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: 12


$\theta=54.9^{\circ}$ $F=10.4kN$

Work Step by Step

We can apply the law of sines to find the angle between the y-axis and $F_A$, ($\theta$) $\sin(90^{\circ}-\theta) / 6 = \sin 50^{\circ} / 8$ Solving for $\theta$, we obtain: $\theta=54.9^{\circ}$ By summing the forces and creating a triangle, we can solve for the missing angle in the triangle knowing that the whole triangle is $180^{\circ}$ $\phi=180^{\circ}-(90^{\circ}-54.9^{\circ})=94.93^{\circ}$ Now we can apply the law of cosines to find the magnitude of the resultant force, $F_R$. $c=\sqrt{a^2+b^2-2*a*b*\cos(C)}$ $c=\sqrt{8^2+6^2-2*8*6*\cos(94.93^{\circ})}$ $c=10.4$ $F=10.4kN$
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