Elementary Technical Mathematics

Published by Brooks Cole
ISBN 10: 1285199197
ISBN 13: 978-1-28519-919-1

Chapter 7 - Section 7.3 - Direct Variation - Exercises - Page 285: 48

Answer

$300\;lb$.

Work Step by Step

The given values are $F_l=7500\;lb$ $r_l=15\;in.$ $r_s=3\;in.$ Formula $\Rightarrow \frac{F_l}{F_s}=\frac{r^2_l}{r^2_s}$ Substitute all the values. $\Rightarrow \frac{7500\;lb}{F_s}=\frac{(15\;in.)^2}{(3\;in.)^2}$ Simplify. $\Rightarrow (7500\;lb)\frac{(3\;in.)^2}{(15\;in.)^2}=F_s$ $\Rightarrow F_s=300\;lb$ Hence, the $F_s$ is $300\;lb$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.