Question Number 28835 by ajfour last updated on 30/Jan/18
Commented by ajfour last updated on 30/Jan/18
$${One}\:{sphere}\:{over}\:{another},\:{the} \\ $$$${two}\:{confined}\:{between}\:{ground} \\ $$$${and}\:{two}\:{inclined}\:{walls}. \\ $$$${Find}\:{contact}\:{force}\:{at}\:{each}\:{contact} \\ $$$${point}\::\:\left({Assume}\:{surfaces}\:{frictionless}\right). \\ $$$${N}_{{G}} \:\:\:\::\left({Normal}\:{reaction}\:{from}\:{ground}\right) \\ $$$${N}_{{S}} \:\::\:{Normal}\:{force}\:{between}\:{spheres} \\ $$$${N}_{{L}} \:\::\:\:\:{Normal}\:{from}\:{left}\:{incline} \\ $$$${N}_{{R}} \:\:\::\:\:\:{from}\:{right}\:{incline}\:. \\ $$
Commented by ajfour last updated on 31/Jan/18
Commented by ajfour last updated on 31/Jan/18
![rsin α+(R+r)cos θ+Rsin α =a+y_1 cot α+y_2 cot α ...(i) y_1 =r(1−cos α) r+(R+r)sin θ−Rcos α=y_2 (y_1 +y_2 )=2r+(R+r)sin θ −(R+r)cos α substituting in (i): ⇒ (r+R)sin α+(R+r)cos θ= a+ 2rcot α+[(((R+r)cos α)/(sin α))]sin θ −(((R+r)cos^2 α)/(sin α)) or (((R+r))/(sin α))[sin αcos θ−cos αsin θ] = a+2rcot α−(((R+r))/(sin α)) sin (α−θ)=((asin α)/(R+r))+((2rcos α)/((R+r)))−1 θ= α−sin^(−1) [((asin α)/(R+r))+((2rcos α)/(R+r))−1] Now considering torque about contact point of bigger sphere with right wall: N_S R sin (θ+90°−α)=MgRsin α ⇒ N_S =((Mgsin α)/(cos (α−θ))) and N_S cos θ = N_R sin α so N_L =N_R = ((Mgcos θ)/(cos (α−θ))) considering torque on smaller sphere about contact point with left wall: N_S rsin (α−θ)=(N_G −mg)rsin α ⇒ N_G =mg+((N_S sin (α+θ−90°))/(sin α)) N_G =mg−((Mgsin α)/(cos (α−θ)))×((cos (α+θ))/(sin α)) ⇒ N_G =mg−((Mgcos (θ+α))/(cos (θ−α))) .](Q28862.png)
$${r}\mathrm{sin}\:\alpha+\left({R}+{r}\right)\mathrm{cos}\:\theta+{R}\mathrm{sin}\:\alpha \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={a}+{y}_{\mathrm{1}} \mathrm{cot}\:\alpha+{y}_{\mathrm{2}} \mathrm{cot}\:\alpha\:\:\:\:...\left({i}\right) \\ $$$${y}_{\mathrm{1}} ={r}\left(\mathrm{1}−\mathrm{cos}\:\alpha\right) \\ $$$${r}+\left({R}+{r}\right)\mathrm{sin}\:\theta−{R}\mathrm{cos}\:\alpha={y}_{\mathrm{2}} \\ $$$$\left({y}_{\mathrm{1}} +{y}_{\mathrm{2}} \right)=\mathrm{2}{r}+\left({R}+{r}\right)\mathrm{sin}\:\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\left({R}+{r}\right)\mathrm{cos}\:\alpha \\ $$$${substituting}\:{in}\:\left({i}\right): \\ $$$$\Rightarrow\:\left({r}+{R}\right)\mathrm{sin}\:\alpha+\left({R}+{r}\right)\mathrm{cos}\:\theta=\:{a}+ \\ $$$$\mathrm{2}{r}\mathrm{cot}\:\alpha+\left[\frac{\left({R}+{r}\right)\mathrm{cos}\:\alpha}{\mathrm{sin}\:\alpha}\right]\mathrm{sin}\:\theta \\ $$$$\:\:\:\:\:\:\:\:−\frac{\left({R}+{r}\right)\mathrm{cos}\:^{\mathrm{2}} \alpha}{\mathrm{sin}\:\alpha} \\ $$$$ \\ $$$${or}\:\:\:\frac{\left({R}+{r}\right)}{\mathrm{sin}\:\alpha}\left[\mathrm{sin}\:\alpha\mathrm{cos}\:\theta−\mathrm{cos}\:\alpha\mathrm{sin}\:\theta\right] \\ $$$$\:\:\:\:\:\:\:\:=\:{a}+\mathrm{2}{r}\mathrm{cot}\:\alpha−\frac{\left({R}+{r}\right)}{\mathrm{sin}\:\alpha} \\ $$$$\mathrm{sin}\:\left(\alpha−\theta\right)=\frac{{a}\mathrm{sin}\:\alpha}{{R}+{r}}+\frac{\mathrm{2}{r}\mathrm{cos}\:\alpha}{\left({R}+{r}\right)}−\mathrm{1} \\ $$$$\:\theta=\:\alpha−\mathrm{sin}^{−\mathrm{1}} \left[\frac{{a}\mathrm{sin}\:\alpha}{{R}+{r}}+\frac{\mathrm{2}{r}\mathrm{cos}\:\alpha}{{R}+{r}}−\mathrm{1}\right] \\ $$$${Now}\:{considering}\:{torque}\:{about} \\ $$$${contact}\:{point}\:{of}\:{bigger}\:{sphere} \\ $$$${with}\:{right}\:{wall}: \\ $$$${N}_{{S}} {R}\:\mathrm{sin}\:\left(\theta+\mathrm{90}°−\alpha\right)={MgR}\mathrm{sin}\:\alpha \\ $$$$\Rightarrow\:\:{N}_{{S}} =\frac{{Mg}\mathrm{sin}\:\alpha}{\mathrm{cos}\:\left(\alpha−\theta\right)} \\ $$$${and}\:\:{N}_{{S}} \:\mathrm{cos}\:\theta\:=\:{N}_{{R}} \:\mathrm{sin}\:\alpha \\ $$$${so}\:\:\:{N}_{{L}} ={N}_{{R}} =\:\frac{{Mg}\mathrm{cos}\:\theta}{\mathrm{cos}\:\left(\alpha−\theta\right)}\: \\ $$$${considering}\:{torque}\:{on}\:{smaller} \\ $$$${sphere}\:{about}\:{contact}\:{point}\:{with}\:\:\: \\ $$$${left}\:{wall}: \\ $$$$\:{N}_{{S}} \:{r}\mathrm{sin}\:\left(\alpha−\theta\right)=\left({N}_{{G}} −{mg}\right){r}\mathrm{sin}\:\alpha \\ $$$$\Rightarrow\:{N}_{{G}} ={mg}+\frac{{N}_{{S}} \:\mathrm{sin}\:\left(\alpha+\theta−\mathrm{90}°\right)}{\mathrm{sin}\:\alpha} \\ $$$${N}_{{G}} ={mg}−\frac{{Mg}\mathrm{sin}\:\alpha}{\mathrm{cos}\:\left(\alpha−\theta\right)}×\frac{\mathrm{cos}\:\left(\alpha+\theta\right)}{\mathrm{sin}\:\alpha} \\ $$$$\Rightarrow\:\:{N}_{{G}} ={mg}−\frac{{Mg}\mathrm{cos}\:\left(\theta+\alpha\right)}{\mathrm{cos}\:\left(\theta−\alpha\right)}\:. \\ $$
Commented by mrW2 last updated on 31/Jan/18
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Answered by mrW2 last updated on 30/Jan/18
Commented by mrW2 last updated on 31/Jan/18
)+a=(r/(tan γ))+(r+R)cos θ+(R/(cos β)) r+(r+R)sin θ+a tan α=r tan (α/2) tan α+(r+R)tan α cos θ+(R/(cos α)) sin α cos θ−cos α sin θ=((cos α)/(r+R))[a tan α+r(1− tan (α/2) tan α)−(R/(cos α))] sin (α−θ)=((cos α)/(r+R))[a tan α+r(1− tan (α/2) tan α)−(R/(cos α))] α−θ=sin^(−1) {(1/(r+R))[a sin α+r(cos α− tan (α/2) sin α)−R]} ⇒θ=α−sin^(−1) {(1/(r+R))[a sin α+r(cos α− tan (α/2) sin α)−R]} ⇒θ=α−sin^(−1) {(1/(r+R))[a sin α+r(cos α− ((1−cos α)/(sin α)) sin α)−R]} ⇒θ=α−sin^(−1) [((a sin α+2r cos α)/(r+R))−1] N_R cos β=N_S cos θ ⇒N_S =((cos β)/(cos θ))N_R =((sin α)/(cos θ))N_R N_R sin β+N_S sin θ=Mg N_R (cos α+((sin α)/(cos θ))sin θ)=Mg ⇒N_R =((Mg)/(cos α+sin α tan θ))=((cos θ)/(cos (α−θ)))Mg ⇒N_S =((sin α)/(cos (α−θ)))Mg N_L sin α=N_S cos θ ⇒N_L =((cos θ)/(sin α))×((sin α)/(cos (α−θ)))Mg=((cos θ)/(cos (α−θ)))Mg=N_R N_G +N_L cos α=mg+N_S sin θ ⇒N_G =mg+((sin α sin θ)/(cos (α−θ)))Mg−((cos α cos θ)/(cos (α−θ)))Mg ⇒N_G =mg−((cos (α+θ))/(cos (α−θ)))Mg](Q28842.png)
$$\mathrm{2}\gamma+\alpha=\mathrm{180} \\ $$$$\Rightarrow\gamma=\mathrm{90}−\frac{\alpha}{\mathrm{2}} \\ $$$$\beta=\mathrm{90}−\alpha \\ $$$$ \\ $$$$\left[{r}+\left({r}+{R}\right)\mathrm{sin}\:\theta\right]\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}+{a}=\frac{{r}}{\mathrm{tan}\:\gamma}+\left({r}+{R}\right)\mathrm{cos}\:\theta+\frac{{R}}{\mathrm{cos}\:\beta} \\ $$$${r}+\left({r}+{R}\right)\mathrm{sin}\:\theta+{a}\:\mathrm{tan}\:\alpha={r}\:\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\:\mathrm{tan}\:\alpha+\left({r}+{R}\right)\mathrm{tan}\:\alpha\:\mathrm{cos}\:\theta+\frac{{R}}{\mathrm{cos}\:\alpha} \\ $$$$\mathrm{sin}\:\alpha\:\mathrm{cos}\:\theta−\mathrm{cos}\:\alpha\:\mathrm{sin}\:\theta=\frac{\mathrm{cos}\:\alpha}{{r}+{R}}\left[{a}\:\mathrm{tan}\:\alpha+{r}\left(\mathrm{1}−\:\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\:\mathrm{tan}\:\alpha\right)−\frac{{R}}{\mathrm{cos}\:\alpha}\right] \\ $$$$\mathrm{sin}\:\left(\alpha−\theta\right)=\frac{\mathrm{cos}\:\alpha}{{r}+{R}}\left[{a}\:\mathrm{tan}\:\alpha+{r}\left(\mathrm{1}−\:\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\:\mathrm{tan}\:\alpha\right)−\frac{{R}}{\mathrm{cos}\:\alpha}\right] \\ $$$$\alpha−\theta=\mathrm{sin}^{−\mathrm{1}} \left\{\frac{\mathrm{1}}{{r}+{R}}\left[{a}\:\mathrm{sin}\:\alpha+{r}\left(\mathrm{cos}\:\alpha−\:\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\:\mathrm{sin}\:\alpha\right)−{R}\right]\right\} \\ $$$$\Rightarrow\theta=\alpha−\mathrm{sin}^{−\mathrm{1}} \left\{\frac{\mathrm{1}}{{r}+{R}}\left[{a}\:\mathrm{sin}\:\alpha+{r}\left(\mathrm{cos}\:\alpha−\:\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\:\mathrm{sin}\:\alpha\right)−{R}\right]\right\} \\ $$$$\Rightarrow\theta=\alpha−\mathrm{sin}^{−\mathrm{1}} \left\{\frac{\mathrm{1}}{{r}+{R}}\left[{a}\:\mathrm{sin}\:\alpha+{r}\left(\mathrm{cos}\:\alpha−\:\frac{\mathrm{1}−\mathrm{cos}\:\alpha}{\mathrm{sin}\:\alpha}\:\mathrm{sin}\:\alpha\right)−{R}\right]\right\} \\ $$$$\Rightarrow\theta=\alpha−\mathrm{sin}^{−\mathrm{1}} \left[\frac{{a}\:\mathrm{sin}\:\alpha+\mathrm{2}{r}\:\mathrm{cos}\:\alpha}{{r}+{R}}−\mathrm{1}\right] \\ $$$$ \\ $$$${N}_{{R}} \mathrm{cos}\:\beta={N}_{{S}} \mathrm{cos}\:\theta \\ $$$$\Rightarrow{N}_{{S}} =\frac{\mathrm{cos}\:\beta}{\mathrm{cos}\:\theta}{N}_{{R}} =\frac{\mathrm{sin}\:\alpha}{\mathrm{cos}\:\theta}{N}_{{R}} \\ $$$${N}_{{R}} \mathrm{sin}\:\beta+{N}_{{S}} \mathrm{sin}\:\theta={Mg} \\ $$$${N}_{{R}} \left(\mathrm{cos}\:\alpha+\frac{\mathrm{sin}\:\alpha}{\mathrm{cos}\:\theta}\mathrm{sin}\:\theta\right)={Mg} \\ $$$$\Rightarrow{N}_{{R}} =\frac{{Mg}}{\mathrm{cos}\:\alpha+\mathrm{sin}\:\alpha\:\mathrm{tan}\:\theta}=\frac{\mathrm{cos}\:\theta}{\mathrm{cos}\:\left(\alpha−\theta\right)}{Mg} \\ $$$$\Rightarrow{N}_{{S}} =\frac{\mathrm{sin}\:\alpha}{\mathrm{cos}\:\left(\alpha−\theta\right)}{Mg} \\ $$$$ \\ $$$${N}_{{L}} \mathrm{sin}\:\alpha={N}_{{S}} \mathrm{cos}\:\theta \\ $$$$\Rightarrow{N}_{{L}} =\frac{\mathrm{cos}\:\theta}{\mathrm{sin}\:\alpha}×\frac{\mathrm{sin}\:\alpha}{\mathrm{cos}\:\left(\alpha−\theta\right)}{Mg}=\frac{\mathrm{cos}\:\theta}{\mathrm{cos}\:\left(\alpha−\theta\right)}{Mg}={N}_{{R}} \\ $$$${N}_{{G}} +{N}_{{L}} \mathrm{cos}\:\alpha={mg}+{N}_{{S}} \mathrm{sin}\:\theta \\ $$$$\Rightarrow{N}_{{G}} ={mg}+\frac{\mathrm{sin}\:\alpha\:\mathrm{sin}\:\theta}{\mathrm{cos}\:\left(\alpha−\theta\right)}{Mg}−\frac{\mathrm{cos}\:\alpha\:\mathrm{cos}\:\theta}{\mathrm{cos}\:\left(\alpha−\theta\right)}{Mg} \\ $$$$\Rightarrow{N}_{{G}} ={mg}−\frac{\mathrm{cos}\:\left(\alpha+\theta\right)}{\mathrm{cos}\:\left(\alpha−\theta\right)}{Mg} \\ $$
Commented by mrW2 last updated on 31/Jan/18
![It can also be solved like this: ΣF_x =0: N_L cos β−N_R cos β=0 ⇒N_L =N_R ΣM_A =0: Mg(R+r)cos θ−N_R cos β (R+r)sin θ−N_R sin β (R+r)cos θ=0 ⇒Mgcos θ−N_R cos β sin θ−N_R sin β cos θ=0 ⇒Mgcos θ−N_R sin α sin θ−N_R cos α cos θ=0 ⇒Mgcos θ−N_R cos (α−θ)=0 ⇒N_R =((cos θ)/(cos (α−θ)))Mg=N_L ΣF_y =0: Mg+mg−N_R sin β−N_L sin β−N_G =0 ⇒N_G =Mg+mg−2sin β((cos θMg)/(cos (α−θ))) ⇒N_G =mg+[1−((2cos αcos θ)/(cos (α−θ)))]Mg ⇒N_G =mg−((cos (α+θ))/(cos (α−θ)))Mg ΣF_(x(B)) =0: N_S cos θ−N_R cos β=0 ⇒N_S =((cos β)/(cos θ))×((cos θ)/(cos (α−θ)))Mg ⇒N_S =((sin α)/(cos (α−θ)))Mg](Q28849.png)
$${It}\:{can}\:{also}\:{be}\:{solved}\:{like}\:{this}: \\ $$$$\Sigma{F}_{{x}} =\mathrm{0}: \\ $$$${N}_{{L}} \mathrm{cos}\:\beta−{N}_{{R}} \mathrm{cos}\:\beta=\mathrm{0} \\ $$$$\Rightarrow{N}_{{L}} ={N}_{{R}} \\ $$$$\Sigma{M}_{{A}} =\mathrm{0}: \\ $$$${Mg}\left({R}+{r}\right)\mathrm{cos}\:\theta−{N}_{{R}} \mathrm{cos}\:\beta\:\left({R}+{r}\right)\mathrm{sin}\:\theta−{N}_{{R}} \mathrm{sin}\:\beta\:\left({R}+{r}\right)\mathrm{cos}\:\theta=\mathrm{0} \\ $$$$\Rightarrow{Mg}\mathrm{cos}\:\theta−{N}_{{R}} \mathrm{cos}\:\beta\:\mathrm{sin}\:\theta−{N}_{{R}} \mathrm{sin}\:\beta\:\mathrm{cos}\:\theta=\mathrm{0} \\ $$$$\Rightarrow{Mg}\mathrm{cos}\:\theta−{N}_{{R}} \mathrm{sin}\:\alpha\:\mathrm{sin}\:\theta−{N}_{{R}} \mathrm{cos}\:\alpha\:\mathrm{cos}\:\theta=\mathrm{0} \\ $$$$\Rightarrow{Mg}\mathrm{cos}\:\theta−{N}_{{R}} \mathrm{cos}\:\left(\alpha−\theta\right)=\mathrm{0} \\ $$$$\Rightarrow{N}_{{R}} =\frac{\mathrm{cos}\:\theta}{\mathrm{cos}\:\left(\alpha−\theta\right)}{Mg}={N}_{{L}} \\ $$$$\Sigma{F}_{{y}} =\mathrm{0}: \\ $$$${Mg}+{mg}−{N}_{{R}} \mathrm{sin}\:\beta−{N}_{{L}} \mathrm{sin}\:\beta−{N}_{{G}} =\mathrm{0} \\ $$$$\Rightarrow{N}_{{G}} ={Mg}+{mg}−\mathrm{2sin}\:\beta\frac{\mathrm{cos}\:\theta{Mg}}{\mathrm{cos}\:\left(\alpha−\theta\right)} \\ $$$$\Rightarrow{N}_{{G}} ={mg}+\left[\mathrm{1}−\frac{\mathrm{2cos}\:\alpha\mathrm{cos}\:\theta}{\mathrm{cos}\:\left(\alpha−\theta\right)}\right]{Mg} \\ $$$$\Rightarrow{N}_{{G}} ={mg}−\frac{\mathrm{cos}\:\left(\alpha+\theta\right)}{\mathrm{cos}\:\left(\alpha−\theta\right)}{Mg} \\ $$$$\Sigma{F}_{{x}\left({B}\right)} =\mathrm{0}: \\ $$$${N}_{{S}} \mathrm{cos}\:\theta−{N}_{{R}} \mathrm{cos}\:\beta=\mathrm{0} \\ $$$$\Rightarrow{N}_{{S}} =\frac{\mathrm{cos}\:\beta}{\mathrm{cos}\:\theta}×\frac{\mathrm{cos}\:\theta}{\mathrm{cos}\:\left(\alpha−\theta\right)}{Mg} \\ $$$$\Rightarrow{N}_{{S}} =\frac{\mathrm{sin}\:\alpha}{\mathrm{cos}\:\left(\alpha−\theta\right)}{Mg} \\ $$
Commented by ajfour last updated on 31/Jan/18
$$\mathbb{THANK}\:{you}\:\boldsymbol{{Sir}}. \\ $$