# Problem: Building Anchor

This question varies a bit from the normal Air Teaser, however it still uses some of the same formulas that are used to determine a force to overcome motion.

You have a building 12′ high, 8′ wide, and 20′ long that has a flat roof. The wind is hitting the side of the building straight on (12′ x 20′) with a 60-mph speed. How much would the building need to weigh to keep it from tipping over if it was only anchored on the left side to keep it from sliding? The building is uniformly loaded.

Two useful equations:

HP =

Pounds of Pull x Distance in Feet / Sec.
550

HP =

FA x MPH3
150,000

FA = Square feet of surface area

##### Find Out The Solution

Let’s use a moment equation to solve this problem: weight of the building x the average distance = the force of the wind x average height.

Weight = what we are looking for

Average height = 12′ / 2 or 6 feet vertical distance to pivot

Average width = 8′ / 2 or 4 feet horizontal distance to pivot

Frontal Area = Wind surface area = 12′ x 20′ = 240 sq. ft.

Wind speed = 60 mph = 5,280 / 60 minutes = 88′ per second

Force of wind:  HP = FA x MPH3         and
150,000

HP = Pounds of push or pull x Distance in feet / second
550

HP = 240 x 603         HP = 345.6
150,000

345.6 HP = Pounds of push x 88           Pounds of push = 2160 lbs.
550

Moment Equation:

Weight x 4′ = 2160 lbs. x 6′          Weight = 2160 x 6 / 4 = 3240+ lbs. of weight needed

### Deadline past. Not available for submissions.

Winner:

Kevin Breunig, CFPS, Motion Industries, Inc., Houston, TX

Dallas Anderson, Montrose, MN

By Ernie Parker, AI, AJPP, AJPPCC, S, MT, MM, MIH, MIP, MMH,
Fluid Power Instructor, Hennepin Technical College, EParker@Hennepintech.edu

This teaser is printed in the Fluid Power Journal. Those who submit the correct answer before the deadline will have their names printed in the Society Page newsletter and in Fluid Power Journal. The winners will also be entered into a drawing for a special gift.

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