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: 15

Answer

$F=917lb$ $\theta=31.8^{\circ}$

Work Step by Step

We are asked to determine the magnitude of $F$ and its direction, $\theta$ Given that the force on member AB is 650lb and the force of member BC is 500lb, we can apply the law of cosines to find the magnitude of $F$. $c=\sqrt{a^2+b^2-2*a*b*\cos(C)}$ $c=\sqrt{500^2+650^2-2*500*650*\cos(105^{\circ})}$ $c=917$ $F=917lb$ Using this result, we can apply the law of sines to find the$ \theta$ $\sin\theta / 500 = \sin 105^{\circ} / 917$ Solving for $\theta$, we obtain: $\theta=31.8^{\circ}$
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