# What are inclines?

Surfaces usually aren't perfectly horizontal. Learn how to deal with slopes!

## What are inclines?

Slides at the park, steep driveways, and shipping truck loading ramps are all examples of inclines. Inclines or inclined planes are diagonal surfaces that objects can sit on, slide up, slide down, roll up, or roll down.
Inclines are useful since they can reduce the amount of force required to move an object vertically. They're considered one of the six classical simple machines.

## How do we use Newton's second law when dealing with inclined planes?

In most cases, we solve problems involving forces by using Newton's second law for the horizontal and vertical directions. But for inclines, we're typically concerned with the motion parallel to the surface of an incline so it's often more useful to solve Newton's second law for the directions parallel to and perpendicular to the inclined surface.
This means that we will typically be using Newton's second law for the directions perpendicular $\perp$ and parallel $\parallel$ to the surface of the inclined plane.
$\Large a_\perp=\dfrac{\Sigma F_\perp}{m} \qquad \qquad a_\parallel=\dfrac{\Sigma F_\parallel}{m}$
Since the mass typically slides parallel to the surface of the incline, and does not move perpendicular to the surface of the incline, we can almost always assume that $a_\perp=0$.

## How do we find the $\perp$ and $\parallel$ components of the force of gravity?

Since we're going to be using Newton's second law for the directions perpendicular to and parallel to the surface of the incline, we'll need to determine the perpendicular and parallel components of the force of gravity.
The components of the force of gravity are given in the diagram below. Be careful, people often mix up whether they should use $\text{sine}$ or $\text{cosine}$ for a given component.

## What is the normal force $F_N$ for an object on an incline?

The normal force $F_N$ is always perpendicular to the surface exerting the force. So an inclined plane will exert a normal force perpendicular to the surface of the incline.
If there is to be no acceleration perpendicular to the surface of the incline, the forces must be balanced in the perpendicular direction. Looking at the forces shown below, we see that the normal force must equal the perpendicular component of the force of gravity, to ensure that the net force is equal to zero in the perpendicular direction.
In other words, for an object sitting or sliding on an incline,
$\Large F_N=mg\text{cos}\theta$

## What do solved examples involving inclines look like?

### Example 1: Snowy sliding sled

A child slides down a snowy hill on a sled. The angle the hill makes with respect with horizontal is $\theta=30^o$ and the coefficient of kinetic friction between the sled and the hill is $\mu_k=0.150$. The combined mass of the child and sled is $65.0 \text{ kg}$.
What is the acceleration of the sled down the hill?
We'll start by drawing a force diagram.
We can use Newton's second law in the direction parallel to the incline to get,
$a_\parallel=\dfrac{\Sigma F_\parallel}{m} \quad \text{(use Newton's second law for the parallel direction)}$
$a_\parallel=\dfrac{mg\text{sin}\theta-F_k}{m} \quad \text{(plug in the parallel forces)}$
$a_\parallel=\dfrac{mg\text{sin}\theta-\mu_kF_N}{m} \quad \text{(plug in the formula for the force of kinetic friction)}$
$a_\parallel=\dfrac{mg\text{sin}\theta-\mu_k(mg\text{cos}\theta)}{m} \quad \text{(plug in } mg\text{cos}\theta \text{ for the normal force } F_N)$
$a_\parallel=\dfrac{\cancel mg\text{sin}\theta-\mu_k(\cancel mg\text{cos}\theta)}{\cancel m} \quad \text{(cancel the mass that's in the numerator and denominator)}$
$a_\parallel=g\text{sin}\theta-\mu_k(g\text{cos}\theta)\quad \text{(savor the awe when you realize the acceleration doesn't depend on mass)}$
$a_\parallel=(9.8\dfrac{\text{m}}{\text{s}^2})\text{sin}30^o-(0.150)(9.8\dfrac{\text{m}}{\text{s}^2})\text{cos}30^o\quad \text{(plug in numerical values)}$
$a_\parallel=3.63\dfrac{\text{m}}{\text{s}^2}\quad \text{(calculate and celebrate)}$

### Example 2: Steep driveway

A person is building a house and wants to know how steep she can make her driveway and still park on it without slipping. She knows the coefficient of static friction between her tires and the concrete driveway is $0.75$.
What is the maximum angle from horizontal that the person can make her driveway and still park her car on it without slipping?
We'll start by using Newton's second law for the parallel direction.
$a_\parallel=\dfrac{\Sigma F_\parallel}{m} \quad \text{(use Newton's second law for the parallel direction)}$
$a_\parallel=\dfrac{mg\text{sin}\theta-F_s}{m} \quad \text{(plug in the parallel forces of gravity and static friction)}$
$0=\dfrac{mg\text{sin}\theta-F_s}{m} \quad \text{(since the car isn't slipping, the acceleration is zero)}$
$0=mg\text{sin}\theta-F_s \quad \text{(multiply both sides by }m)$
$0=mg\text{sin}\theta-F_{s \text{ max}} \quad \text{(assume }F_s \text{ is equal to its maximum value }F_{s\text{ max}})$
$0=mg\text{sin}\theta-\mu_s F_N\quad \text{(plug in formula for maximum static friction force})$
$0=mg\text{sin}\theta-\mu_s (mg\text{cos}\theta)\quad \text{(plug in expression for normal force on an incline})$
$0=\cancel {mg}\text{sin}\theta-\mu_s (\cancel {mg}\text{cos}\theta)\quad \text{(divide both sides by }mg)$
$0=\text{sin}\theta-\mu_s (\text{cos}\theta)\quad \text{(savor the awe when you realize the angle doesn't depend on the car's mass)}$
$\text{sin}\theta=\mu_s (\text{cos}\theta)\quad \text{(solve for sin}\theta)$
$\dfrac{\text{sin}\theta}{\text{cos}\theta}=\mu_s \quad \text{(divide both sides by cos}\theta)$
$\text{tan}\theta=\mu_s \quad \text{(replace }\dfrac{\text{sin}\theta}{\text{cos}\theta} \text{ with tan}\theta)$
$\theta=tan^{-1}(\mu_s) \quad \text{(take inverse tangent of both sides)}$
$\theta=tan^{-1}(0.75) \quad \text{(plug in numerical values)}$
$\theta=37^o \quad \text{(calculate and celebrate)}$