#### Answer

$M_O=460N*m \circlearrowright$

#### Work Step by Step

We will sum the torques about point O
$(+ \circlearrowleft) \sum \tau_O$
$-(3/5)F*5m-(4/5)F*2m$
$-(3/5)(100N)*5m-(4/5)(100N)*2m$
$-460N*m$
$460N*m \circlearrowright$

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Published by
Pearson

ISBN 10:
0133915425

ISBN 13:
978-0-13391-542-6

$M_O=460N*m \circlearrowright$

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