An insect is crawling up a hemispherical bowl of radius R. If the coefficient of friction is 1/3, then upto what height will it be able to crawl? (a)R/5 (b)R/10 (c)R/20 (d)R/30
let the insect crawl upto a height h.
let the radius be r.
also let the point on the surface of the hemishphere at the height h be P.
the center of circular upper portion be O and the point directly below it on the hemisphere be S.
then, there must be a point N on OS such that NS = h. such that ON = (r-h).
also a triangle is formed by O , N and P such that OP = r. let the vertical angle be x in this rigth angled triangle. that is angle PON is x. and ONP = 90 degrees.
also since the insect at the heighest point will be in equilibrium so, there Wsinx = k(Wcosx) where x is the coefficient of friction or 1/3.
so, on solving we get, tanx = k = 1/3
but tanx = (PN)/(ON)
but, if ON = r-h and PN = sqrt[r2 - (r-h)2]
on putting the values and solving we get,
9[r2-(r-h)2] = (r-h)2
or, 3r = sqrt(r-h)
or, (sqrt10r - 3r)/sqrt = h
so, h = 0.05r
or, h = r/20, answer.