# Question #afa31

May 4, 2016

We are discussing inclined plane here which makes an angle $\theta$ with the horizontal.

#### Explanation:

The weight $W = m g$ of the body acts vertically downwards as shown in the figure.

The Normal reaction force $R$ which is perpendicular to the surface of the inclined plane and is therefore, equal and opposite to the $c o s i n e$ component of the weight of the body. See the geometry, and angle $\theta$

$\therefore \vec{R} = - m \cdot \vec{g} \cos \theta$, or considering only the scalar parts

$R = m \cdot g \cos \theta$