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

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> 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       Log On


   

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) About Me  (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) About Me  (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.