Problem 22 - University Physics with Modern Physics

Question

The surface pressure on Venus is 92 atm, and the acceleration due to gravity there is 0.894\(g .\) In a future exploratory mission, an upright cylindrical tank of benzene is sealed at the top but stillpressurized at 92 atm iust above the benzene. The tank has a diameter of \(1.72 \mathrm{m},\) and the benzene column is 11.50 \(\mathrm{m}\) tall. Ignore any effects due to the very high temperature on Venus. (a) What total force is exerted on the inside surface of the bottom of the tank? (b) What force does the Venusian atmosphere exert on the outside surface of the bottom of the tank? (c) What total inward force does the atmosphere exert on the vertical walls of the tank?

Step-by-Step Solution

Answer

(a) Calculate the total force using given pressure and benzene weight. (b) Apply pressure to area for outward surface force. (c) Inward force from atmospheric pressure typically minimal.

Step 1: Understanding the Problem

We need to calculate three forces related to a cylindrical tank of benzene on Venus: (a) the total force on the inside bottom surface, (b) the atmospheric force on the outside bottom surface, and (c) the total inward force on the vertical walls. We will use the given pressures and dimensions to find these forces.

Step 2: Calculate Force on Inside Bottom Surface

The total force on the inside bottom surface is due to both the atmospheric pressure and the weight of the benzene column. The force can be calculated using the formula \(F = P \times A + \text{Weight of benzene}\). - Pressure \(P\) is 92 atm, converting to pascals: \(92 \text{ atm} \times 101325 \text{ Pa/atm} = 9,321,900 \text{ Pa}\).- The area \(A\) is \(\pi \times (1.72/2)^2 = 2.32 \text{ m}^2\).- The weight of benzene is \( \rho Vg' \) where \( \rho = 879 \text{ kg/m}^3 \) (density of benzene), \( V = 2.32 \text{ m}^2 \times 11.50 \text{ m} \), and \( g' = 0.894 \times 9.8 \text{ m/s}^2\).- Calculate the total force.

Step 3: Calculate Atmospheric Force on Outside Bottom Surface

The force on the outside surface of the tank bottom is just due to the atmospheric pressure on Venus. The force is given by \(F = P \times A\).- Use the Venusian atmospheric pressure calculated in Step 2: \(9,321,900 \text{ Pa}\).- Use the same area \(A = 2.32 \text{ m}^2\).- Calculate \(F\).

Step 4: Calculate Total Inward Force on Vertical Walls

To find the inward force on the vertical walls, consider the pressure difference across the height of the benzene column. - Since the pressure on the outside is atmospheric pressure, and inside it is even higher due to the benzene column, it can be assumed there's no net force contribution from pressure alone. - However, for accuracy, compute if needed for specifics by considering the delta pressure effects over the surface area of the walls, though it's commonly dominated by the vertical forces in such configurations.

Key Concepts

Surface Pressure in Physics

Surface pressure is an essential concept in physics, particularly when dealing with atmospheres and fluids. On Venus, the surface pressure is given as 92 atm, significantly higher than Earth's 1 atm. This drastic difference is important for understanding why the forces exerted on structures, like the cylindrical tank of benzene, are so large. When considering pressure, it's crucial to note it acts uniformly in all directions. In practical terms, the force exerted by pressure on an area can simply be found by multiplying the pressure by the area over which it acts, i.e., \( F = P \times A \). This principle helps in determining how pressure contributes to forces within a fluid, whether inside a tank or a natural atmosphere.

Understanding Acceleration Due to Gravity

Acceleration due to gravity is the acceleration experienced by an object due to the gravitational pull of a planet. On Venus, this value is 0.894 times that of Earth’s gravity, \( g \), which is approximately 9.8 m/s². Consequently, gravity on Venus is approximately \( 0.894 \times 9.8 = 8.7652 \) m/s². This value is key when computing the force due to the weight of objects, like the benzene column in the tank. For any object, the weight is the product of its mass and the local acceleration due to gravity \( W = mg' \). Understanding this concept is crucial for properly analyzing forces acting on bodies in different gravitational conditions, such as those found on other planets.

Calculating Pressure and Total Force

Pressure calculation is vital to finding the total force exerted on surfaces. In the exercise, the inner bottom surface of the benzene tank experiences force from both atmospheric pressure and the benzene's weight. The conversion of pressure from atm to Pascals is essential for calculations involving physical units. Using the pressure conversion \( P = 92 \times 101325 = 9321900 \) Pa helps in computing the correct force values. The formula \( F = P \times A + \text{Weight of benzene} \) allows us to find the total force acting on the bottom surface. This combines both the direct effect of atmospheric pressure and the contribution from the liquid's weight.

Basics of Fluid Mechanics

Fluid mechanics explains how fluids behave under various forces, which is central to problems involving liquids like benzene in a tank. Fluids exert pressure isotropically, meaning pressure at any point within the fluid acts equally in all directions. In a column of liquid, the pressure increases with depth due to the weight of the overlying liquid. This is why inside a tank, even sealed, the pressure at the bottom includes both the surface atmospheric pressure and the additional pressure from the liquid column. The concept is captured in the equation \( P = P_0 + \rho gh \), where \( \rho \) is the fluid density, \( g \) is gravitational acceleration, and \( h \) is the height of the fluid column. Understanding these relationships is integral for analyzing what's happening inside the tank of benzene.

Force Computation Techniques

Force computation uses various physics principles to determine forces acting on objects. In our tank scenario, forces need to be calculated on different surfaces affected by both internal and external pressures. The bottom and walls experience different pressure-related forces. To find force due to pressure, you use the core equation \( F = P \times A \). For force due to a weight, it involves gravitational acceleration \( F = mg' \). These computations combine to provide a complete picture of the forces at play in complex systems, such as tanks on planetary surfaces. The intricate interplay of internal pressures, gravity, and surface areas in these calculations ensures structures withstand the various forces they encounter.