Engineering Mechanics: Statics & Dynamics (14th Edition)

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

Chapter 2 - Force Vectors - Section 2.4 - Addition of a System of Coplanar Forces - Problems - Page 43: 58



Work Step by Step

We find the force components: $F_x=8cos 30^{\circ}=6.93KN$ and $F_y=8 sin 30^{\circ}=4KN$ Thus $F_x=6+Fsin\theta-4sin 15^{\circ}=6.93$ $\implies Fsin\theta=1.9653$ and $F_y=Fcos\theta+4cos 15^{\circ}=4$ $\implies Fcos\theta=0.1363$ Now $tan\theta=\frac{1.9653}{0.1363}$ $\implies \theta=86.03^{\circ}$ and $F=\frac{1.9653}{sin\theta}$ $\implies F=1.97KN$
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