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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..
$${very}\:{meticulous}\:{Sir},\:{I}\:{am}\:{trying} \\ $$$${to}\:{follow},\:{the}\:{outer}\:{sphere}\:{is} \\ $$$${massless},\:{have}\:{you}\:{used}\:{to}\:{fact} \\ $$$${to}\:{simplify}\:{a}\:{bit}.. \\ $$
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.
$${this}\:{is}\:{the}\:{method}\:{i}\:{can}\:{use}.\:{vector} \\ $$$${method}\:{may}\:{have}\:{more}\:{simple} \\ $$$${expressions}\:{than}\:{this}. \\ $$
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.
$${but}\:{actually}\:{the}\:{mass}\:{of}\:{the}\:{outer}\:{sphere} \\ $$$${plays}\:{no}\:{role},\:{it}\:{doesn}'{t}\:{affect}\:{the} \\ $$$${position}\:{of}\:{the}\:{equilibrium}. \\ $$
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.
$${if}\:{we}\:{treat}\:{the}\:{three}\:{solid}\:{spheres}\:{as} \\ $$$${a}\:{single}\:{object}\:{whose}\:{mass}\:{is}\:{M}\:{and} \\ $$$${center}\:{of}\:{mass}\:{is}\:{point}\:{M},\:{we}\:{can}\:{see} \\ $$$${that}\:{N},\:{Mg}\:{and}\:{mg}\:{must}\:{be}\:{on}\:{the} \\ $$$${same}\:{perpendicular}\:{line}\:{when}\:{all} \\ $$$${things}\:{are}\:{in}\:{equilibrium}.\:{m}\:{is}\:{the} \\ $$$${mass}\:{of}\:{the}\:{outer}\:{sphere}.\:{since}\:{mg} \\ $$$${lies}\:{at}\:{the}\:{point}\:{S},\:{it}\:{only}\:{affects}\:{the} \\ $$$${size}\:{of}\:{the}\:{force}\:{N},\:{but}\:{doesn}'{t}\:{affect} \\ $$$${the}\:{position}\:{of}\:{the}\:{solid}\:{spheres}. \\ $$
Commented by ajfour last updated on 20/Oct/19
Thanks for the surplus explanation,  Sir.
$${Thanks}\:{for}\:{the}\:{surplus}\:{explanation}, \\ $$$${Sir}. \\ $$

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