Question 2.CA.1: A force of 800 N is exerted on a bolt A as shown in Fig. 2.1......

In order to obtain the correct sign for the scalar components F_x \text { and } F_y, we could substitute the value 180° – 35° = 145° for θ in Eqs. (2.8). However, it is often more practical to determine by inspection the signs of F_x \text { and } F_y (Fig. 2.18b) and then use the trigonometric functions of the angle α = 35°. Therefore,

F_x=F \cos \theta \quad F_y=F \sin \theta (2.8)

\begin{array}{l}F_x=-F \cos \alpha=-(800 \mathrm{~N}) \cos 35^{\circ}=-655 \mathrm{~N} \\F_y=+F \sin \alpha=+(800 \mathrm{~N}) \sin 35^{\circ}=+459 \mathrm{~N}\end{array}

The vector components of F are thus

\mathbf{F}_x=-(655 \mathrm{~N}) \mathbf{i} \quad \mathbf{F}_y=+(459 \mathrm{~N}) \mathbf{j}

and we may write F in the form

\mathbf{F}=-(655 \mathrm{~N}) \mathbf{i}+(459 \mathrm{~N}) \mathbf{j}