Loan arranger and Solution to Problem 176. A penny saved is not necessarily a penny earned, as this problem by Gene Likins, of Duncan, Okla., demonstrates

Loan arranger Problem 177 — A penny saved is not necessarily a penny earned, as this month’s problem by Gene Likins, of Duncan, Okla., demonstrates.

The Mucho Peso Mexican restaurant has long been the favorite haunt of loan sharks, lawyers, and business people of questionable repute.

“ ... and, as I was saying, (more water, please!) the funds from the Hula Ballou account will be available next week,” whispered Milton McSnead. “That’ll more than cover our ‘travel and entertainment’ expenses.”

“Splendid!” retorted Morton Pettifogg. “You can pay me the $30 you personally owe me as well (This food is awful!)”

“As a token of my good faith,” announced McSnead, “I’ll get the bill.”

To his horror, McSnead discovered a $1 bill in his wallet, hardly enough to cover the check. To Pettifogg’s outraged demand for an explanation, Mc- Snead offered the following:

“I started out with plenty of money to pay you and the restaurant bill. Unfortunately my wife demanded half of what I had before I got out the door. I gave her half and 50¢ as well. I then met up with a very dear friend of mine — with whom I had some business. I gave him half of what I had left and 50¢ on top of it. As if that weren’t enough, I then ran into my bookie. I gave him half of what remained and another 50¢ to go away. All I have left is $1.”

How much money did McSnead really start out with (in whole dollars)?

Send your answer to:

Fun With Fundamentals

POWER TRANSMISSION DESIGN

1100 Superior Ave.

Cleveland, OH 44114-2543

*Technical consultant, Jack Couillard, Menasha, Wis.*

** Solution to last month’s problem 176 —** You don’t tip the scales if you answered **no**. Here’s the estimated time of arrival: Let:

*l*_{1} = Length of Conveyor 1 from largewheel centerpoint to small-wheel centerpoint, given as 120 in.

l_{2} = Length of Conveyor 2, from largewheel centerpoint to small-wheel centerpoint, given as 150 in.

ω_{1} = Drive wheel velocity of Conveyor 1, given as 10 rpm

ω_{2} = Drive wheel velocity of Conveyor 2, given as 20 rpm*r _{a}* = Radius of large wheel, given as 15 in.

*= Radius of small wheel, given as 7.5 in.*

r

r

_{b}*t*

_{1}= Time for package on Conveyor 1 to travel length

*l*

_{1}, min

t

t

_{2}= Time for package on Conveyor 2 to travel length

*l*

_{2}, min

Knowing Tipitt’s track record for innovative solutions, we’d better check to see if the packages on Conveyor 1 slide or stay where they’re placed. In addition to being tilted 15 deg, the packages are also raised by the belt angle due to the different radii of the wheels.

Let:

*a* = Belt angle due to difference between small and large wheels

θ = Total angle that package is inclined, or 15 deg + *a*

The friction acting on the package is tan^{-1}(θ + *a*) or 0.34, which is below the given coefficient of 0.41. Therefore, the package does not slide. The time that packages from both conveyors arrive depends on conveyor speed and distance.

The conveyors are still out of synch, and Tipitt needs a new angle.