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 30: 9


$F=616lb$ $\theta=46.9^{\circ}$

Work Step by Step

We are given that in the figure, that the magnitude of the resultant force is 1200N. We are asked to determine the magnitude of the force, F, and its direction, measured counterclockwise from the positive x axis. We can apply the law of cosines to find the magnitude of F. $c=\sqrt{a^2+b^2-2*a*b*\cos(C)}$ $c=\sqrt{900^2+1200^2-2*900*1200*\cos(30^{\circ})}$ $c=616lb$ $F=616lb$ Using this result, we can apply the law of sines to find the angle F makes to the x-axis, $\theta$ $\sin\theta / 900lb = \sin 30^{\circ} / 616lb$ Solving for $\theta$, we obtain: $\theta=46.9^{\circ}$
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