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Question Number 86260 by ajfour last updated on 27/Mar/20

Commented by ajfour last updated on 27/Mar/20

$$Findthemaximumforceapplied$$$$inupturningtheeql.△,expending$$$$minimumenergy.$$

Answered by mr W last updated on 27/Mar/20

Commented by mr W last updated on 27/Mar/20

$$\tan \alpha =\frac{\sqrt{3}a}{2(b+\frac{a}{2})}=\frac{\sqrt{3}a}{a+2b}$$$$d=b\sin \alpha =\frac{\sqrt{3}ab}{2\sqrt{a^2+ab+b^2}}$$$$Fd-mg\frac{a}{2}=I\frac{d\omega }{dt}=0$$$$\Rightarrow \Rightarrow F=\frac{mg\sqrt{a^2+ab+b^2}}{\sqrt{3}b}=mg\sqrt{\frac{a^2+ab+b^2}{3b^2}}$$

Commented by ajfour last updated on 27/Mar/20

$$yessir,understood,thanks!$$

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