This article presents a single problem solution in fluid mechanics. Again, my goal is to demonstrate how easily we can solve physics problems when we consult basic principles. The problem is solved using Archimedes’ principle and Newton’s second law.
Because the text editor does not accept certain mathematical symbols, I must use some rather unusual notation. It is: (a) Almost all variables are represented by capital letters. (b) Variables are capitalized so lower case letters can be used for subscripts. (c) Powers are represented with a ^. For example, “A squared” is written A^2. (d) Variables normally represented by Greek letters are presented as italicized capital English letters. For example, the Greek “rho” is represented by P.
Problem. An oak block of density Po= 600 kg/m^3 is held 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’ll assume that the volume of the block is V. (i) The block’s mass and weight 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 release, the only forces on the block are the buoyancy and its weight. With 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(Fy) = MAy
……………………………………………….B – W = MAy
……………………………………….PwVG – PoVG = PoVAy.
After canceling the common factor V and rearranging terms, we find that the acceleration of the block is
……………………………………………Ay = ((Pw – Po)/Po)G.
You can do the arithmetic. You’ll find that Ay = 6.5 m/s^2.
We have another demonstration of how easily physics problems are solved when they are attacked by starting with one or more fundamental principles. Here, Archimedes’ principle from fluid mechanics and Newton’s second law from Newtonian mechanics were combined to solve the problem. In the next article, I will show two more problem solutions that depend on combining Bernoulli’s equation with (1) the equation of continuity and (2) the pressure variation with height.
