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
Answered Correctly:
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.
