#### Answer

167 N

#### Work Step by Step

$T=Frsin\theta$
For force to be minimum, $sin\theta$ needs to be maximum i.e. 1
So, $F_{min}=\frac{T}{r}=\frac{25}{0.15}=166.7N\approx167N$

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Published by
Pearson

ISBN 10:
0-32160-183-1

ISBN 13:
978-0-32160-183-4

167 N

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