## Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson

# Chapter 3 - Equilibrium of a Particle - Section 3.3 - Coplanar Force Systems - Problems - Page 98: 1

#### Answer

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

#### Work Step by Step

First we will sum the forces horizontally $(\rightarrow +) \sum \overrightarrow{F}_{x} =0$ $F_2\sin70^{\circ}+F_1\cos60^{\circ}−5\cos30^{\circ}−4/5(7)=0$ Now we will sum the forces vertically $(\uparrow +) \sum \overrightarrow{F}_{y} =0$ $F_2\cos70^{\circ}+5\sin30^{\circ}−F_1\sin60^{\circ}−3/5(7)=0$ Solving the system of two equations, we obtain : $F_2=9.60kN$ $F_1=1.83kN$

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