# Chapter 2 - Force Vectors - Section 2.4 - Addition of a System of Coplanar Forces - Problems - Page 40: 37

$F_{R} = 1.96 KN$ $\theta = 4.12 °$

#### Work Step by Step

Projecting the three forces on the x axis: $F_{R}x = 900 + 750 \times \cos 45 ° + 650 \times 4/5 = 1950.33N$ Projecting the three forces on the y axis: $F_{R}y = 750 \times \sin 45 ° - 650 \times 3/5 = 140.33N$ The magnitude is obtained: $F_{R} = \sqrt {F_{R}y^{2}+F_{R}x^{2}} = 1955N \approx 1.96KN$ The direction of the force: $\theta = \tan^{-1} (\frac{F_{R}y}{F_{R}x}) = 4.12°$

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