# It is known that a cable with a cross-sectional area of 0.70 sq in. has a capacity to hold 2500...

## Question:

It is known that a cable with a cross-sectional area of 0.70 sq in. has a capacity to hold 2500 lb. If the capacity of the cable is proportional to its cross-sectional area, what size cable is needed to hold 7000 lb?

## Using Proportionality to Find Cable Size

The question reads that the weight holding capacity of a cable is proportional to its cross-sectional area. Using this fact and the given data we calculate the constant of proportionality. Using the constant of proportionality we then find the size of the cable needed to hold a much heavier load.

## Answer and Explanation:

Let the cable of cross sectional area A hold load W. Then because of the proportionality we can state that

{eq}W = k A \qquad (1) {/eq}

Given that W = 2500 lb when A = 0.70 sq. in., we can substitute in (1) to get that

{eq}\displaystyle 2500 = k (0.7) \implies k = \frac {2500}{0.7} \qquad (2) {/eq}

This give us the formula as

{eq}\displaystyle W = \frac {2500}{0.7} A \qquad (3) {/eq}

To find the size of the cable's cross-sectional area that is able to hold a load of W = 7000 lb we can substitute in (3) to get

{eq}\displaystyle 7000 = \frac {2500}{0.7} A \implies A = \frac {7000(0.7)}{2500} = 1.96 \; sq. \; in. {/eq}

**We need a cable of cross-sectional area 1.96 square inches to hold the 7000 lb load.**

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from 6th-8th Grade Math: Practice & Review

Chapter 50 / Lesson 13