Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 4 - Force System Resultants - Section 4.4 - Principle of Moments - Problems - Page 138: 4

Answer

$M_{F1}=-3000\:lb.ft.$ $M_{F2}=-5600\:lb.ft.$ $M_{F1}=-2593\:lb.ft.$

Work Step by Step

Counterclockwise is positive. For F1; $M_{F1}=-375\:lb\:(8\:ft)=-3000\:lb.ft.\;or\;3000\:lb.ft.\;clockwise$ For F2; $M_{F2}=-500\:lb(\frac{4}{5})\:(14\:ft)=-5600\:lb.ft.\;or\;5600\:lb.ft.\;clockwise$ For F3; $M_{F2}=-160\:lb(cos\:30)\:(19\:ft)+160\:lb(sin\:30)\:(0.5\:ft)=-2593\:lb.ft.\;or\;2593\:lb.ft.\;clockwise$
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