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Question Number 131767 by ajfour last updated on 08/Feb/21

Commented by ajfour last updated on 08/Feb/21

Find force of Normal reaction between the cylinders hanging on a belt wrap.

$${Find}\:{force}\:{of}\:{Normal}\:{reaction} \\ $$$${between}\:{the}\:{cylinders}\:{hanging} \\ $$$${on}\:{a}\:{belt}\:{wrap}. \\ $$

Answered by mr W last updated on 08/Feb/21

Commented by mr W last updated on 08/Feb/21

T=tension in belt N=normal contact force C=com of both cylinders CD is perpendicular to horizontal AB=R+r AC=((M(R+r))/(M+m)) CB=((m(R+r))/(M+m)) let a=DE=DF 2(tan^(−1) (r/a)+tan^(−1) (R/a))=45° (given) tan^(−1) (r/a)+tan^(−1) (R/a)=((45°)/2) (((r/a)+(R/a))/(1−(r/a)×(R/a)))=tan ((45°)/2)=(√2)−1 ((√2)−1)a^2 −(R+r)a−((√2)−1)Rr=0 ⇒a=((R+r+(√(R^2 +r^2 +2(7−4(√2))Rr)))/(2((√2)−1))) α=tan^(−1) (a/r)=(π/2)−tan^(−1) (r/a) β=tan^(−1) (a/R)=(π/2)−tan^(−1) (R/a) ((sin γ)/(BC))=((sin (γ+β))/(DB)) ((DB)/(BC))=((sin (γ+β))/(sin γ))=cos β+((sin β)/(tan γ)) (((M+m)(√(a^2 +R^2 )))/(m(R+r)))=(R/( (√(a^2 +R^2 ))))+(a/(tan γ(√(a^2 +R^2 )))) ⇒tan γ=(a/((((M+m)(a^2 +R^2 ))/(m(R+r)))−R)) ⇒γ=tan^(−1) ((m(R+r)a)/((M+m)(a^2 +R^2 )−m(R+r)R)) φ=(π/2)+sin^(−1) ((R−r)/(R+r)) δ=π−φ−2 tan^(−1) (R/a)=(π/2)−sin^(−1) ((R−r)/(R+r))−2 tan^(−1) (R/a) ϕ=(π/2)−β−γ ⇒ϕ=(π/2)−tan^(−1) (a/R)−tan^(−1) ((m(R+r)a)/((M+m)(a^2 +R^2 )−m(R+r)R)) θ=(π/2)−(π/8)−δ ⇒θ=sin^(−1) ((R−r)/(R+r))+2 tan^(−1) (R/a)−(π/8) 2T cos ((45°)/2)=(M+m)g ⇒T=(((√(2(2−(√2))))(M+m)g)/2) N cos ϕ=T(sin ((45°)/2)+cos θ) N=(√(2(2−(√2))))(M+m)g×(((((√2)−1)/( (√(2(2−(√2))))))+cos (sin^(−1) ((R−r)/(R+r))+2 tan^(−1) (R/a)−(π/8)))/(2 sin (tan^(−1) (a/R)+tan^(−1) ((m(R+r)a)/((M+m)(a^2 +R^2 )−m(R+r)R))))) ⇒(N/((M+m)g))=(((√2)−1+(√(2(2−(√2)))) cos (sin^(−1) ((R−r)/(R+r))+2 tan^(−1) (R/a)−(π/8)))/(2 cos (tan^(−1) (R/a)−tan^(−1) ((m(R+r)a)/((M+m)(a^2 +R^2 )−m(R+r)R)))))

$${T}={tension}\:{in}\:{belt} \\ $$$${N}={normal}\:{contact}\:{force} \\ $$$${C}={com}\:{of}\:{both}\:{cylinders} \\ $$$${CD}\:{is}\:{perpendicular}\:{to}\:{horizontal} \\ $$$${AB}={R}+{r} \\ $$$${AC}=\frac{{M}\left({R}+{r}\right)}{{M}+{m}} \\ $$$${CB}=\frac{{m}\left({R}+{r}\right)}{{M}+{m}} \\ $$$${let}\:{a}={DE}={DF} \\ $$$$\mathrm{2}\left(\mathrm{tan}^{−\mathrm{1}} \frac{{r}}{{a}}+\mathrm{tan}^{−\mathrm{1}} \frac{{R}}{{a}}\right)=\mathrm{45}°\:\left({given}\right) \\ $$$$\mathrm{tan}^{−\mathrm{1}} \frac{{r}}{{a}}+\mathrm{tan}^{−\mathrm{1}} \frac{{R}}{{a}}=\frac{\mathrm{45}°}{\mathrm{2}} \\ $$$$\frac{\frac{{r}}{{a}}+\frac{{R}}{{a}}}{\mathrm{1}−\frac{{r}}{{a}}×\frac{{R}}{{a}}}=\mathrm{tan}\:\frac{\mathrm{45}°}{\mathrm{2}}=\sqrt{\mathrm{2}}−\mathrm{1} \\ $$$$\left(\sqrt{\mathrm{2}}−\mathrm{1}\right){a}^{\mathrm{2}} −\left({R}+{r}\right){a}−\left(\sqrt{\mathrm{2}}−\mathrm{1}\right){Rr}=\mathrm{0} \\ $$$$\Rightarrow{a}=\frac{{R}+{r}+\sqrt{{R}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{7}−\mathrm{4}\sqrt{\mathrm{2}}\right){Rr}}}{\mathrm{2}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)} \\ $$$$\alpha=\mathrm{tan}^{−\mathrm{1}} \frac{{a}}{{r}}=\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \frac{{r}}{{a}} \\ $$$$\beta=\mathrm{tan}^{−\mathrm{1}} \frac{{a}}{{R}}=\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \frac{{R}}{{a}} \\ $$$$\frac{\mathrm{sin}\:\gamma}{{BC}}=\frac{\mathrm{sin}\:\left(\gamma+\beta\right)}{{DB}} \\ $$$$\frac{{DB}}{{BC}}=\frac{\mathrm{sin}\:\left(\gamma+\beta\right)}{\mathrm{sin}\:\gamma}=\mathrm{cos}\:\beta+\frac{\mathrm{sin}\:\beta}{\mathrm{tan}\:\gamma} \\ $$$$\frac{\left({M}+{m}\right)\sqrt{{a}^{\mathrm{2}} +{R}^{\mathrm{2}} }}{{m}\left({R}+{r}\right)}=\frac{{R}}{\:\sqrt{{a}^{\mathrm{2}} +{R}^{\mathrm{2}} }}+\frac{{a}}{\mathrm{tan}\:\gamma\sqrt{{a}^{\mathrm{2}} +{R}^{\mathrm{2}} }} \\ $$$$\Rightarrow\mathrm{tan}\:\gamma=\frac{{a}}{\frac{\left({M}+{m}\right)\left({a}^{\mathrm{2}} +{R}^{\mathrm{2}} \right)}{{m}\left({R}+{r}\right)}−{R}} \\ $$$$\Rightarrow\gamma=\mathrm{tan}^{−\mathrm{1}} \frac{{m}\left({R}+{r}\right){a}}{\left({M}+{m}\right)\left({a}^{\mathrm{2}} +{R}^{\mathrm{2}} \right)−{m}\left({R}+{r}\right){R}} \\ $$$$\phi=\frac{\pi}{\mathrm{2}}+\mathrm{sin}^{−\mathrm{1}} \frac{{R}−{r}}{{R}+{r}} \\ $$$$\delta=\pi−\phi−\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \frac{{R}}{{a}}=\frac{\pi}{\mathrm{2}}−\mathrm{sin}^{−\mathrm{1}} \frac{{R}−{r}}{{R}+{r}}−\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \frac{{R}}{{a}} \\ $$$$\varphi=\frac{\pi}{\mathrm{2}}−\beta−\gamma \\ $$$$\Rightarrow\varphi=\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \frac{{a}}{{R}}−\mathrm{tan}^{−\mathrm{1}} \frac{{m}\left({R}+{r}\right){a}}{\left({M}+{m}\right)\left({a}^{\mathrm{2}} +{R}^{\mathrm{2}} \right)−{m}\left({R}+{r}\right){R}} \\ $$$$\theta=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{8}}−\delta \\ $$$$\Rightarrow\theta=\mathrm{sin}^{−\mathrm{1}} \frac{{R}−{r}}{{R}+{r}}+\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \frac{{R}}{{a}}−\frac{\pi}{\mathrm{8}} \\ $$$$ \\ $$$$\mathrm{2}{T}\:\mathrm{cos}\:\frac{\mathrm{45}°}{\mathrm{2}}=\left({M}+{m}\right){g} \\ $$$$\Rightarrow{T}=\frac{\sqrt{\mathrm{2}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)}\left({M}+{m}\right){g}}{\mathrm{2}} \\ $$$${N}\:\mathrm{cos}\:\varphi={T}\left(\mathrm{sin}\:\frac{\mathrm{45}°}{\mathrm{2}}+\mathrm{cos}\:\theta\right) \\ $$$${N}=\sqrt{\mathrm{2}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)}\left({M}+{m}\right){g}×\frac{\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\:\sqrt{\mathrm{2}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)}}+\mathrm{cos}\:\left(\mathrm{sin}^{−\mathrm{1}} \frac{{R}−{r}}{{R}+{r}}+\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \frac{{R}}{{a}}−\frac{\pi}{\mathrm{8}}\right)}{\mathrm{2}\:\mathrm{sin}\:\left(\mathrm{tan}^{−\mathrm{1}} \frac{{a}}{{R}}+\mathrm{tan}^{−\mathrm{1}} \frac{{m}\left({R}+{r}\right){a}}{\left({M}+{m}\right)\left({a}^{\mathrm{2}} +{R}^{\mathrm{2}} \right)−{m}\left({R}+{r}\right){R}}\right)} \\ $$$$\Rightarrow\frac{{N}}{\left({M}+{m}\right){g}}=\frac{\sqrt{\mathrm{2}}−\mathrm{1}+\sqrt{\mathrm{2}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)}\:\mathrm{cos}\:\left(\mathrm{sin}^{−\mathrm{1}} \frac{{R}−{r}}{{R}+{r}}+\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \frac{{R}}{{a}}−\frac{\pi}{\mathrm{8}}\right)}{\mathrm{2}\:\mathrm{cos}\:\left(\mathrm{tan}^{−\mathrm{1}} \frac{{R}}{{a}}−\mathrm{tan}^{−\mathrm{1}} \frac{{m}\left({R}+{r}\right){a}}{\left({M}+{m}\right)\left({a}^{\mathrm{2}} +{R}^{\mathrm{2}} \right)−{m}\left({R}+{r}\right){R}}\right)} \\ $$

Commented by ajfour last updated on 08/Feb/21

Looks a fabulous design Sir. I couldn′t dare attempt, let me some time following...

$${Looks}\:{a}\:{fabulous}\:{design}\:{Sir}. \\ $$$${I}\:{couldn}'{t}\:{dare}\:{attempt},\:{let}\:{me} \\ $$$${some}\:{time}\:{following}... \\ $$

Commented by otchereabdullai@gmail.com last updated on 10/Feb/21

You are gifted prof W

$$\mathrm{You}\:\mathrm{are}\:\mathrm{gifted}\:\mathrm{prof}\:\mathrm{W} \\ $$

Answered by ajfour last updated on 08/Feb/21

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