Engineering Mechanics: Statics & Dynamics (14th Edition)

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

Chapter 2 - Force Vectors - Section 2.3 - Vector Addition of Forces - Problems: 22

Answer

$F'=161N$ $F_R=257N$ $\phi=163^{\circ}$

Work Step by Step

$c=\sqrt{a^2+b^2-2*a*b*\cos(C)}$ $F'=\sqrt{300^2+200^2-2*300*200*\cos(30^{\circ})}$ $F'=161.48N$ $\sin30^{\circ} / 161.48 = \sin \theta'/ 200$ $\theta'= 38.26^{\circ}$ $c=\sqrt{a^2+b^2-2*a*b*\cos(C)}$ $F_R=\sqrt{400^2+161.48^2-2*400*161.48*\cos(21.74^{\circ})}$ $F_R=257N$ $\sin 21.74^{\circ} / 257.05 = \sin \theta/ 161.48$ $\theta= 13.45^{\circ}$ $\phi=90^{\circ}+60^{\circ}+13.45^{\circ}=163^{\circ}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.