PROBLEM    44- 1251:       The elongation, E, of a steel wire when a mass, m, is hung

from its free end varies jointly as m and the length, x, of the

wire and inversely as the cross sectional area A, of the wire.

Given that E = 0.001 inches when m = 20 pounds, x = 10

inches, and A = 0.01 square inches, find E when m = 40

pounds, x = 15.5 inches, and A = 0.015 square inches.

Solution:                    If E is directly proportional to m and x and inversely proportional to

A, our equation is

E = kmx / A

where k is a constant which we can determine from the given information.

Since, when m = 20, x = 10, and A = .01, E = .001, we can use this

information to solve for k. Substituting we obtain:

0.001 = {k(20)(10)} / .01,  and multiplying both

sides by .01/{(20)(10)}         we obtain:

Multiplying numerator and denominator by 100,000 (move the decimal 5

places to the right) we obtain:

Hence,               E = (1 / 20,000,000) (mx/A)

and when m= 40, x= 15.5, and A = 0.015, we have:

E =[1 / 20,000,000] [{40(15.5)} / 0.015]

= 0.00207 inches (approximately).
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