Statics and Hydrostatics: Principles of Equilibrium and Buoyancy

We know that a condition of equilibrium of a body is that the resultant of the forces acting on it is zero. If the moment of the system (which is the sum of the moments of the forces acting on it) is zero, the body does not acquire rotational motion.

The principles of statics can be observed in everyday structures in cities, such as buildings, houses, and bridges, among others. Mechanical structures like cranes, winches, and overhead cranes would not be easily designed without the use of static equilibrium equations. These examples illustrate how important the principles of statics are, associated with the equilibrium of rigid bodies.

Source: General and Experimental Physics I. Luis Ricardo Arruda de Andrade. Maringá-PR.: Unicesumar, 2019. p. 277-280.

Hydrostatics studies the intrinsic characteristics of fluids, such as their properties, one of which is the density of the fluid under static equilibrium conditions. One of the most important studies in hydrostatics is that of Archimedes' principle, which states that a solid body submerged in a fluid will be subject to a vertical force, upward, called buoyancy, whose intensity is equal to the weight of the fluid displaced by the body.

Archimedes' principle has applications in various areas of engineering, such as in the design of ships and aircraft, measurement of liquid density, and design of flow meters.

Source: General and Experimental Physics I. Luis Ricardo Arruda de Andrade. Maringá-PR.: Unicesumar, 2019. p. 293-295.

The two concepts mentioned above are put to the test by you: imagine that a piece of equipment overheats in your company and some nuts and bolts are loosened and deformed.

1) To request a repurchase, it is necessary to know the exact mass of the parts. You have available in the virtual laboratories the Statics experiment, which specifically addresses the equilibrium of rigid bodies. After reading the experiment presentation, the theoretical summary, the script, and taking the pre-test, perform the virtual experiment.

Imagine that each of those test bodies in the virtual experiment is a loose part of your equipment and that the mass needs to be determined. Place only one test body on the plate and balance the scale. Repeat the procedure for each test body and, for each balance, fill in a row of the following table

Note: columns 2, 3, 4, and 5 of the table are obtained from experimental data, column 6 of the table is obtained by calculations.

When performing the experiment, it is noted that the mass of the plate and the distance from its center to the central pivot are:

• \( m_{plate} = 200 \, g \)
• \( d_{plate} = 14.5 \, cm \)

The mass of the counterweight is:

• \( m_{counterweight} = 500 \, g \)

According to the problem statement, the balance must be in equilibrium for each of the observed weights. Thus, once equilibrium is restored, by summing the torques, the mass of each weight is determined by:

\[ \sum \tau = 0 \\ \tau_{Plate+Weight} - \tau_{Counterweight} = 0 \\ \tau_{Plate+Weight} = \tau_{Counterweight} \\ (m_{weight} + m_{plate}) \cdot g \cdot d_{plate} = m_{Counterweight} \cdot g \cdot d_{Counterweight} \\ (m_{weight} + m_{plate}) \cdot d_{plate} = m_{Counterweight} \cdot d_{Counterweight} \\ m_{weight} + m_{plate} = \frac{m_{Counterweight} \cdot d_{Counterweight}}{d_{plate}} \\ m_{weight} = \frac{m_{Counterweight} \cdot d_{Counterweight}}{d_{plate}} - m_{plate} \]

Using the above expression for each of the weights, the following results are obtained:

Distance from the center of rotation to the counterweight	Mass of the counterweight	Distance from the center of rotation to the center of the plate	Mass of the plate	Mass of the test body	Mass of the Test Body
Weight 1	7.3	500	14.5	200	51.724
Weight 2	7.8	500	14.5	200	68.966
Weight 3	8.8	500	14.5	200	103.448
Weight 4	10.2	500	14.5	200	151.724

2) To request the repurchase of another specific part, you changed experiments and will determine the mass of the part using hydrostatics. You have available in the virtual laboratories the Hydrostatics experiment, which specifically addresses buoyancy. After reading the experiment presentation, the theoretical summary, the script, and taking the pre-test, perform the virtual experiment.

Imagine that the cylinder (test body of the virtual experiment) is the loose part of your equipment and that the mass needs to be determined. Follow all the steps of the experiment indicated in the script and fill in the table below:

a) Using only volume and density data of the cylinder, provided in the experiment footer, what is the mass of this solid?

The volume and density of the cylinder are:

• \( Volume = 50 \, cm^3 \)
• \( Density = 1.86 \, g/cm^3 \)

Thus, the mass of the cylinder is:

\[ Mass = Density \cdot Volume \\ Mass = (1.86 \, g/cm^3) \cdot (50 \, cm^3) \\ Mass = 93 \, g \]

b) Using only weight data (values provided by the dynamometer in row I) and g=9.81 m/s², what is the mass of this solid?

Using only weight data (values provided by the dynamometer in row I) and \( g = 9.81 \, m/s^2 \), what is the mass of this solid?

When placing the cylinder on the dynamometer, its weight is:

• \( Weight \, of \, the \, cylinder = 0.9091 \, N \)

Thus, the mass of the cylinder can be calculated using the following relationship:

\[ Weight = Mass \cdot Gravity \\ Mass = \frac{Weight}{Gravity} \\ Mass = \frac{0.9091 \, N}{9.81 \, m/s^2} \\ Mass = 0.09267 \, kg \\ Mass = 92.67 \, g \]

c) Using only theoretical values (Volume of the cylinder: 50 cm³, density of water 1000 kg/m³, g=9.81 m/s², what is the buoyancy force on this cylinder?

d) Using only weight data (values provided by the dynamometer in rows I, II, III, and IV), what are the two values that can be found for the buoyancy force on this cylinder?

The buoyant force equals the weight of the displaced water. Therefore, there is a buoyant force corresponding to the situation in row II and another value corresponding to row IV.

• Buoyant force corresponding to row II (Solid cylinder inside water):

The force diagram of the cylinder when submerged is as follows:

In row I of the table, the weight of the solid cylinder outside the water is:

• \( Weight \, of \, the \, cylinder = 0.9091 \, N \)

The reading on the dynamometer when the solid cylinder is submerged is:

• \( Tension = 0.4184 \, N \)

Applying Newton's second law, the buoyant force in this situation is:

\[ \sum F_y = 0 \\ T + E - P = 0 \\ E = P - T \\ E = 0.9091 - 0.4184 \\ E = 0.4907 \, N \]

• Buoyant force corresponding to row IV (Solid cylinder + filled water, inside water):

The force diagram of the cylinder when submerged is as follows:

In row III of the table, the weight of the solid cylinder + filled water outside the water is:

• \( Weight \, of \, the \, cylinder = 0.9091 \, N \)

The reading on the dynamometer when the solid cylinder + filled water is submerged is:

• \( Tension = 0.8976 \, N \)

Applying Newton's second law, the buoyant force in this scenario is:

\[ \sum F_y = 0 \\ T + E - P = 0 \\ E = P - T \\ E = 0.9091 - 0.8976 \\ E = 0.0115 \, N \]

e) Justify the reason why we use the expression "apparent decrease in the weight of the body" and not "decrease in the weight of the body".