# Fun with Fundamentals: Problem 166

Oct. 1, 2000
Current affairs and Solution to Problem 165. The path not taken makes no difference, as this month’s problem by Frank Risko of Livonia, Mich., demonstrates.

### Current affairs

Problem 166 — The path not taken makes no difference, as this month’s problem by Frank Risko of Livonia, Mich., demonstrates.

An aroma of pencil shavings was in the air as the competition for the Zapp Annual Electrical Engineering Scholarship began. The key problem was Zapp’s own “Resistive Cube” puzzle.

The cube consists of an electrically balanced parallelepiped for which each leg has a 120- V resistor. There are 12 resistors. What is the equivalent resistance from Point A to Point B?

Fun With Fundamentals
POWER TRANSMISSION DESIGN
Penton Publishing
1100 Superior Ave.
Cleveland, OH 44114-2543

Technical consultant: Jack Couillard, Menasha, Wis.

Solution to last month’s problem 164 — You landed on your feet if you answered yes. Here’s how things fell out:

Let:
F
= Force needed to stop the 250-lb rider within 20 ft, lb
m
= Mass of rider, cable, and drum, slug
a
= Acceleration of rider, ft/sec2
v
= Velocity of rider, given as 22 fps
μ
= Coefficient of friction between brake pad and drum, given as 0.30
s
= Vertical distance rider travels between brake application and complete stop, given as 20 ft

We know that F = ma, and we can calculate both the acceleration and mass from the data given:

And so,

F = 23.x slugs 3 12.1 ft/sec2 = 281.9 lb

This is the tangential force applied to the winch drum. The axial cylinder force, Fc = F ÷ μ. But the winch drum has a radius of 1.5 ft, and the brake drum, 1.0 ft. Therefore,

0 = 2bx1 + x12 - 2aa1 + a12 (4)

0 = 2bx2 + x22 - 2aa2 + a22 (5)

Now we can calculate the bore required to provide a force of 1,409.5 lb at 112 psi. Let r = radius of bore, in.

1,409.5 lb/in.2 = 112 lb/in.2 × π × r

r2 = 4.00 in.
r = 2.0 in.

Bore = 2r = 4.0 in.

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