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: 6

Answer

$F_R=8.03kN$ $\phi=1.22^{\circ}$

Work Step by Step

We are asked to determine the magnitude of the force, $F_R$, and its direction, measured clockwise from the positive u axis. We can treat the forces like a triangle and use the law of cosines: $c^2=a^2+b^2−2a∗b∗\cos C$ $c=\sqrt{a^2+b^2−2a∗b∗\cos C}$ $c=\sqrt{(4kN)^2+(6kN)^2−2∗4kN∗6kN∗\cos(105^{\circ})}$ $c=8.03kN$ $F_R=8.03kN$ Now we can use the law of sines to find the angle between $F$ and the x-axis. $\sin\alpha/6kN=\sin105^{\circ}/8.03kN$ $\alpha=\arcsin(\sin105^{\circ}/8.03kN∗6kN)$ $\alpha=46.22^{\circ}$ We need to subtract 45 from our angle to get the angle as measure from the u-axis, $\phi$ $\phi=46.22^{\circ}-45^{\circ}=1.22^{\circ}$
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