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: 50

Answer

$81\;lb$.

Work Step by Step

The given values are $F_l=8100\;lb$ $r_l=30\;in.$ $r_s=3\;in.$ Formula $\Rightarrow \frac{F_l}{F_s}=\frac{r^2_l}{r^2_s}$ Substitute all the values. $\Rightarrow \frac{8100\;lb}{F_s}=\frac{(30\;in.)^2}{(3\;in.)^2}$ Simplify. $\Rightarrow (8100\;lb)\frac{(3\;in.)^2}{(30\;in.)^2}=F_s$ $\Rightarrow F_s=81\;lb$ Hence, the $F_s$ is $81\;lb$.
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