Originally posted by talzamir
A glass is shaped like a straight cylinder with a round base, height h, radius r. If the glass is half full and tilted so that the liquid just touches the rim of the glass, how high from the bottom of the glass is the surface on the opposite side of the glass?
Taking the cross section, and doing it in 2d for simplicity, which shouldn't change the answer:
Area of water when glass is half full = rh
When glass is tilted, the required point is x from the bottom. The shape the water is now in is a composition of two shapes:
a) A rectangle, area 2rx;
b) And a right angled triangle, with non-hypotenuse sides 2r, and h-x
the area of this triangle is r(h-x)
This gives a total water area of r(2x + h - x)
Tilting the glass does not change the area of the water, so
r(2x + h - x) = rh
2rx + rh - rx = rh
rx = 0
So, assuming r is not 0, the edge of the water now touches the bottom of the glass.