Question Number 70395 by ajfour last updated on 04/Oct/19
Commented by ajfour last updated on 04/Oct/19
$${Q}.\mathrm{70159}\:\:\:\:\left({attempt}\:{to}\:{resolve}\right) \\ $$
Answered by ajfour last updated on 04/Oct/19
Commented by ajfour last updated on 04/Oct/19
![let θ/2=φ As moment about left end is zero; R(2Lsin φ)=2MgLsin φ+mg(L+b)sin φ ⇒ R=Mg+((mg)/2)(1+(b/L)) As moment about right end is zero; N=Mg+((mg)/2)(1−(b/L)) Torque on left half about hinge: NLsin φ−Mg((L/2)sin φ)=mga ⇒ sin φ=((2m(a/L))/(M+m(1−b/L))) but bcos 2φ=a ; let M=μm ⇒ (μ+1−(b/L))sin φ=2((b/L))cos 2φ let sin φ=t , b/L=f 1−2t^2 −(((μ+1)/f)−1)(t/2)=0 ⇒ t^2 +(((μ+1)/f)−1)(t/4)−(1/2)=0 ..(I) ________________________ Also considering torque of right half about hinge: RLsin φ=2mgbcos 2φ+Mg((L/2))sin φ (R−((Mg)/2))Lsin φ=2mgbcos 2φ [M+m(1+(b/L))]t=4m((b/L))(1−2t^2 ) ⇒ (μ+1+(b/L))t=4((b/L))(1−2t^2 ) ⇒ 1−2t^2 −(((μ+1)/f)+1)(t/4)=0 t^2 +(((μ+1)/(2f))+(1/2))(t/4)−(1/2)=0 For agreement of (I)& (II) ((μ+1)/f)−1=((μ+1)/(2f))+(1/2) ⇒ ((μ+1)/f)=3 (b/L)=((μ+1)/3) = (1/3)(1+(M/m)) ; Now t^2 +(t/2)−(1/2)=0 ⇒ t=sin φ=(1/2) , θ=2φ =60° .](Q70404.png)
$${let}\:\theta/\mathrm{2}=\phi \\ $$$${As}\:{moment}\:{about}\:{left}\:{end}\:{is}\:{zero}; \\ $$$${R}\left(\mathrm{2}{L}\mathrm{sin}\:\phi\right)=\mathrm{2}{MgL}\mathrm{sin}\:\phi+{mg}\left({L}+{b}\right)\mathrm{sin}\:\phi \\ $$$$\Rightarrow\:{R}={Mg}+\frac{{mg}}{\mathrm{2}}\left(\mathrm{1}+\frac{{b}}{{L}}\right)\:\:\: \\ $$$${As}\:{moment}\:{about}\:{right}\:{end}\:{is}\:{zero}; \\ $$$$\:\:\:\:\:\:{N}={Mg}+\frac{{mg}}{\mathrm{2}}\left(\mathrm{1}−\frac{{b}}{{L}}\right) \\ $$$${Torque}\:{on}\:{left}\:{half}\:{about}\:{hinge}: \\ $$$$\:\:\:\:{NL}\mathrm{sin}\:\phi−{Mg}\left(\frac{{L}}{\mathrm{2}}\mathrm{sin}\:\phi\right)={mga} \\ $$$$\Rightarrow\:\mathrm{sin}\:\phi=\frac{\mathrm{2}{m}\left({a}/{L}\right)}{{M}+{m}\left(\mathrm{1}−{b}/{L}\right)} \\ $$$${but}\:\:{b}\mathrm{cos}\:\mathrm{2}\phi={a}\:;\:\:{let}\:{M}=\mu{m} \\ $$$$\Rightarrow\:\:\left(\mu+\mathrm{1}−\frac{{b}}{{L}}\right)\mathrm{sin}\:\phi=\mathrm{2}\left(\frac{{b}}{{L}}\right)\mathrm{cos}\:\mathrm{2}\phi \\ $$$${let}\:\:\mathrm{sin}\:\phi={t}\:,\:\:{b}/{L}={f} \\ $$$$\mathrm{1}−\mathrm{2}{t}^{\mathrm{2}} −\left(\frac{\mu+\mathrm{1}}{{f}}−\mathrm{1}\right)\frac{{t}}{\mathrm{2}}=\mathrm{0}\:\: \\ $$$$\Rightarrow\:\:{t}^{\mathrm{2}} +\left(\frac{\mu+\mathrm{1}}{{f}}−\mathrm{1}\right)\frac{{t}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0}\:\:..\left({I}\right) \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$${Also}\:{considering}\:{torque}\:{of} \\ $$$${right}\:{half}\:{about}\:{hinge}: \\ $$$$\:\:\:{RL}\mathrm{sin}\:\phi=\mathrm{2}{mgb}\mathrm{cos}\:\mathrm{2}\phi+{Mg}\left(\frac{{L}}{\mathrm{2}}\right)\mathrm{sin}\:\phi \\ $$$$\:\:\:\left({R}−\frac{{Mg}}{\mathrm{2}}\right){L}\mathrm{sin}\:\phi=\mathrm{2}{mgb}\mathrm{cos}\:\mathrm{2}\phi \\ $$$$\left[{M}+{m}\left(\mathrm{1}+\frac{{b}}{{L}}\right)\right]{t}=\mathrm{4}{m}\left(\frac{{b}}{{L}}\right)\left(\mathrm{1}−\mathrm{2}{t}^{\mathrm{2}} \right) \\ $$$$\Rightarrow \\ $$$$\:\:\left(\mu+\mathrm{1}+\frac{{b}}{{L}}\right){t}=\mathrm{4}\left(\frac{{b}}{{L}}\right)\left(\mathrm{1}−\mathrm{2}{t}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\:\mathrm{1}−\mathrm{2}{t}^{\mathrm{2}} −\left(\frac{\mu+\mathrm{1}}{{f}}+\mathrm{1}\right)\frac{{t}}{\mathrm{4}}=\mathrm{0}\:\: \\ $$$$\:\:{t}^{\mathrm{2}} +\left(\frac{\mu+\mathrm{1}}{\mathrm{2}{f}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\frac{{t}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$${For}\:{agreement}\:{of}\:\left({I}\right)\&\:\left({II}\right) \\ $$$$\:\:\frac{\mu+\mathrm{1}}{{f}}−\mathrm{1}=\frac{\mu+\mathrm{1}}{\mathrm{2}{f}}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\frac{\mu+\mathrm{1}}{{f}}=\mathrm{3} \\ $$$$\:\:\frac{{b}}{{L}}=\frac{\mu+\mathrm{1}}{\mathrm{3}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}+\frac{{M}}{{m}}\right)\:\:;\:{Now} \\ $$$$\:\:{t}^{\mathrm{2}} +\frac{{t}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\:\Rightarrow\:{t}=\mathrm{sin}\:\phi=\frac{\mathrm{1}}{\mathrm{2}}\:,\:\:\theta=\mathrm{2}\phi\:=\mathrm{60}°\:. \\ $$
Commented by mr W last updated on 04/Oct/19
$${answer}\:{is}\:{correct}\:{sir}! \\ $$$${it}'{s}\:{obvious}\:{that}\: \\ $$$$\mathrm{90}°−\theta=\frac{\theta}{\mathrm{2}}\:\Rightarrow\:\theta=\mathrm{60}° \\ $$
Commented by mr W last updated on 04/Oct/19
$${interesting}\:{to}\:{see}\:{that}\:\theta\:{is}\:{always}\:\mathrm{60}°. \\ $$$$\frac{{b}}{{L}}=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}+\frac{{M}}{{m}}\right)\geqslant\frac{\mathrm{1}}{\mathrm{3}},\:\leqslant\mathrm{1} \\ $$$${with}\:{M}=\mathrm{0}:\:\frac{{b}}{{L}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${with}\:{M}=\mathrm{2}{m}:\:\frac{{b}}{{L}}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{0}\leqslant{M}\leqslant\mathrm{2}{m} \\ $$