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: 49

Answer

$640\;lb$.

Work Step by Step

The given values are $F_s=40\;lb$ $r_l=28\;in.$ $r_s=7\;in.$ Formula $\Rightarrow \frac{F_l}{F_s}=\frac{r^2_l}{r^2_s}$ Substitute all the values. $\Rightarrow \frac{F_l}{40\;lb}=\frac{(28\;in.)^2}{(7\;in.)^2}$ Simplify. $\Rightarrow F_l=(40\;lb)\frac{(28\;in.)^2}{(7\;in.)^2}$ $\Rightarrow F_l=640\;lb$ Hence, the $F_l$ is $640\;lb$.
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