## Problem on projection of gravitational force down an inclined plane

A block rests on an inclined plane of angle $\theta$, as shown. Gravity provides a force $\bfF = (0, -m g)$ on the block.
(a) What is the component of $\bfF$ in the downhill direction?
(b) What is the projection of $\bfF$ in the same direction?
• ## Solution

We begin by drawing the relevant vectors.
We note that $\bfF = (0, -m g)$. A vector pointing downhill as shown is $\bfd = (\cos \theta, - \sin \theta)$.
Recall that
We now compute that
Hence, $$\text{comp}_\bfd \bfF = mg \sin \theta.$$
Recall that
Having already computed $\bfF \cdot \bfd$ and $|\bfd|$, we can write that $$\text{proj}_{\bfd}\ \bfF = mg \sin \theta (\cos \theta, - \sin \theta).$$