Question

On the afternoon of January 15, 1919, an unusually warm day in Boston, a 17.7-mhigh, 27.4-m-diameter cylindrical metal tank used for storing molasses ruptured. Molasses flooded into the streets in a 5-mdeep stream, killing pedestrians and horses and knocking down buildings. The molasses had a density of 1600 kg/m\(^3\). If the tank was full before the accident, what was the total outward force the molasses exerted on its sides? (\(Hint:\) Consider the outward force on a circular ring of the tank wall of width \({dy}\) and at a depth \(y\) below the surface. Integrate to find the total outward force. Assume that before the tank ruptured, the pressure at the surface of the molasses was equal to the air pressure outside the tank.)

Short answer

The total outward force exerted on the tank's sides was approximately 65.75 million N.

Step-by-step solution

Understand the Problem

The problem involves calculating the total outward force that the molasses exerted on the sides of a cylindrical tank before it ruptured. The tank's relevant dimensions are given, and we have to consider a differential element at a depth to integrate for total force.

Determine Differential Pressure

At any depth \(y\) below the molasses surface, the pressure exerted on the tank wall is due to the molasses' weight above it. This pressure is given by \( p(y) = \rho g y \), where \(\rho = 1600\, \text{kg/m}^3\) is the density of molasses and \(g = 9.81\, \text{m/s}^2\) is the acceleration due to gravity.

Calculate Differential Force

Consider a small circular strip (ring) of the tank wall at depth \(y\) with thickness \(dy\). The force \(dF\) on this ring due to the pressure at depth \(y\) is \(dF = p(y) \times A_{ring}\), where \(A_{ring}\) is the area of the circular strip, \(A_{ring} = 2\pi r\, dy\), and \(r=13.7\, \text{m}\) is the tank's radius.

Integrate to Find Total Outward Force

Substitute \(p(y) = \rho g y \) and \(A_{ring} = 2\pi r\, dy\) into the differential force equation: \(dF = \rho g y \times 2\pi r \, dy\). The total force is \(F = \int_0^{17.7} \rho g y \times 2\pi r\, dy\). Evaluating the integral, we get:\[ F = 2\pi r \rho g \int_0^{17.7} y \, dy = 2\pi r \rho g \left[ \frac{y^2}{2} \right]_0^{17.7}\].

Compute the Integral

Compute the definite integral: \(\int_0^{17.7} y \, dy = \frac{1}{2}(17.7^2 - 0) = \frac{1}{2}(313.29) = 156.645 \). Substituting this back into the force equation: \( F = 2\pi (13.7) (1600) (9.81) (156.645) \).

Calculate Total Outward Force

Calculate the total force using the values provided:\[ F = 2\pi (13.7) (1600) (9.81) (156.645) \approx 6.575 \times 10^7 \,\text{N}.\] Thus, the total outward force exerted by molasses on the tank’s sides was approximately 65.75 million newtons.

Key concepts

Pressure Underneath the Surface

Pressure is a critical concept in fluid mechanics. It is the force exerted per unit area. This concept is vital when analyzing fluids in containers like tanks or large vessels. In this problem, we are looking at how molasses exerts pressure on a cylindrical tank's walls.

For any fluid at rest, like the molasses in the tank, pressure increases with depth due to the gravitational force acting on the fluid. This is mathematically described by the equation:

• \( p(y) = \rho g y \)

Here, \( \rho \) represents density, \( g \) is the gravitational acceleration, and \( y \) is the depth from the surface. As the depth increases, so does the pressure, meaning the deeper molasses puts more force against the tank walls. By understanding pressure, we better understand how forces build up against the sides of large containers.

Cylindrical Tanks

Cylindrical tanks are common structures used for storing liquids such as water, oil, and in our case, molasses. These tanks are sturdy and designed to hold vast amounts of fluid.

Understanding the cylindrical shape is crucial because the way pressure distributes over this shape's curved walls affects the forces acting upon it. The tank's geometry dictates that forces due to fluid pressure act perpendicular to the walls. This impact is spread over the cylinder's height, and each ring or strip at a specific depth experiences different pressure.

For the Boston molasses tank, knowing its height (17.7 m) and diameter (27.4 m) provides the necessary framework to calculate the pressure distribution and total force on the tank as pressure acts in a radial pattern evenly along the wall's circumference.

Understanding Force Integration

Integrating force allows us to sum up contributions of small differential forces over a certain dimension. It turns a complex problem with various pressures into one solvable equation.

To determine the total force on the tank's surface, we first consider a narrow strip, a ring, at depth \( y \). The force on this ring is derived from the pressure at that depth. The ring's differential force \( dF \) is given by:

• \( dF = p(y) \times A_{ring} \)
• \( A_{ring} = 2\pi r \, dy \)

Substituting these into the equation gives:

• \( dF = \rho g y \times 2\pi r \, dy \)

We then integrate this expression over the tank's height (from 0 to 17.7 m) to find the total force. Integration simplifies our understanding by summing the pressure effects from each thin ring. This technique highlights how calculus, specifically integration, is used in fluid mechanics to manage and calculate the forces exerted by fluids in containers.