Solutions of problems in fluid mechanics II

This article presents the solution of a unique problem in fluid mechanics. Once again, my goal is to show how easily we can solve physics problems when we refer to basic principles. The problem is solved using Archimedes’ principle and Newton’s second law.

Because the text editor doesn’t accept certain mathematical symbols, I have to use a rather unusual notation. It is: (a) Almost all variables are represented by capital letters. (b) Variables are capitalized so that lowercase letters can be used for subscripts. (c) Powers are represented by a ^. For example, “A squared” is written A^2. (d) Variables normally represented by Greek letters are presented as capital letters in English italics. For example, the Greek “rho” is represented by p.

Problem. A density oak block Po= 600 kg/m^3 remains below the surface of the water (Pw = 1000 kg/m^3) in a beaker. What is the acceleration of the block at the instant it is released?

Analysis. We will assume that the volume of the block is V. (i) The mass and weight of the block are then M = PoV and W = MG = PoVG. (ii) Since the block displaces a volume V of water, the buoyant force on it is

………………………………………..Archimedes’ principle

………………………………………………….. .. ….. B= PwVG.

Once the block has acquired an upward velocity, it experiences viscous forces (friction) due to the water. But at the instant of launch, the only forces on the block are buoyancy and its weight. Using Newton’s second law and the help of a free-body diagram, we have at the instant the block is released

…………………………………………………Newton’s Second Law

………………………………………………….. …SUM(year) = May

………………………………………………….. .. …..B – W = May

………………………………………………….. ………..PwVG- PoVG = PoVAy.

After canceling the common factor V and rearranging the terms, we find that the acceleration of the block is

………………………………………………….. …ay = ((Pw- Peither)/Po)G.

You can do the arithmetic. You will find that Ay = 6.5 m/s^2.

We have another demonstration of the ease with which physics problems are solved when they are attacked by starting with one or more fundamental principles. Here, Archimedes’ principle of fluid mechanics and Newton’s second law of Newtonian mechanics were combined to solve the problem. In the next article, I’ll show two more problem solutions that depend on combining Bernoulli’s equation with (1) the continuity equation and (2) the variation of pressure with height.