Chapter 7 Fluids And Chapter 8 Motion Of Fluids Question 1a

Chapter 7 Fluids Chapter 8 Motion Of Fluidsquestion 1a Container Con

Analyze the problem involving two vertical cylindrical columns of different diameters connected by a narrow horizontal section, both topped with light movable plates, filled with oil of density 0.820 g/cm³. A 125-kg object is placed on the larger plate at A, and the question asks for the mass to be placed on the smaller plate at B to balance it.

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The problem involves fluid equilibrium and hydrostatics principles, specifically communicating vessels and Pascal's law. The goal is to find the mass that must be added to the smaller cylinder to counteract the weight of the object on the larger side, considering the fluid's pressure transmission and the areas involved.

Given the density of oil (0.820 g/cm³), convert to SI units: 0.820 g/cm³ = 820 kg/m³. The cross-sectional area for each cylinder is calculated using A = πr². For A, with diameter 35.0 cm, r = 17.5 cm = 0.175 m; for B, diameter 10.2 cm, r = 5.1 cm = 0.051 m.

The pressure exerted by the weight in the larger cylinder causes a fluid column that exerts an equal pressure at the connecting section, which translates to the smaller cylinder, balancing the system. Applying hydrostatic pressure equations and Pascal's law, the key relation is:

Pressure difference = ρg(h_A - h_B)

Force balance involves considering the weight of the mass at A (W_A = m_A g) and the added mass at B (m_B g).

Using the principle of communicating vessels, the pressures at the connecting points are equal, leading to the relation:

ρg h_A + m_A g / A_A = ρg h_B + m_B g / A_B

Rearranged to solve for m_B:

m_B = (A_B / A_A) * m_A

Given m_A = 125 kg, compute areas, then determine m_B.

The calculation shows that the mass necessary on the smaller plate is approximately 17.1 kg to balance the object on the larger plate, considering the different cross-sectional areas and the fluid's density. This illustrates the principles of hydrostatics and the transmission of pressure in communicating vessels.

Analysis of the Second Problem: Fluid Flow and Pressure Difference

The second problem involves water flowing through a pipeline with given cross-sectional areas and velocities at two points, with the change in height and velocity. Using Bernoulli's equation, the pressure difference P₁ - P₂ can be computed as:

ΔP = ½ ρ (v₂² - v₁²) + ρg(h₁ - h₂)

Given the velocities, areas, and height difference, insert values to find the pressure difference, which reflects the fluid's dynamic and hydrostatic pressure variations.

Additional Questions Covering Fluid Mechanics Concepts

The remaining questions explore basic concepts such as pressure calculations, buoyancy, fluid densities, flow rates, viscosity, and energy conversions, consistent with principles in fluid mechanics. For example, calculating the pressure exerted by a person on the ground, buoyant force from weights in air and water, the density of materials, and flow rates using the continuity equation. In each case, applying the fundamental equations—Archimedes' principle, Bernoulli's equation, the ideal gas law, and viscosity formulas—is essential to derive accurate results. The questions about air resistance on a parachuter and the forces on a jumper involve dynamic fluid resistance and Newtonian mechanics, illustrating the interplay of gravity, drag, and thrust during motion.

Most other questions deal with the concepts of equilibrium (static and dynamic), how forces distribute on levers and structures, and the energy transformations in motion scenarios like jumping or running. For example, the analysis of a standing jump involves kinetic and potential energy conversion, while the stability questions examine the role of torque and the position of the center of mass. The detailed calculations require applying summation of forces and moments, understanding of equilibrium conditions, and energy conservation principles.

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