A prism whose cross section is an equilateral triangle with sides of length l floats in a pool of liquid of weight density p with its height parallel to the vertical axis. The prism is set in motion by displacing it from its equilibrium position (see Fig. 14-4) and giving it an initial velocity.

Question

A prism whose cross section is an equilateral triangle with sides of length l floats in a pool of liquid of weight density p with its height parallel to the vertical axis. The prism is set in motion by displacing it from its equilibrium position (see Fig. 14-4) and giving it an initial velocity. Determine the differential equation governing the subsequent motion of this prism.

Accepted Answer

Equilibrium occurs when the buoyant force of the displaced liquid equals the force of gravity on the body. The area of an equilateral triangle with sides of length l is [latex]A=\sqrt{3} l^{2} / 4[/latex]. For the prism depicted in Fig. 14-4, with h units of height submerged at equilibrium, the volume of water displaced at equilibrium is [latex]\sqrt{3} l^{2} h / 4[/latex], providing a buoyant force of [latex]\sqrt{3} l^{2} h \rho / 4[/latex]. By Archimedes' principle, this buoyant force at equilibrium must equal the weight of the prism mg; hence,

[latex]\sqrt{3} l^{2} h \rho / 4=m g[/latex] (1)

We arbitrarily take the upward direction to be the positive x-direction. If the prism is raised out of the water by x(t) units, as shown in Fig. 14-4, then it is no longer in equilibrium. The downward or negative force on such a body remains mg but the buoyant or positive force is reduced to [latex]\sqrt{3} l^{2}[h-x(t)] \rho / 4[/latex]. It now follows from Newton's second law that

[latex]m \ddot{x}=\frac{\sqrt{3} l^{2}[h-x(t)] \rho}{4}-m g[/latex]

Substituting (1) into this last equation, we simplify it to

[latex]\ddot{x}+\frac{\sqrt{3} l^{2} \rho}{4 m} x=0[/latex]

Question 14.25: A prism whose cross section is an equilateral triangle with ...

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