# Fun With Fundamentals: Problem 215

May 1, 2000
Might makes right, as this month's problem by Mike Williams of Waynesboro, Va., demonstrates

PROBLEM 215 -- Might makes right, as this month's problem by Mike Williams of Waynesboro, Va., demonstrates.

"We're sorry about your pasture, but as you know, the road must go through," stated the official from the State Bureau of Transportation. "You'll receive the usual compensation for 45,000 ft2 of property."

"Why, certainly sir," grinned McFarley McTavish.

The flat pasture measures 400 ft by 300 ft, and the proposed 100-ft-wide road cuts a diagonal swath through it. Compute the area of the road to see if McTavish should have any qualms of conscience about cashing the check.

SOLUTION TO LAST MONTH'S PROBLEM214 -- You know the swing of things if you answered 29.16 ft. Here's how Buford Buff saved himself from a financial soaking:

Refering to the diagram, we can use the formula for the time of a single revolution for a conical pendulum to solve for θ

where:

r = Radius of circle from pole, ft
T = Time for a single revolution, sec
l
= Length of cable, ft
θ Angle between the cable and vertical, deg
g = Acceleration due to gravity, 32.2 ft/sec2

We know that T is 10 rev/60 sec, or 6 sec/rev. Plug this back into (1) and solve for θ.

Now use the formula for the radius to solve for r.

There is a trick here. The problem said to treat the rider and chair as if they were a 4-ft-diameter sphere. Since (2) computes the distance from the center of the sphere to the pole, we have to add an additional 2 ft to r to obtain the distance of the edge of the sphere from the pole. The correct distance is then 29.16 ft.

Buff moved the waterfall anyway!

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