question_answer

Calculate the height upto which an insect can crawl up a fixed bowl in the form of a hemisphere of radius r. Given coefficient of friction is \[1/\sqrt{3}\].

Answer:

                  In Fig. 3(HT) .11, suppose the insect can crawl up the bowl upto a pt. P from the bottom B. In doing so, the insect rises through a height BA = h, above the bottom B of the bowl of radius r. As is clear from Fig. 3(HT). 11, \[R=W\cos \alpha \] \[f=W\,\sin \alpha \], where W is weight of insect and f is force of friction. \[\mu =\frac{f}{R}=\frac{W\,\sin \alpha }{W\,\cos \alpha }=\tan \alpha =\frac{1}{\sqrt{3}}\] In \[\Delta OPA\], \[\tan \alpha =\frac{AP}{OA}=\frac{\sqrt{{{r}^{2}}-{{y}^{2}}}}{y}=\frac{1}{\sqrt{3}}\]                 or                \[\frac{{{r}^{2}}-{{y}^{2}}}{{{y}^{2}}}=\frac{1}{3}\] or            \[{{y}^{2}}=\frac{3{{r}^{2}}}{4},y=\sqrt{3}r/2\] \[h=BA=OB-OA=r-y\] \[h=r-\frac{\sqrt{3}}{2}\]              \[r=0.134\,r=13.4%\]of r