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Question-86260




Question Number 86260 by ajfour last updated on 27/Mar/20
Commented by ajfour last updated on 27/Mar/20
Find the maximum force applied  in upturning the eql. △, expending  minimum energy.
$${Find}\:{the}\:{maximum}\:{force}\:{applied} \\ $$$${in}\:{upturning}\:{the}\:{eql}.\:\bigtriangleup,\:{expending} \\ $$$${minimum}\:{energy}. \\ $$
Answered by mr W last updated on 27/Mar/20
Commented by mr W last updated on 27/Mar/20
tan α=(((√3)a)/(2(b+(a/2))))=(((√3)a)/(a+2b))  d=b sin α=(((√3)ab)/(2(√(a^2 +ab+b^2 ))))  Fd−mg(a/2)=I(dω/dt)=0  ⇒⇒F=((mg(√(a^2 +ab+b^2 )))/( (√3)b))=mg(√((a^2 +ab+b^2 )/(3b^2 )))
$$\mathrm{tan}\:\alpha=\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}\left({b}+\frac{{a}}{\mathrm{2}}\right)}=\frac{\sqrt{\mathrm{3}}{a}}{{a}+\mathrm{2}{b}} \\ $$$${d}={b}\:\mathrm{sin}\:\alpha=\frac{\sqrt{\mathrm{3}}{ab}}{\mathrm{2}\sqrt{{a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} }} \\ $$$${Fd}−{mg}\frac{{a}}{\mathrm{2}}={I}\frac{{d}\omega}{{dt}}=\mathrm{0} \\ $$$$\Rightarrow\Rightarrow{F}=\frac{{mg}\sqrt{{a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} }}{\:\sqrt{\mathrm{3}}{b}}={mg}\sqrt{\frac{{a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} }{\mathrm{3}{b}^{\mathrm{2}} }} \\ $$
Commented by ajfour last updated on 27/Mar/20
yes sir, understood, thanks!
$${yes}\:{sir},\:{understood},\:{thanks}!\: \\ $$

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