You have rectangular pot of water 10cm by 10cm at the bottom and 6 cm deep. It is filled to a depth of 3cm.
You have 7 solid steel shapes in front of you. The question is to find the new level of the water after each shape is put in the pot in the orientation described. The previous shape is removed before adding the next.
1. A cube 5cm on a side.
2. A prism in the shape of a right triangle with legs 5cm long. It is 5cm high but is placed on one of its legs with its hypotenuse sloping out of the water.
3. Another prism, this one having an equilateral triangle of sides 5cm and height 5cm. It is to be placed on its side with two faces sloping up out of the water.
4. A regular hexagonal prism. Each edge of the hexagon is 4cm and the height is 5cm. It is to be placed on its side.
5. A right square pyramid. Its base is 6cm on a side. It is 5cm high. It is to be placed base down.
6. A right cylinder of radius 3cm and length 5cm. It is to be placed on its side.
7a. A right cone of radius 3cm and height 5cm. It is to be placed base down.
7b. The same cone as 7a. This time placed on its side.
The height of the water after placing the cone on its side is approximately 3.4124 cm.
The calculation involved was tedious (and was subject to error). The volume of the conical wedge that remained above the surface of the water was determined using the equation:
Vol_{wedge} = 1/3 * A_{0}*h_{0}  A_{1}*h_{1}, where A_{0} is the segment area of the base of the cone above the surface of the water; h_{0} is the cone's height [5 cm]; A_{1} is the area of the parabolic segment created by the intersection of the plane [water surface] and the cone; h_{1} is the height of the water.
The segment area of the base of the cone was calculated using the following:
A_{0} = (R^{2}*a)/2, where R is the radius of the cone's base and a is the apothem to the chord that is created by the intersection of the water surface and the cone's base. The length of the apothem was determined as a function of the height of the water and of the angle of the cone's base to the floor of the pot:
a = (h_{1}/sin(tan^{1}(5/3))  3).
The second wedge area at the water's surface was calculated using the following:
A_{1} = 4/3*(P_{width/2})*P_{height}, where P_{width/2} is half the width of the parbolic segment at the point of intersection with the cone's base and P_{height} is the distance from the base to the apex of the parabolic segment. (P_{width} is the chord that is created by the intersection of the water surface and the cone's base.)
P_{width/2} =
[SQRT(6*h_{1*}sin(tan^{1}(5/3))  h_{1}^{2})]/[sin(tan^{1}(5/3))]
P_{height} =
[SQRT(R^{2} + h_{0}^{2})  h_{1}* (3/5 + 8/15)].
The Volume of the water [300 cm^{3}] and the volume of the solid below the water's surface [15*pi  Volume_{wedge}] divided by the rectangular pot's base surface area [100 cm^{2}] equals the height of the water.
Thank you Mindrod for your collaboration.
Thank you Charlie for pointing out that I must have had an error in my equations.
Edited to correct the angle used in calculation of the apothem and the equation for the segment area of the base of the cone, P_{width/2}, and to correct a sign error in the calculation of P_{height}.
Edited on July 5, 2006, 3:09 am

Posted by Dej Mar
on 20060702 21:45:19 