__Category__: Algebra, Strength of Materials**"Published in Newark, California, USA"**

The elongation of any metal is directly proportional to the product of the applied force F and the length L, and inversely proportional to the beam cross-sectional area. A steel bar 2 square inches in cross section and 20 inches in length is elongated by 0.03 inches when a force of 10,000 lbs. was applied to it. A certain member of the same metal whose length is 3 feet is allowed a maximum elongation of 0.5 inch when subjected to a force of 18,500 lbs. Compute the minimum permissible area of the member.

__Solution__:

To illustrate the problem, let's draw the figure as follows

Photo by Math Principles in Everyday Life |

As you can see in the figure that when you applied a force F at the end of a metal with length L, there will be an elongation with length ∆L. From the first statement of the word problem, "the elongation of any metal is directly proportional to the product of the applied force F and the length L, and inversely proportional to the beam cross-sectional area," the working equation can be written as follows

and

Combining the two equations above, we have

or

where

∆L = elongation of a metal

k = proportionality constant

F = applied force

L = length of a metal

A = cross sectional area of a metal

From the second statement of a word problem, "a steel bar 2 square inches in cross section and 20 inches in length is elongated by 0.03 inches when a force of 10,000 lbs. was applied to it," substitute the given items to the working equation in order to get the value of k as follows

or

From the third statement of a word problem, "a certain member of the same metal whose length is 3 feet is allowed a maximum elongation of 0.5 inch when subjected to a force of 18,500 lbs," substitute the given items to the working equation in order to get the value of A as follows

Therefore, the minimum permissible area of the member is

**0.1998 in**.

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