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 - Page 31: 20

Answer

$\phi=38.3^{\circ}$

Work Step by Step

Given that $\theta=30^{\circ}$ and the component of the 400lb force which acts up and to the left in the direction of AB is 600lb $c=\sqrt{a^2+b^2-2*a*b*\cos(C)}$ $c=\sqrt{400^2+600^2-2*400*600*\cos(30^{\circ})}$ $c=322.97$ $F_{AC}=322.97lb$ Using this result, we can apply the law of sines to find $ \phi $ $\sin30^{\circ} / 322.97 = \sin \phi / 400$ Solving for $\phi$, we obtain: $\phi=38.3^{\circ}$
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