Engineering Mechanics: Statics & Dynamics (14th Edition)

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

Chapter 2 - Force Vectors - Section 2.4 - Addition of a System of Coplanar Forces - Problems - Page 43: 54


$-1.98KN; 184.53^{\circ}$

Work Step by Step

We know that $F_x=-30sin 30^{\circ}-26(\frac{5}{13})$ $\implies F_x=-25KN$ and $F_y=-30 cos30^{\circ}+26(\frac{12}{13})$ $\implies F_y=-1.98KN$ Now the resultant force can be calculated as $F_R=\sqrt{(F_x)^2+(F_y)^2}$ We plug in the known values to obtain: $F_R=\sqrt{(-25KN)^2+(-1.98KN)^2}$ $\implies F_R=25.08KN$ The orientation can be determined as $\phi=tan^{-1}(\frac{F_y}{F_x})$ $\phi=(\frac{-1.98KN}{-25KN})=4.53^{\circ}$ $\implies \theta=184.53^{\circ}$
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