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Physics library

Course: Physics library > Unit 3

Lesson 5: Inclined planes and friction
Science
Physics library
Forces and Newton's laws of motion
Inclined planes and friction
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What are inclines?

Google Classroom
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, start subscript, \perp, end subscript, equals, start fraction, \Sigma, F, start subscript, \perp, end subscript, divided by, m, end fraction, a, start subscript, \parallel, end subscript, equals, start fraction, \Sigma, F, start subscript, \parallel, end subscript, divided by, m, end fraction
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, start subscript, \perp, end subscript, equals, 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 start text, s, i, n, e, end text or start text, c, o, s, i, n, e, end text for a given component.

What is the normal force F, start subscript, N, end subscript for an object on an incline?

The normal force F, start subscript, N, end subscript 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,
F, start subscript, N, end subscript, equals, m, g, cosine, 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, equals, 30, start superscript, o, end superscript and the coefficient of kinetic friction between the sled and the hill is mu, start subscript, k, end subscript, equals, 0, point, 150. The combined mass of the child and sled is 65, point, 0, start text, space, k, g, end text.
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∥=ΣF∥m(use Newton’s second law for the parallel direction)a∥=mgsin⁡θ−Fkm(plug in the parallel forces)a∥=mgsin⁡θ−μkFNm(plug in the formula for the force of kinetic friction)a∥=mgsin⁡θ−μk(mgcos⁡θ)m(plug in mgcos⁡θ for the normal force FN)a∥=mgsin⁡θ−μk(mgcos⁡θ)m(cancel the mass that’s in the numerator and denominator)a∥=gsin⁡θ−μk(gcos⁡θ)(savor the awe when you realize the acceleration doesn’t depend on mass)a∥=(9.8ms2)sin⁡30∘−(0.150)(9.8ms2)cos⁡30∘(plug in numerical values)a∥=3.63ms2(calculate and celebrate)\begin{aligned} a_\parallel&=\dfrac{\Sigma F_\parallel}{m} \quad \text{(use Newton's second law for the parallel direction)}\\\\ a_\parallel&=\dfrac{mg\sin\theta-F_k}{m} \quad \text{(plug in the parallel forces)}\\\\ a_\parallel&=\dfrac{mg\sin\theta-\mu_kF_N}{m} \quad \text{(plug in the formula for the force of kinetic friction)}\\\\ a_\parallel&=\dfrac{mg\sin\theta-\mu_k(mg\cos\theta)}{m} \quad \text{(plug in } mg\cos\theta \text{ for the normal force } F_N)\\\\ a_\parallel&=\dfrac{\cancel mg\sin\theta-\mu_k(\cancel mg\cos\theta)}{\cancel m} \quad \text{(cancel the mass that's in the numerator and denominator)}\\\\ a_\parallel&=g\sin\theta-\mu_k(g\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})\sin30^\circ-(0.150)(9.8\dfrac{\text{m}}{\text{s}^2})\cos30^\circ\quad \text{(plug in numerical values)}\\\\ a_\parallel&=3.63\dfrac{\text{m}}{\text{s}^2}\quad \text{(calculate and celebrate)} \end{aligned}

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, point, 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∥=ΣF∥m(use Newton’s second law for the parallel direction)a∥=mgsin⁡θ−Fsm(plug in the parallel forces of gravity and static friction)0=mgsin⁡θ−Fsm(since the car isn’t slipping, the acceleration is zero)0=mgsin⁡θ−Fs(multiply both sides by m)0=mgsin⁡θ−Fs max(assume Fs is equal to its maximum value Fs max)\begin{aligned} a_\parallel&=\dfrac{\Sigma F_\parallel}{m} \quad \text{(use Newton's second law for the parallel direction)}\\\\ a_\parallel&=\dfrac{mg\sin\theta-F_s}{m} \quad \text{(plug in the parallel forces of gravity and static friction)}\\\\ 0&=\dfrac{mg\sin\theta-F_s}{m} \quad \text{(since the car isn't slipping, the acceleration is zero)}\\\\ 0&=mg\sin\theta-F_s \quad \text{(multiply both sides by }m)\\\\ 0&=mg\sin\theta-F_{s \text{ max}} \quad \text{(assume }F_s \text{ is equal to its maximum value }F_{s\text{ max}}) \end{aligned}
0=mgsin⁡θ−μsFN(plug in formula for maximum static friction force)0=mgsin⁡θ−μs(mgcos⁡θ)(plug in expression for normal force on an incline)0=mgsin⁡θ−μs(mgcos⁡θ)(divide both sides by mg)0=sin⁡θ−μs(cos⁡θ)(savor the awe when you realize the angle doesn’t depend on the car’s mass)sin⁡θ=μs(cos⁡θ)(solve for sinθ)sin⁡θcos⁡θ=μs(divide both sides by cosθ)tan⁡θ=μs(replace sin⁡θcos⁡θ with tanθ)θ=tan⁡−1(μs)(take inverse tangent of both sides)θ=tan⁡−1(0.75)(plug in numerical values)θ=37∘(calculate and celebrate)\begin{aligned} 0&=mg\sin\theta-\mu_s F_N\quad \text{(plug in formula for maximum static friction force})\\\\ 0&=mg\sin\theta-\mu_s (mg\cos\theta)\quad \text{(plug in expression for normal force on an incline})\\\\ 0&=\cancel {mg}\sin\theta-\mu_s (\cancel {mg}\cos\theta)\quad \text{(divide both sides by }mg)\\\\ 0&=\sin\theta-\mu_s (\cos\theta)\quad \text{(savor the awe when you realize the angle doesn't depend on the car's mass)}\\\\ \sin\theta&=\mu_s (\cos\theta)\quad \text{(solve for sin}\theta)\\\\ \dfrac{\sin\theta}{\cos\theta}&=\mu_s \quad \text{(divide both sides by cos}\theta)\\\\ \tan\theta&=\mu_s \quad \text{(replace }\dfrac{\sin\theta}{\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^\circ \quad \text{(calculate and celebrate)} \end{aligned}

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