Eureka! Consider a body of mass m submerged either partially or totally in a liquid of weight density p (in units of Newtons per cubic meter, or kg/ (m's )). Such a body experiences two forces, a downward force due to gravity (my) and a counter force governed by Archimedes" law: A body in liquid experiences a buoyant upward force equal to the weight of the liquid displaced by that body. Evidently, equilibrium occurs when the buoyancy force of the displaced liquid equals, in magnitude, the force of the gravity on the body. Supprese that a prism of height is h and with cross section an equilateral triangle of side length f is floating, partially submerged, in a pool of liquid of weight density p with its height parallel to the vertical axis as shown in the figure. (a) How far below the surface is the end of the prism at equilibrium? Your anewer should depend on E,p.m.g. (b) Let r(f) be the difference between the depth of the bottom of the prism and the value you found in part (a). Assuming that the prican is set in motion by displacing it from equilibrium and giving it initial velocity, write a differential equation governing the motion of the prism. (c) Find the general solution r(f) for the equation you found in part (b). (d) From your solution of part (c). explain how the motion of the partially submerged prism would be different in a denser liquid than in a less-dense liquid