Answer
$50 \ lb$
Work Step by Step
By substitution, we get
$$
\frac{F_l}{F_s}=\frac{r_l^2}{r_s^2}\Longrightarrow \frac{6050}{F_s}=\frac{22^2}{2^2}
.$$
That is
$$
F_s=\frac{6050\times2^2}{22^2}=50 \ lb
.$$
Elementary Technical Mathematics
by Nelson, Robert; Eren, Dale;
Published by
Brooks Cole
ISBN 10:
1285199197
ISBN 13:
978-1-28519-919-1
Chapter 7 - Review - Page 290: 27
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