Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 3 - Equilibrium of a Particle - Section 3.3 - Coplanar Force Systems - Preliminary Problems: 1

Answer

$F_1=1.83 kN$ $F_2=9.60 kN$

Work Step by Step

First we will sum the forces horizontally $(\rightarrow +) \sum \overrightarrow{F}_{x} =0 $ $F_2 \sin 70^{\circ} + F_1 \cos 60^{\circ} - 5 \cos 30^{\circ} - 4/5 (7) = 0$ Now we will sum the forces vertically $(\uparrow +) \sum \overrightarrow{F}_{y} =0 $ $F_2 \cos 70^{\circ} + 5 \sin 30^{\circ} - F_1 \sin 60^{\circ} - 3/5 (7) = 0$ Solving the system of two equations, we obtain : $F_2=9.60 kN$ $F_1=1.83 kN$
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