Question

Find the total force exerted by the fluid against the lateral surface of a right circular cylinder of height 6 feet, which stands on its circular base of radius 5 feet, if it is filled with oil \((\delta=50\) pounds per cubic foot).

Short answer

The total force exerted by the oil is approximately 28274 pounds.

Step-by-step solution

Understand the Problem

We are given a right circular cylinder with a height of 6 feet and a radius of 5 feet. The cylinder is filled with oil which has a density of \(\delta = 50\) pounds per cubic foot. We need to find the total force exerted by the oil on the lateral surface of the cylinder.

Set Up the Integral for Force Calculation

The force exerted by the fluid at any depth is given by the formula \(F = \delta \cdot A \cdot d\), where \(A\) is the area of a differential strip of the lateral surface and \(d\) is the depth of that strip. To find the total force, we need to integrate this expression over the entire height of the cylinder.

Define the Area of a Differential Strip

Consider a horizontal strip at a depth \(y\) below the surface. The area of this strip is the circumference of the base (which is \(2\pi r\)) times an infinitesimal thickness \(dy\). Thus, \(dA = 2\pi r\, dy\).

Calculate the Force on a Differential Strip

Using the formula \(F = \delta \cdot A \cdot d\), the force on a small strip at depth \(y\) is \(dF = 50 \cdot 2\pi \cdot 5 \cdot y \cdot dy = 500\pi y\, dy\).

Integrate Over the Entire Height to Find Total Force

Integrate the expression for \(dF\) from 0 to 6: \[ F = \int_0^6 500\pi y\, dy \].

Perform the Integration

Calculate the integral: \[ F = 500\pi \int_0^6 y\, dy \].The antiderivative of \(y\) is \(\frac{1}{2}y^2\), so:\[ F = 500\pi \left[ \frac{1}{2}y^2 \right]_0^6 = 500\pi \left( \frac{1}{2} \times 36 \right) = 9000\pi \].

Compute the Final Answer

Multiply by \(\pi\) to get the numerical value of the force:\[ F = 9000 \pi \approx 28274 \text{ pounds} \].

Key concepts

Force Calculation

Calculating the force exerted by a fluid involves understanding how pressure acts across a surface. When dealing with a cylinder, the force due to a fluid is considered across the entire lateral surface. This is because pressure acts perpendicularly, and its magnitude varies with depth.

In our problem, we are using a formula for force based on pressure, given by the equation:

• \( F = \delta \cdot A \cdot d \)

where \( F \) is the force, \( \delta \) is the density of the fluid, \( A \) is the surface area element, and \( d \) is the depth from the surface.

The force calculation involves summing up the components of force exerted by the fluid over the surface area, so understanding each contributing factor is crucial.

Integral Setup

The integral setup is essential for solving problems involving varying pressure across a surface. Here, we express the force exerted by the fluid as an integral over a range.

In this exercise, the oil exerts a pressure that varies with depth, y, in the cylinder. To find the total force, set up an integral over the lateral surface:

• \( F = \int_0^6 500\pi y \, dy \)

where \( 6 \) is the height of the cylinder.

The idea is to add up all the little forces acting on each horizontal strip of the cylinder. Each strip has an influence that can be summed from the top (\( y = 0 \)) to the bottom (\( y = 6 \)). This technique accurately captures the total force by accounting for the changeable nature of pressure with depth.

Right Circular Cylinder

A right circular cylinder is a three-dimensional shape with a circular base and specific height. The cylinder in our problem has a height of 6 feet and a radius of 5 feet.

Understanding the geometry of the cylinder helps visualize the lateral surface area, which is crucial in calculating fluid forces. The surface area of such a cylinder can be imagined as a rectangle when "unrolled," with:

• Width: \( 2\pi r \) (the circumference of the base)
• Height: the cylinder's height, 6 feet

Understanding these dimensions is key because it affects how we calculate area and integrate forces exerted by the fluid.

Density and Pressure

Density and pressure are fundamental concepts in fluid mechanics, impacting how fluids exert force. Density, \( \delta \), indicates how much mass a fluid has in a given volume. Here, the fluid is oil with a density of 50 pounds per cubic foot.

Pressure is the force per unit area exerted by a fluid perpendicular to a surface. In this exercise, the pressure increases with depth within the cylinder as described by:

• Pressure at a depth \( y \) is \( \delta y \)

This implies that deeper surfaces encounter higher force from the fluid because they experience more pressure. The density value is crucial as it scales the pressure and overall force exerted by the fluid.