Fun with Fundamentals: Problem 161

Base rate

Problem 161 — A mound of dirt by any other name can also smell as sweet, as this month’s problem by James White of Henderson, Ky., demonstrates.

“Mr. Wurme, do you have anything further to add to your testimony?” The case of Lee Key Hydraulic Co. vs. Finagel J. Wurme was now in its fifth week. The case centered on the company’s on-site landfill, which had overflowed into a nearby river. Eight years ago, Wurme had been in charge of digging the landfill and assuring it could handle Lee Key’s needs for the next ten years. Wurme contended it was Lucius Bluff’s inept management of waste disposal that caused the mishap.

Wurme’s design of the landfill called for a base layer covered by a top layer. The base layer is composed of filter cake (FC) and bottom ash (BA); and the top layer, FC and fly ash (FA).

“Your honor, I present the following statement that proves the landfill should have been able to last us for ten years,” snipped Wurme. “When I made the calculations for this landfill eight years ago, it could have lasted us for ten years easily.”

Wurme’s statement gave the following information:
• 1,750,000 yd³ had been the total space available for both layers.
• 387,000 yd³ was the space available for the top layer.
• Lee Key produces FC at a rate of 288,000 tons per year. FC has a density of 87.5 lb/ft³.
• The company produces 30,000 tons per year of BA. It has a density of 60.0 lb/ft³.
• Of the space available for top layer, 356,122 tons were allotted for FC.
• The production rate of FA is inconsequential, since any amount over the landfill’s capacity is sold.

Ladies and gentlemen of the readership, you have the facts before you. Given the above yearly production rates and capacities of the landfill, for how many years could the landfill have been used? Is Wurme guilty or innocent of an EPA violation?

Remember, the FC goes to both base and top layer.

Technical consultant: Jack Couillard, Menasha, Wis.

Solution to last month’s problem 160 — You know what your leverage is if you answered 80 ft. Here are the pivotal forces:

Divide the problem into two parts, and let F₁ be the force on the rope exiting the large pulley and F₂, the 10-lb upward force on the long end of the lever. First, solve for F₁:

F₁ = w × R_a

w = weight of water, given as 833.6 lb (8.336 lb/gal 3100 gal).

R_a = mechanical advantage of pulley system. In this case it is the small pulley diameter divided by the large, or ⁴/₂₅.

Thus,

F₁ = 833.6 lb × ⁴/₂₅ = 133.4 lb

Now, let:

s₁ = length of short end of lever, given as 6 ft.

s₂ = length of long end of lever, ft. F₁s₁ = F₂s₂

Solving for s₂:

Thanks to Wurme, Bluff’s presentation was all wet!