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.
a=ΣFma=ΣFm\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=0a_\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 sine\text{sine} or cosine\text{cosine} for a given component.

What is the normal force FNF_N for an object on an incline?

The normal force FNF_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,
FN=mgcosθ\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 θ=30o\theta=30^o and the coefficient of kinetic friction between the sled and the hill is μk=0.150\mu_k=0.150. The combined mass of the child and sled is 65.0 kg65.0 \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=ΣFm(use 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=mgsinθFkm(plug in the parallel forces)a_\parallel=\dfrac{mg\text{sin}\theta-F_k}{m} \quad \text{(plug in the parallel forces)}
a=mgsinθμkFNm(plug in the formula for the force of kinetic friction)a_\parallel=\dfrac{mg\text{sin}\theta-\mu_kF_N}{m} \quad \text{(plug in the formula for the force of kinetic friction)}
a=mgsinθμk(mgcosθ)m(plug in mgcosθ for the normal force FN)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=mgsinθμk(mgcosθ)m(cancel the mass that’s in the numerator and denominator)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=gsinθμk(gcosθ)(savor the awe when you realize the acceleration doesn’t depend on mass)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=(9.8ms2)sin30o(0.150)(9.8ms2)cos30o(plug in numerical values)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=3.63ms2(calculate and celebrate)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 0.750.75.
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=ΣFm(use 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=mgsinθFsm(plug in the parallel forces of gravity and static friction)a_\parallel=\dfrac{mg\text{sin}\theta-F_s}{m} \quad \text{(plug in the parallel forces of gravity and static friction)}
0=mgsinθFsm(since the car isn’t slipping, the acceleration is zero)0=\dfrac{mg\text{sin}\theta-F_s}{m} \quad \text{(since the car isn't slipping, the acceleration is zero)}
0=mgsinθFs(multiply both sides by m)0=mg\text{sin}\theta-F_s \quad \text{(multiply both sides by }m)
0=mgsinθFs max(assume Fs is equal to its maximum value Fs max)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=mgsinθμsFN(plug in formula for maximum static friction force)0=mg\text{sin}\theta-\mu_s F_N\quad \text{(plug in formula for maximum static friction force})
0=mgsinθμs(mgcosθ)(plug in expression for normal force on an incline)0=mg\text{sin}\theta-\mu_s (mg\text{cos}\theta)\quad \text{(plug in expression for normal force on an incline})
0=mgsinθμs(mgcosθ)(divide both sides by mg)0=\cancel {mg}\text{sin}\theta-\mu_s (\cancel {mg}\text{cos}\theta)\quad \text{(divide both sides by }mg)
0=sinθμs(cosθ)(savor the awe when you realize the angle doesn’t depend on the car’s mass)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)}
sinθ=μs(cosθ)(solve for sinθ)\text{sin}\theta=\mu_s (\text{cos}\theta)\quad \text{(solve for sin}\theta)
sinθcosθ=μs(divide both sides by cosθ)\dfrac{\text{sin}\theta}{\text{cos}\theta}=\mu_s \quad \text{(divide both sides by cos}\theta)
tanθ=μs(replace sinθcosθ with tanθ)\text{tan}\theta=\mu_s \quad \text{(replace }\dfrac{\text{sin}\theta}{\text{cos}\theta} \text{ with tan}\theta)
θ=tan1(μs)(take inverse tangent of both sides)\theta=tan^{-1}(\mu_s) \quad \text{(take inverse tangent of both sides)}
θ=tan1(0.75)(plug in numerical values)\theta=tan^{-1}(0.75) \quad \text{(plug in numerical values)}
θ=37o(calculate and celebrate)\theta=37^o \quad \text{(calculate and celebrate)}
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