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 - Problems - Page 98: 5


$T=7.20kN$ $F=5.40kN$

Work Step by Step

Given that $\theta=90^{\circ}$, we can solve for the angle that C makes with the x-axis, $\phi$. $\phi=90^{\circ}-\arctan(3/4)=53.13^{\circ}$ First we will sum the forces horizontally $(\rightarrow +) \sum \overrightarrow{F}_{x} =0 $ $T\cos53.13^{\circ}−4/5(F)=0$ Now we will sum the forces vertically $(\uparrow +) \sum \overrightarrow{F}_{y} =0 $ $9-T\sin53.13^{\circ}-3/5(F)=0$ Solving the system of two equations, we obtain : $T=7.20kN$ $F=5.40kN$
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