplify a bit..

verymeticulousSir,Iamtryingtofollow,theoutersphereismassless,haveyouusedtofacttosimplifyabit..

Commented by mr W last updated on 20/Oct/19

S=center of outer sphere M=center of total mass of spheres A,B,C G=contact point on ground M must lie in the plane of ΔABC S,M,G must be collinear and the line is perpendicular to the ground. we take ΔABC as xy−plane, A as the origin, AB as the x−axis let a=q+r, b=r+p, c=p+q let M(s,t,0), S(u,v,w) let area of ΔABC=Δ using Heron formula we will get Δ=(√((p+q+r)pqr)) (1/2)ah_A =Δ ⇒h_A =((2(√((p+q+r)pqr)))/(q+r)) (1/2)bh_B =Δ ⇒h_B =((2(√((p+q+r)pqr)))/(r+p)) (1/2)ch_C =Δ ⇒h_C =((2(√((p+q+r)pqr)))/(p+q)) A(0, 0, 0) B(p+q, 0, 0) C(x_C , y_C , 0) y_C =h_C =((2(√((p+q+r)pqr)))/(p+q)) x_C =(√(b^2 −h_C ^2 ))=(√((r+p)^2 −((4(p+q+r)pqr)/((p+q)^2 ))))=((√((p+q)^2 (r+p)^2 −4(p+q+r)pqr))/(p+q)) t=(r^3 /(p^3 +q^3 +r^3 ))h_C =((2r^3 (√((p+q+r)pqr)))/((p^3 +q^3 +r^3 )(p+q))) s=((q^3 (p+q)+r^3 x_C )/(p^3 +q^3 +r^3 ))=((q^3 (p+q)^2 +r^3 (√((p+q)^2 (r+p)^2 −4(p+q+r)pqr)))/((p^3 +q^3 +r^3 )(p+q))) u^2 +v^2 +w^2 =(R−p)^2 ...(i) (u−p−q)^2 +v^2 +w^2 =(R−q)^2 ...(ii) (u−x_C )^2 +(v−y_C )^2 +w^2 =(R−r)^2 ...(iii) (i)−(ii): 2(p+q)u−(p+q)^2 =−(2R−p−q)(p−q) ⇒u=((p(p+q)−R(p−q))/(p+q)) (i)−(iii): 2ux_C −x_C ^2 +2vy_C −y_C ^2 =(2R−p−r)(r−p) 2ux_C +2vy_C =(r+p)^2 +(2R−p−r)(r−p) [((p(p+q)−R(p−q))/(p+q))]((√((p+q)^2 (r+p)^2 −4(p+q+r)pqr))/(p+q))+v((2(√((p+q+r)pqr)))/(p+q))=(1/2)[(r+p)^2 +(2R−p−r)(r−p)] ⇒v=(((p+q)^2 [p(r+p)+R(r−p)]−[p(p+q)−R(p−q)](√((p+q)^2 (r+p)^2 −4(p+q+r)pqr)))/(2(p+q)(√((p+q+r)pqr)))) ⇒w=(√((R−p)^2 −u^2 −v^2 )) let δ=MS=(√((u−s)^2 +(v−t)^2 +w^2 ))=(√((R−p)^2 +s^2 +t^2 −2(su+tv))) GS=R ((x_G −s)/(s−u))=(R/δ) ⇒x_G =(1+(R/δ))s−(R/δ)u similarly ⇒y_G =(1+(R/δ))t−(R/δ)v ⇒z_G =−(R/δ)w eqn. of ground plane with MS as normal: normal MS=(u−s,v−t,w)=δ eqn. of ground plane: (u−s)(x−x_G )+(v−t)(y−y_G )+w(z−z_G )=0 distance of point A,B,C to ground plane: d_A =((∣(u−s)x_G +(v−t)y_G +wz_G ∣)/δ) d_B =((∣(u−s)(x_G −p−q)+(v−t)y_G +wz_G ∣)/δ) d_C =((∣(u−s)(x_G −x_C )+(v−t)(y_G −y_C )+wz_G )∣)/δ)

S=centerofoutersphereM=centeroftotalmassofspheresA,B,CG=contactpointongroundMmustlieintheplaneofΔABCS,M,Gmustbecollinearandthelineisperpendiculartotheground.wetakeΔABCasxy−plane,Aastheorigin,ABasthex−axisleta=q+r,b=r+p,c=p+qletM(s,t,0),S(u,v,w)letareaofΔABC=ΔusingHeronformulawewillgetΔ=(p+q+r)pqr12ahA=Δ⇒hA=2(p+q+r)pqrq+r12bhB=Δ⇒hB=2(p+q+r)pqrr+p12chC=Δ⇒hC=2(p+q+r)pqrp+qA(0,0,0)B(p+q,0,0)C(xC,yC,0)yC=hC=2(p+q+r)pqrp+qxC=b2−hC2=(r+p)2−4(p+q+r)pqr(p+q)2=(p+q)2(r+p)2−4(p+q+r)pqrp+qt=r3p3+q3+r3hC=2r3(p+q+r)pqr(p3+q3+r3)(p+q)s=q3(p+q)+r3xCp3+q3+r3=q3(p+q)2+r3(p+q)2(r+p)2−4(p+q+r)pqr(p3+q3+r3)(p+q)u2+v2+w2=(R−p)2...(i)(u−p−q)2+v2+w2=(R−q)2...(ii)(u−xC)2+(v−yC)2+w2=(R−r)2...(iii)(i)−(ii):2(p+q)u−(p+q)2=−(2R−p−q)(p−q)⇒u=p(p+q)−R(p−q)p+q(i)−(iii):2uxC−xC2+2vyC−yC2=(2R−p−r)(r−p)2uxC+2vyC=(r+p)2+(2R−p−r)(r−p)[p(p+q)−R(p−q)p+q](p+q)2(r+p)2−4(p+q+r)pqrp+q+v2(p+q+r)pqrp+q=12[(r+p)2+(2R−p−r)(r−p)]⇒v=(p+q)2[p(r+p)+R(r−p)]−[p(p+q)−R(p−q)](p+q)2(r+p)2−4(p+q+r)pqr2(p+q)(p+q+r)pqr⇒w=(R−p)2−u2−v2letδ=MS=(u−s)2+(v−t)2+w2=(R−p)2+s2+t2−2(su+tv)GS=RxG−ss−u=Rδ⇒xG=(1+Rδ)s−Rδusimilarly⇒yG=(1+Rδ)t−Rδv⇒zG=−Rδweqn.ofgroundplanewithMSasnormal:normalMS=(u−s,v−t,w)=δeqn.ofgroundplane:(u−s)(x−xG)+(v−t)(y−yG)+w(z−zG)=0distanceofpointA,B,Ctogroundplane:dA=∣(u−s)xG+(v−t)yG+wzG∣δdB=∣(u−s)(xG−p−q)+(v−t)yG+wzG∣δdC=∣(u−s)(xG−xC)+(v−t)(yG−yC)+wzG)∣δ

Commented by mr W last updated on 19/Oct/19

this is the method i can use. vector method may have more simple expressions than this.

thisisthemethodicanuse.vectormethodmayhavemoresimpleexpressionsthanthis.

Commented by mr W last updated on 19/Oct/19

but actually the mass of the outer sphere plays no role, it doesn′t affect the position of the equilibrium.

butactuallythemassoftheoutersphereplaysnorole,itdoesn′taffectthepositionoftheequilibrium.

Commented by mr W last updated on 20/Oct/19

Commented by mr W last updated on 20/Oct/19

if we treat the three solid spheres as a single object whose mass is M and center of mass is point M, we can see that N, Mg and mg must be on the same perpendicular line when all things are in equilibrium. m is the mass of the outer sphere. since mg lies at the point S, it only affects the size of the force N, but doesn′t affect the position of the solid spheres.

ifwetreatthethreesolidspheresasasingleobjectwhosemassisMandcenterofmassispointM,wecanseethatN,Mgandmgmustbeonthesameperpendicularlinewhenallthingsareinequilibrium.misthemassoftheoutersphere.sincemgliesatthepointS,itonlyaffectsthesizeoftheforceN,butdoesn′taffectthepositionofthesolidspheres.

Commented by ajfour last updated on 20/Oct/19

Thanks for the surplus explanation, Sir.

Thanksforthesurplusexplanation,Sir.

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