Answer
$M_O=460N*m \circlearrowright$
Work Step by Step
We will sum the torques about point O
$(+ \circlearrowleft) \sum \tau_O$
$-(3/5)F*5m-(4/5)F*2m$
$-(3/5)(100N)*5m-(4/5)(100N)*2m$
$-460N*m$
$460N*m \circlearrowright$
Engineering Mechanics: Statics & Dynamics (14th Edition)
by Hibbeler, Russell C.
Published by
Pearson
ISBN 10:
0133915425
ISBN 13:
978-0-13391-542-6
Chapter 4 - Force System Resultants - Section 4.4 - Principle of Moments - Fundamental Problems - Page 136: 1
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