Question

A force of 4.48 Newtons is applied horizontally while a force of 3.36 N is applied vertically to an object. Find the resultant force (magnitude and angle  from the horizontal).

Answer 1

The magnitude of resultant force

$$F=\sqrt{F_x^2+F_y^2}=\sqrt{4.48^2+3.36^2}=5.6\:\rm N$$

The direction of resultant force

$$\theta=\tan^{-1}\frac{F_y}{F_x}=\tan^{-1}\frac{3.36}{4.48}=36.9^{\circ}$$