Problem Solutions in Fluid Mechanics I

The fluid mechanics taught in introductory physics is an ideal topic for demonstrating how to teach in terms of basic ideas. There are four important principles taught in introductory fluid mechanics. They are: (i) pressure variation with height; (ii) Archimedes’ principle; (iii) equation of continuity; and (iv) Bernoulli’s equation. These four can be combined with mechanics principles such as Newton’s second law and the conditions for static equilibrium in a wide range of interesting problems. I will cover my problem-solving approach for fluids in three separate articles. As usual, I describe the method in terms of problem solutions.

The unusual notation I use is summarized in the Ezine article “Teaching Rotational Dynamics”. I do have to introduce some new notation here, the symbols for volume and area mass density. Volume mass density will be represented by an italicized capital P with a subscript (lower case letters) representing a substance. For example the volume mass density of sea water is represented by Psw. The surface mass density is represented by an italicized capital S with a subscript (lower case letters) representing a substance. For example, the surface mass density of copper is Scu.

Problem. A girl holds a string attached to a helium-filled balloon of volume V= 0.320 m**3. The mass of the rubber of the balloon is Mr = 14.0 g, the mass of the string is negligible, the density of the helium is Ph = 0.18 kg/m**3, and the density of the air is Pa = 1.2 kg/m**3. What is the tension in the string?

Analysis. We investigate the static equilibrium of the balloon. (i) The balloon is touching the air, which exerts an upward buoyant force B on it. The buoyant force is equal to the weight of the air displaced:

……………………………….Archimedes’ Principle

………………………………………..B = PaVG

(ii) The balloon is also touching the string, which exerts a downward tension T on it. (iii) The gravitational force W on the balloon is the weight of the rubber plus the weight of the enclosed helium:

…………………………………….W = MrG + PhVG.

Since the balloon is stationary, it in static equilibrium. We now have with the help of a free-body diagram

………………………..Conditions for Static Equilibrium

……………………………………..SUM(Fy) = 0

……………………………………..B – T – W = 0

………………………….PaVG – T – MrG – PhVG = 0,

so………………………..T = (Pa – Ph)VG – MrG.

You can do the arithmetic. You’ll find that T = 3.2 N.

Problem. A hollow cubical box is made of thin sheet metal whose mass density per unit area is Ssm = 30 kg/m**2. What minimum size must the box have if it is to float in sea water (Psw = 1025 kg/m**3)? Do you see why an ocean liner made of thick metal sheets can float?

Analysis. Let’s consider a cubical box with sides of length L. We’ll assume the box floats with its vertical sides submerged to a depth H and then calculate H. If H < L, the box floats; if H > L, the box sinks. Independent of size, the only forces on the box are (i) its buoyancy and (ii) its weight. The buoyancy is the weight of the sea water displaced. With the box submerged to a depth H, the volume submerged is V = HL**2, and the buoyant force B is

……………………………Archimedes’ Principle

………………………….B = PswVG = (PswHL**2)G.

The weight of the box is the sum of the weights of six sheets of area L**2:

………………………..W = MG = (6SsmL**2)G.

Assuming the box floats, we have with the help of a free-body diagram

……………………Conditions for Static Equilibrium

………………………………….SUM(Fy) = 0

…………………………………….B – W = 0

………………………….(PswHL**2)G – (6SsmL**2)G = 0,

and…………………………… H = 6Ssm/Psw.

With the given values for Ssm and Psw, we find from this equation that H = 0.18 m = 18 cm. Thus the box will be submerged to a depth of 18 cm when it floats, and the sides of the box must be greater than 18 cm — anything less and the box sinks. The same reasoning can be applied to actual sea vessels. Their hulls are much thicker than our box so Ssm, and therefore minimum size for floating, is much larger.

I’ve just covered two problem solutions in fluid mechanics. These problems are considered fairly difficult for introductory physics students. However, they are, in fact, quite easy if they are approached in terms of what I call basic principles. In both cases, straight-forward applications of Archimedes’ principle and one of the conditions for static equilibrium lead to a very simple solution.