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: 47

Answer

$400\;lb$.

Work Step by Step

The given values are $F_l=6400\;lb$ $r_l=16\;in.$ $r_s=4\;in.$ Formula $\Rightarrow \frac{F_l}{F_s}=\frac{r^2_l}{r^2_s}$ Substitute all the values. $\Rightarrow \frac{6400\;lb}{F_s}=\frac{(16\;in.)^2}{(4\;in.)^2}$ Simplify. $\Rightarrow (6400\;lb)\frac{(4\;in.)^2}{(16\;in.)^2}=F_s$ $\Rightarrow F_s=400\;lb$ Hence, the $F_s$ is $400\;lb$.
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