In this lesson, we will learn:

- How to solve force problems when force is applied at an angle
- How to solve force problems with inclines

__Notes:__- When looking at forces in two dimensions, a force can point along the x or y axis, or at any angle in between. The net force acting on an object is found by adding all the forces acting on that object using vector addition.
- When solving for net force it can be helpful to break angled forces into x and y components so that the forces is the x and y directions can be added separately.

- When an object is on a slope, it tends to be pulled down the slope by gravity. We can understand why gravity pulls the object down the slope if we break the force of gravity into two components: one that is parallel to the slope, and one that is perpendicular.
- We can redefine the x direction to be parallel to the slope and the y direction to be perpendicular to the slope for a particular problem. Essentially, we "tilt" the axes to line up with the slope. The components can then be solved like normal x and y components.
- $\vec{F}_{x}$ represents the amount of $\vec{F}_{g}$ which is pulling the object down the slope.
- $\vec{F}_{y}$ represents the amount of $\vec{F}_{g}$ pushing into the slope. It is balanced by the normal force from the slope pushing back on the box.

**Newton's Second Law**

$\Sigma \vec{F} = \vec{F}_{net} = m\vec{a}$

$\Sigma \vec{F}:$ sum of all forces, in newtons (N)

$\vec{F}_{net}:$ net force, in newtons (N)

$m:$ mass, in kilograms (kg)

$\vec{a}:$ acceleration, in meters per second squared $(m/s^{2})$

**x and y Components of Force**

$\vec{F}_{x or y} = \vec{F}\sin(\theta)$ (For the component opposite to $\theta$)

$\vec{F}_{x or y} = \vec{F}\cos(\theta)$ (For the component adjacent to $\theta$)