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: 46

Answer

$6561\;lb$.

Work Step by Step

The given values are $F_s=81\;lb$ $r_l=9\;in.$ $r_s=1\;in.$ Formula $\Rightarrow \frac{F_l}{F_s}=\frac{r^2_l}{r^2_s}$ Substitute all the values. $\Rightarrow \frac{F_l}{81\;lb}=\frac{(9\;in.)^2}{(1\;in.)^2}$ Simplify. $\Rightarrow F_l=(81\;lb)\frac{(9\;in.)^2}{(1\;in.)^2}$ $\Rightarrow F_l=6561\;lb$ Hence, the $F_l$ is $6561\;lb$.
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