SOLUTION: 24. The owner of a highway construction company knows that the maximum load a cylindrical column of a circular cross section can support varies directly as the fourth power of the

Question 55212This question is from textbook Applied College Algebra
: 24. The owner of a highway construction company knows that the maximum load a cylindrical column of a circular cross section can support varies directly as the fourth power of the diameter and inversely as the square of the height.
If a column 2 feet in diameter and 10 feet high supports up to 6 tons, how much of a load can a column 3 feet in diameter and 14 feet high support? This question is from textbook Applied College Algebra

Found 2 solutions by venugopalramana, stanbon:
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
The owner of a highway construction company knows that the maximum load (L SAY)a cylindrical column of circular cross section can support varies directly as the fourth power of the diameter(D SAY) and inversely as the square of the height(H SAY). If a column 2 feet in diameter and 10 feet high supports up to 6 tons, how much of a load can a column 3 feet in diameter and 14 feet high support?
L = K*D^4*1/H^2...WHERE K IS A CONSTANT
CASE 1..
6=K*2^4/10^2=16K/100
K=6*100/16=75/2=37.5
CASE 2
L=37.5*3^4/14^2 = 37.5*81/196 = 15.5 T.

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The owner of a highway construction company knows that the maximum load a cylindrical column of a circular cross section can support varies directly as the fourth power of the diameter and inversely as the square of the height.
If a column 2 feet in diameter and 10 feet high supports up to 6 tons, how much of a load can a column 3 feet in diameter and 14 feet high support?
--------
L=k(d^4/h^2)
need to find "k"
6 tons lbs. = k(2^4/10^2)
k=600/16=37.5
EQUATION:
Load = (37.5)(d^4/h^2)
Load = (37.5)(3^4/14^2)
=(37.5)(81/196)
=15.497... tons
Cheers,
Stan H.
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