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

The unusual notation I use is summarized in the ezine article “Teaching Rotational Dynamics”. I have to introduce a new notation here, the symbols for volume and areal mass density. The volumetric mass density will be represented with a capital letter in italics. P with a subscript (lowercase letters) representing a substance. For example, the volumetric mass density of seawater is represented by PSOUTHWEST. Surface mass density is represented by an italic capital letter. S with a subscript (lowercase letters) representing a substance. For example, the areal 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 balloon rubber is Mr = 14.0 g, the mass of the string is negligible, the density of helium is Ph = 0.18 kg/m**3, and the density of air is Pennsylvania = 1.2kg/m**3. What is the tension in the string?

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

……………………………….Archimedes’ principle

………………………………………………………. ……..B = Paverage

(ii) The balloon also touches 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 is in static equilibrium. We now have with the help of a free body diagram

………………………..Conditions for static equilibrium

……………………………………..SUM(Year) = 0

……………………………………..B – T – W = 0

……………………………….PaVG-T-MrG- PhVG = 0,

then……………..T = (PTO – Ph) VG-MrG.

You can do the arithmetic. You will find that T = 3.2 N.

Problem. A hollow cubic box is made of a thin sheet of metal whose mass density per unit area is Sm2 = 30kg/m**2. What minimum size must the box be in order for it to float in seawater (Psw = 1025 kg/m**3)? Do you see why an ocean liner made of thick sheets of metal can float?

Analysis. Consider a cubic box with sides of length L. We will assume that the box floats with its vertical sides submerged at a depth H and then calculate H. If H < L, la caja flota; si H > L, the box sinks. Regardless of size, the only forces on the box are (i) its buoyancy and (ii) its weight. Buoyancy is the weight of displaced seawater. With the box submerged to a depth H, the submerged volume 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 = (6Ssmall**2)G.

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

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

………………………………….SUM(Year) = 0

……………………………………..B – W = 0

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

and ………………………….. H = 6SYE/PSOUTHWEST.

With the values ​​given for Sand Psw, we find from this equation that H = 0.18 m = 18 cm. So 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 real sea vessels. Their hulls are much thicker than our box, so Ssm, and therefore the minimum size to float, is much larger.

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