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Chapter 6: Problem 48

A solid cone of height \(H\) and base radius \(H / 2\) floats in a liquid of density \(\rho\). It is hanging from the ceiling with the help of a string. The force by the fluid on the curved surface of the cone is \(\left(P_{0}=\right.\) atmospheric pressure \()\) (1) \(\pi H^{2}\left(\frac{P_{0}}{4}+\frac{\rho g H}{3}\right)\) (2) \(\pi H^{2}\left(\frac{P_{0}}{4}+\frac{\rho g H}{6}\right)\) (3) \(\frac{\pi H^{2}}{4}\left(\frac{P_{0}}{4}+\rho g H\right)\) (4) \(\frac{\pi H^{2}}{4}\left(P_{0}+\rho g H\right)\)

Short Answer

Expert verified
The force by the fluid on the curved surface is option (2): \(\pi H^{2}\left(\frac{P_{0}}{4}+\frac{\rho g H}{6}\right)\).

Step by step solution

01

Understanding the Problem

We have a cone floating in a liquid with known height and base radius. The task is to calculate the force exerted by the fluid on the curved surface of the cone. The critical factor is understanding how pressure varies along the height of the cone to determine the force.
02

Analyze the Pressure on Curved Surface

The pressure on any area along the curved surface of the cone increases linearly from the top (atmospheric pressure, \(P_0\)) to the base due to the weight of the liquid above. Hence, it can be represented as \(P = P_0 + \rho g z\), where \(z\) is the depth from the surface.
03

Setup Integration for Total Force Calculation

To find the total force on the curved surface due to liquid pressure, we must integrate the pressure over the surface area, which involves expressing a small area element \(dA\) in terms of \(z\).
04

Determine Expression for Area Element

For a cone, a small area \(dA\) on the curved surface at height \(z\) can be computed using the slant height. The slant height is \(s = \sqrt{r^2 + H^2}\). The radius at height \(z\), \(r(z) = \frac{H-z}{H} \cdot \frac{H}{2}\).
05

Integrate Pressure over Curved Surface

The total force \(F\) by the fluid is given by \(F = \int_{0}^{H} (P_0 + \rho g z) \, dA \). Substitute the expression for \(dA\) in terms of \(z\) and integrate from \(0\) to \(H\).
06

Simplify Integration Result

Solving the integral will yield the expression for the force, leading to \(\pi H^2 \left( \frac{P_{0}}{4} + \frac{\rho g H}{6} \right) \). This matches with option (2).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hydrostatic Pressure
When a solid object is submerged or partially submerged in a fluid, the fluid exerts a force on its surface due to pressure. This phenomenon primarily relates to hydrostatic pressure, which is the pressure exerted by a fluid due to gravity. It varies with depth in the fluid.

The formula for pressure at any depth in a fluid is given by:
  • Initial Pressure: The surface pressure of the liquid, which is the atmospheric pressure \(P_0\).
  • Additional Pressure: This accounts for the increase in pressure as you move deeper into the fluid. It's calculated as \(\rho g z\), where \(\rho\) is the fluid's density, \(g\) is the gravity, and \(z\) is the depth.
Therefore, the total pressure at a depth \(z\) can be represented as \(P = P_0 + \rho g z\).

This principle of pressure variation is essential for understanding how forces are distributed over submerged surfaces, such as the curved surface of a floating cone in our exercise.
Integration in Calculus
Integration is a mathematical tool used to find areas, volumes, central points, and many useful things, especially in a scenario where variables continuously change along a path or boundary. Here, it is used to calculate the total force exerted by the fluid on the cone's curved surface.

To determine the total force, we integrate the varying pressure over the surface area. This involves:
  • Expressing the small differential area element \(dA\) in terms of the variable of integration \(z\). For a cone, this depends on the geometry of the cone.
  • Setting the limits of integration from 0 (at the apex of the cone) to \(H\) (at the base).
The integral for total force is then set up as \(F = \int_{0}^{H} (P_0 + \rho g z) \, dA \), representing the sum of all infinitesimal forces over the cone's height. After evaluating this integral, the solution provides the correct expression for the force.
Cone Geometry
The geometry of a cone is pivotal in determining the surface area over which the fluid's pressure is acting. A standard cone is characterized by its height \(H\) and base radius \(R\). In the context of our exercise, the geometry helps calculate the small element \(dA\) needed for our integration approach.

Important geometric properties include:
  • Slant Height \(s\): Given by \(s = \sqrt{R^2 + H^2}\). This represents the direct line length from the apex to the base edge.
  • Radius with Depth \(r(z)\): The radius at any height \(z\) is calculated by the linear relation \(r(z) = \frac{H-z}{H} \cdot \frac{H}{2}\).
These geometric calculations allow us to define the small surface element \(dA\) and integrate over this surface. This ensures that we apply the correct mathematical model to solve for the force exerted by the fluid on the cone's surface.

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Most popular questions from this chapter

A uniform \(\operatorname{rod} A B, 12 \mathrm{~m}\) long weighing \(24 \mathrm{~kg}\), is supported at end \(B\) by a flexible light string and a lead weight (of very small size) of \(12 \mathrm{~kg}\) attached at end \(A\). The rod floats in water with one-half of its length submerged. Find the volume of the rod (in \(\times 10^{-3} \mathrm{~m}^{3}\) ) [Take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\), density of water \(\left.=1000 \mathrm{~kg} / \mathrm{m}^{3}\right]\)

For Problems \(18-20\) A liquid having density \(6000 \mathrm{~kg} / \mathrm{m}^{3}\) stands to a height of \(4 \mathrm{~m}\) in a sealed tank as shown in the figure. The tank contains compressed air at a gauge pressure of \(3 \mathrm{~atm}\). The horizontal outlet pipe has a cross-sectional area of \(6 \mathrm{~cm}^{2}\) and \(3 \mathrm{~cm}^{2}\) at larger and smaller sections. Atmospheric pressure \(=1 \mathrm{~atm}, g=10 \mathrm{~m} / \mathrm{s}^{2}\), \(1 \mathrm{~atm}=10^{5} \mathrm{~N} / \mathrm{m}^{2}\). Assume that depth of water in the tank remains constant due to \(s\) very large base and air pressure above it remains constant. Based on the above information, answer the following questions. The discharge rate from the outlet is (1) \(4.02 \times 10^{-3} \mathrm{~m}^{3} / \mathrm{s}\) (2) \(6 \times 10^{-5} \mathrm{~m}^{3} / \mathrm{s}\) (3) \(7.56 \times 10^{-3} \mathrm{~m}^{3} / \mathrm{s}\) (4) \(4.02 \times 10^{-5} \mathrm{~m}^{3} / \mathrm{s}\)

A solid block of volume \(V=10^{-3} \mathrm{~m}^{3}\) and density \(d=800\) \(\mathrm{kg} / \mathrm{m}^{3}\) is tied to one end of a string, the other end of which is tied to the bottom of the vessel. The vessel contains two immiscible liquids of densities \(\rho_{1}=1000 \mathrm{~kg} / \mathrm{m}^{3}\) and \(\rho_{2}=\) \(1500 \mathrm{~kg} / \mathrm{m}^{3}\). The solid block is immersed with \(2 / 5\) th of its volume in the liquid of higher density and \(3 / 5\) th in the liquid of lower density. The vessel is placed in an elevator which is moving up with an acceleration of \(a=g / 2\). Find the tension in the string (in N). [g=10 m/s \(^{2}\) ].

For Problems \(13-15\) An object of density \(\sigma(<\rho)\) is dropped from rest at a height \(h\) into a liquid of density \(\rho\) kept in a tall vertical cylindrical container. Neglect all dissipative effects and assume that there is no spilling of the liquid. The acceleration of the object while it is inside the liquid will be (1) \(\frac{\sigma g}{\rho}\) upwards (2) \(\frac{\sigma g}{\rho}\) downwards (3) \(g\left(\frac{\sigma}{\rho}-1\right)\) upwards (4) \(g\left(\frac{\rho}{\sigma}-1\right)\) upwards

A liquid is kept in a cylindrical vessel which is rotated along its axis. The liquid rises at the side. If the radius of the vessel is \(5 \mathrm{~cm}\) and the speed of rotation is \(4 \mathrm{rev} / \mathrm{s}\), then the difference in the height of the liquid at the centre of the vessel and its sides is (1) \(8 \mathrm{~cm}\) (2) \(2 \mathrm{~cm}\) (3) \(40 \mathrm{~cm}\) (4) \(4 \mathrm{~cm}\)
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