A man pulls with a force of 300 N on a rope attached to the top of a building, as shown in Fig. 2.19a. What are the horizontal and vertical components of the force exerted by the rope at point A?
You can see from Fig. 2.19b that
F_x=+(300 \mathrm{~N}) \cos \alpha \quad F_y=-(300 \mathrm{~N}) \sin \alpha
Observing that AB = 10 m, we find from Fig. 2.19a
\cos \alpha=\frac{8 \mathrm{~m}}{A B}=\frac{8 \mathrm{~m}}{10 \mathrm{~m}}=\frac{4}{5} \quad \sin \alpha=\frac{6 \mathrm{~m}}{A B}=\frac{6 \mathrm{~m}}{10 \mathrm{~m}}=\frac{3}{5}
We thus obtain
F_x=+(300 \mathrm{~N}) \frac{4}{5}=+240 \mathrm{~N} \quad F_y=-(300 \mathrm{~N}) \frac{3}{5}=-180 \mathrm{~N}
This gives us a total force of
\mathbf{F}=(240 \mathrm{~N}) \mathbf{i}-(180 \mathrm{~N}) \mathbf{j}