一螞蟻在半徑R之碗壁沿壁上爬

Let a be the angle substends at the centre of the semi-spherical bowl with the earth at which the ant begins to slide. Then,
mg.sin(a) = u.r
where m is the mass of the ant, g is the acceleration due to gravity, r is the normal reaction
but r = mg.cos(a)
hence, mg.sin(a) = u.mg.cos(a)
u = tan(a)

Let h be the height reached by the ant.
hence, tan(a) = square-root[R^2 - (R-h)^2]/(R-h)
i.e. u = square-root[2Rh - h^2]/(R-h)
solve for h