Moment of truth
Problem 204
— It pays to know where you stand relative to the break-even point, as this month’s problem by Earl Link of Cincinnati demonstrates.

It was the day before the unveiling of the new offices of McBean, Bluster, and Fogg, Certified Public Accounts, and there were a myriad of last-minute details.

“Why it’s just that upper portion of the wall that needs touching up with paint,” stated Barney McBean. “I’ve put a ladder just underneath it.”

“Very well,” said junior partner Marius McSquibb. “This isn’t in the job description.”

The ladder is 20 ft long and weighs 80 lb, which is uniformly distributed along its length. It rests with one end on a rough floor and the other against the smooth wall. The ladder makes an angle of 60 deg with the floor, and the coefficient of friction between the ladder and floor is 0.3. How far can the 150-lb McSquibb go up the ladder before it begins to slip?

Send your answer to:

Fun With Fundamentals
1100 Superior Ave.
Cleveland, OH 44114-2543

Deadline is May 10. Good luck!

Technical consultant, Jack Couillard, Menasha, Wis.

Solution to last month’s problem 203 — You know when to turn it off if you answered 17 hr. Here’s how Bluff derived a large cleaning bill:

Let: v = Velocity of water out of port, fps
h = Head of water in tank, ft
= Gravitational constant, 32.2 ft/sec2
= Diameter of tank, given as 8 ft
= Area of port opening
= Discharge from port, ft3/min
= Discharge from spigot, given as 65 gpm or 8.69 ft3/min
= Discharge coefficient, 0.61
t = Time to fill tank to 90% capacity, min

We can use Bernoulli’s Theorem to compute how long it takes the float switch to turn off the spigot. We are told the port opening is 1 in. in diameter. Therefore the area is π(1/24 ft)250.0055 ft2

h varies with time and also with the flow rates from the port and spigot. The change in head is the difference in flow rates divided by the tank area.