Question-71645

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.

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

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

Commented by ajfour last updated on 19/Oct/19

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

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)

$${S}={center}\:{of}\:{outer}\:{sphere} \\ $$$${M}={center}\:{of}\:{total}\:{mass}\:{of}\:{spheres}\:{A},{B},{C} \\ $$$${G}={contact}\:{point}\:{on}\:{ground} \\ $$$${M}\:{must}\:{lie}\:{in}\:{the}\:{plane}\:{of}\:\Delta{ABC} \\ $$$${S},{M},{G}\:{must}\:{be}\:{collinear}\:{and}\:{the}\:{line} \\ $$$${is}\:{perpendicular}\:{to}\:{the}\:{ground}. \\ $$$$ \\ $$$${we}\:{take}\:\Delta{ABC}\:{as}\:{xy}−{plane},\:{A}\:{as}\:{the} \\ $$$${origin},\:{AB}\:{as}\:{the}\:{x}−{axis} \\ $$$${let}\:{a}={q}+{r},\:{b}={r}+{p},\:{c}={p}+{q} \\ $$$${let}\:{M}\left({s},{t},\mathrm{0}\right),\:{S}\left({u},{v},{w}\right) \\ $$$$ \\ $$$${let}\:{area}\:{of}\:\Delta{ABC}=\Delta \\ $$$${using}\:{Heron}\:{formula}\:{we}\:{will}\:{get} \\ $$$$\Delta=\sqrt{\left({p}+{q}+{r}\right){pqr}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{ah}_{{A}} =\Delta\:\Rightarrow{h}_{{A}} =\frac{\mathrm{2}\sqrt{\left({p}+{q}+{r}\right){pqr}}}{{q}+{r}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{bh}_{{B}} =\Delta\:\Rightarrow{h}_{{B}} =\frac{\mathrm{2}\sqrt{\left({p}+{q}+{r}\right){pqr}}}{{r}+{p}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{ch}_{{C}} =\Delta\:\Rightarrow{h}_{{C}} =\frac{\mathrm{2}\sqrt{\left({p}+{q}+{r}\right){pqr}}}{{p}+{q}} \\ $$$$ \\ $$$${A}\left(\mathrm{0},\:\mathrm{0},\:\mathrm{0}\right) \\ $$$${B}\left({p}+{q},\:\mathrm{0},\:\mathrm{0}\right) \\ $$$${C}\left({x}_{{C}} ,\:{y}_{{C}} ,\:\mathrm{0}\right) \\ $$$${y}_{{C}} ={h}_{{C}} =\frac{\mathrm{2}\sqrt{\left({p}+{q}+{r}\right){pqr}}}{{p}+{q}} \\ $$$${x}_{{C}} =\sqrt{{b}^{\mathrm{2}} −{h}_{{C}} ^{\mathrm{2}} }=\sqrt{\left({r}+{p}\right)^{\mathrm{2}} −\frac{\mathrm{4}\left({p}+{q}+{r}\right){pqr}}{\left({p}+{q}\right)^{\mathrm{2}} }}=\frac{\sqrt{\left({p}+{q}\right)^{\mathrm{2}} \left({r}+{p}\right)^{\mathrm{2}} −\mathrm{4}\left({p}+{q}+{r}\right){pqr}}}{{p}+{q}} \\ $$$${t}=\frac{{r}^{\mathrm{3}} }{{p}^{\mathrm{3}} +{q}^{\mathrm{3}} +{r}^{\mathrm{3}} }{h}_{{C}} =\frac{\mathrm{2}{r}^{\mathrm{3}} \sqrt{\left({p}+{q}+{r}\right){pqr}}}{\left({p}^{\mathrm{3}} +{q}^{\mathrm{3}} +{r}^{\mathrm{3}} \right)\left({p}+{q}\right)} \\ $$$${s}=\frac{{q}^{\mathrm{3}} \left({p}+{q}\right)+{r}^{\mathrm{3}} {x}_{{C}} }{{p}^{\mathrm{3}} +{q}^{\mathrm{3}} +{r}^{\mathrm{3}} }=\frac{{q}^{\mathrm{3}} \left({p}+{q}\right)^{\mathrm{2}} +{r}^{\mathrm{3}} \sqrt{\left({p}+{q}\right)^{\mathrm{2}} \left({r}+{p}\right)^{\mathrm{2}} −\mathrm{4}\left({p}+{q}+{r}\right){pqr}}}{\left({p}^{\mathrm{3}} +{q}^{\mathrm{3}} +{r}^{\mathrm{3}} \right)\left({p}+{q}\right)} \\ $$$$ \\ $$$${u}^{\mathrm{2}} +{v}^{\mathrm{2}} +{w}^{\mathrm{2}} =\left({R}−{p}\right)^{\mathrm{2}} \:\:\:\:…\left({i}\right) \\ $$$$\left({u}−{p}−{q}\right)^{\mathrm{2}} +{v}^{\mathrm{2}} +{w}^{\mathrm{2}} =\left({R}−{q}\right)^{\mathrm{2}} \:\:\:…\left({ii}\right) \\ $$$$\left({u}−{x}_{{C}} \right)^{\mathrm{2}} +\left({v}−{y}_{{C}} \right)^{\mathrm{2}} +{w}^{\mathrm{2}} =\left({R}−{r}\right)^{\mathrm{2}} \:\:\:…\left({iii}\right) \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$$\mathrm{2}\left({p}+{q}\right){u}−\left({p}+{q}\right)^{\mathrm{2}} =−\left(\mathrm{2}{R}−{p}−{q}\right)\left({p}−{q}\right) \\ $$$$\Rightarrow{u}=\frac{{p}\left({p}+{q}\right)−{R}\left({p}−{q}\right)}{{p}+{q}} \\ $$$$\left({i}\right)−\left({iii}\right): \\ $$$$\mathrm{2}{ux}_{{C}} −{x}_{{C}} ^{\mathrm{2}} +\mathrm{2}{vy}_{{C}} −{y}_{{C}} ^{\mathrm{2}} =\left(\mathrm{2}{R}−{p}−{r}\right)\left({r}−{p}\right) \\ $$$$\mathrm{2}{ux}_{{C}} +\mathrm{2}{vy}_{{C}} =\left({r}+{p}\right)^{\mathrm{2}} +\left(\mathrm{2}{R}−{p}−{r}\right)\left({r}−{p}\right) \\ $$$$\left[\frac{{p}\left({p}+{q}\right)−{R}\left({p}−{q}\right)}{{p}+{q}}\right]\frac{\sqrt{\left({p}+{q}\right)^{\mathrm{2}} \left({r}+{p}\right)^{\mathrm{2}} −\mathrm{4}\left({p}+{q}+{r}\right){pqr}}}{{p}+{q}}+{v}\frac{\mathrm{2}\sqrt{\left({p}+{q}+{r}\right){pqr}}}{{p}+{q}}=\frac{\mathrm{1}}{\mathrm{2}}\left[\left({r}+{p}\right)^{\mathrm{2}} +\left(\mathrm{2}{R}−{p}−{r}\right)\left({r}−{p}\right)\right] \\ $$$$\Rightarrow{v}=\frac{\left({p}+{q}\right)^{\mathrm{2}} \left[{p}\left({r}+{p}\right)+{R}\left({r}−{p}\right)\right]−\left[{p}\left({p}+{q}\right)−{R}\left({p}−{q}\right)\right]\sqrt{\left({p}+{q}\right)^{\mathrm{2}} \left({r}+{p}\right)^{\mathrm{2}} −\mathrm{4}\left({p}+{q}+{r}\right){pqr}}}{\mathrm{2}\left({p}+{q}\right)\sqrt{\left({p}+{q}+{r}\right){pqr}}} \\ $$$$\Rightarrow{w}=\sqrt{\left({R}−{p}\right)^{\mathrm{2}} −{u}^{\mathrm{2}} −{v}^{\mathrm{2}} } \\ $$$$ \\ $$$${let}\:\delta={MS}=\sqrt{\left({u}−{s}\right)^{\mathrm{2}} +\left({v}−{t}\right)^{\mathrm{2}} +{w}^{\mathrm{2}} }=\sqrt{\left({R}−{p}\right)^{\mathrm{2}} +{s}^{\mathrm{2}} +{t}^{\mathrm{2}} −\mathrm{2}\left({su}+{tv}\right)} \\ $$$${GS}={R} \\ $$$$\frac{{x}_{{G}} −{s}}{{s}−{u}}=\frac{{R}}{\delta} \\ $$$$\Rightarrow{x}_{{G}} =\left(\mathrm{1}+\frac{{R}}{\delta}\right){s}−\frac{{R}}{\delta}{u} \\ $$$${similarly} \\ $$$$\Rightarrow{y}_{{G}} =\left(\mathrm{1}+\frac{{R}}{\delta}\right){t}−\frac{{R}}{\delta}{v} \\ $$$$\Rightarrow{z}_{{G}} =−\frac{{R}}{\delta}{w} \\ $$$$ \\ $$$${eqn}.\:{of}\:{ground}\:{plane}\:{with}\:{MS}\:{as}\:{normal}: \\ $$$${normal}\:{MS}=\left({u}−{s},{v}−{t},{w}\right)=\delta \\ $$$${eqn}.\:{of}\:{ground}\:{plane}: \\ $$$$\left({u}−{s}\right)\left({x}−{x}_{{G}} \right)+\left({v}−{t}\right)\left({y}−{y}_{{G}} \right)+{w}\left({z}−{z}_{{G}} \right)=\mathrm{0} \\ $$$$ \\ $$$${distance}\:{of}\:{point}\:{A},{B},{C}\:{to}\:{ground}\:{plane}: \\ $$$${d}_{{A}} =\frac{\mid\left({u}−{s}\right){x}_{{G}} +\left({v}−{t}\right){y}_{{G}} +{wz}_{{G}} \mid}{\delta} \\ $$$${d}_{{B}} =\frac{\mid\left({u}−{s}\right)\left({x}_{{G}} −{p}−{q}\right)+\left({v}−{t}\right){y}_{{G}} +{wz}_{{G}} \mid}{\delta} \\ $$$${d}_{{C}} =\frac{\left.\mid\left({u}−{s}\right)\left({x}_{{G}} −{x}_{{C}} \right)+\left({v}−{t}\right)\left({y}_{{G}} −{y}_{{C}} \right)+{wz}_{{G}} \right)\mid}{\delta} \\ $$

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

$${this}\:{is}\:{the}\:{method}\:{i}\:{can}\:{use}.\:{vector} \\ $$$${method}\:{may}\:{have}\:{more}\:{simple} \\ $$$${expressions}\:{than}\:{this}. \\ $$

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

$${but}\:{actually}\:{the}\:{mass}\:{of}\:{the}\:{outer}\:{sphere} \\ $$$${plays}\:{no}\:{role},\:{it}\:{doesn}'{t}\:{affect}\:{the} \\ $$$${position}\:{of}\:{the}\:{equilibrium}. \\ $$

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

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

$${if}\:{we}\:{treat}\:{the}\:{three}\:{solid}\:{spheres}\:{as} \\ $$$${a}\:{single}\:{object}\:{whose}\:{mass}\:{is}\:{M}\:{and} \\ $$$${center}\:{of}\:{mass}\:{is}\:{point}\:{M},\:{we}\:{can}\:{see} \\ $$$${that}\:{N},\:{Mg}\:{and}\:{mg}\:{must}\:{be}\:{on}\:{the} \\ $$$${same}\:{perpendicular}\:{line}\:{when}\:{all} \\ $$$${things}\:{are}\:{in}\:{equilibrium}.\:{m}\:{is}\:{the} \\ $$$${mass}\:{of}\:{the}\:{outer}\:{sphere}.\:{since}\:{mg} \\ $$$${lies}\:{at}\:{the}\:{point}\:{S},\:{it}\:{only}\:{affects}\:{the} \\ $$$${size}\:{of}\:{the}\:{force}\:{N},\:{but}\:{doesn}'{t}\:{affect} \\ $$$${the}\:{position}\:{of}\:{the}\:{solid}\:{spheres}. \\ $$

Commented by ajfour last updated on 20/Oct/19

$${Thanks}\:{for}\:{the}\:{surplus}\:{explanation}, \\ $$$${Sir}. \\ $$

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