Wouldn't the driveway be a case where we can't round up? At 37 degrees, the car would slip down the driveway, since the maximum was 36.86... degrees.

If I were to build my drive way based on this math I would choose to round down though ;-)

I don't understand the angle part. Is the angle of incline the same as the angle pictured between the mg and mgcos?

Yes , the angle of incline and the angle between mg and mgcos are the same. Lets take a simple example. Say , let the angle of incline be 30 , then draw the mg force downwards so as it forms a right angled triangle including angle of incline 30 . In that triangle the other two angles will be 90 ( the right angled part ) and 60. Now draw the mgcos force from the same point as mg. We see the mgcos is perpendicular to the surface of incline , where one angle we already found out was 60 , therefore the angle between mg and mgcos has to be 90-60 = 30 , which is equal to the angle of incline. draw the forces on the body on the incline , take this example and solve , your doubt will be dispelled. Thank you !

In example 1, wouldn't all the snow piling up in front of the sled change things?

It would but there is no indication in the question about *snow piling up* . So, you don't need to think about that when solving this problem.

Why do you cancel both the masses in the numerator for 1 mass in the denominator?

When you have something of the form (a*x + a*y)/(a*z) this is equivelent to ((a*x)/(a*z)) + ((a*y)/(a*z)) and you can see that the a's in the numerator and denominator to the left of the + sign cancel as do the numerator and denominator to the right of the + sign giving you (x/z) + (y/z)) which is also equivalent to (x + y)/z

Which way does the force of friction act on an inclined plane if the force acting on the mass is in a horizontal direction and the mass is moving?

friction always opposes motion.

can somebody explain why there should be max. steepness for max. static friction!

Because when you go past that point, the object starts to slide, and that's not static friction anymore

I am literally so lost on the last example. Like how did sine theta get on the other side without being negative? Wouldn't you have had to subtract it from the side it was on with the cosine theta and the friction coefficient? Also why did we do tan negative one all of a sudden? I don't feel like there was a good explanation for that.

For the first question you asked, let's take a look. 0=sin(theta)-coefficient (cos theta) (hopefully this makes sense to you :/ ) -sin(theta)=-coefficient (cos theta) sin(theta)=coefficient (cos theta) Is this makes sense to you? It should! :) The next question was that why did (sin theta) / (cos theta) became tan(theta). right? Sine theta is equal to opposite/hypotenuse. Cosine theta is equal to adjacent/hypotenuse. If you do (sin theta) / (cos theta), the hypotenuse thing will cancel out and it is left with opposite/adjacent, which is same as tangent(theta)! Yay! Hopefully, this makes sense to you. It took a long time to type all these.

Ok I am lost on question 1...so we draw the force diagram...then we have the equation a= F/M ..then you start subsiting cos and sin and im lost at that point...please help

You should probably watch the videos on vectors and splitting up a vector into component parts. If you don't understand that, you won't understand this video.

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Physics library

Course: Physics library > Unit 3

Lesson 5: Inclined planes and friction

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.

[What are "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 $⊥$ ‍ and parallel $∥$ ‍ to the surface of the inclined plane.

a_{⊥} = \frac{Σ F_{⊥}}{m} a_{∥} = \frac{Σ F_{∥}}{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_{⊥} = 0

[Huh? Why is perpendicular acceleration zero?]

How do we find the $⊥$ ‍ and $∥$ ‍ 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

cosine

for a given component.

[Whoa, please walk me through how we find these components.]

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,

F_{N} = m g \cos θ

[Wait, please explain why this is true?]

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 = 30^{o}

and the coefficient of kinetic friction between the sled and the hill is

μ_{k} = 0.150

. The combined mass of the child and sled is

65.0 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,

\begin{aligned} a_{∥} & = \frac{Σ F_{∥}}{m} (use Newton’s second law for the parallel direction) \\ a_{∥} & = \frac{m g \sin θ - F_{k}}{m} (plug in the parallel forces) \\ a_{∥} & = \frac{m g \sin θ - μ_{k} F_{N}}{m} (plug in the formula for the force of kinetic friction) \\ a_{∥} & = \frac{m g \sin θ - μ_{k} (m g \cos θ)}{m} (plug in m g \cos θ for the normal force F_{N}) \\ a_{∥} & = \frac{m g \sin θ - μ_{k} (m g \cos θ)}{m} (cancel the mass that’s in the numerator and denominator) \\ a_{∥} & = g \sin θ - μ_{k} (g \cos θ) (savor the awe when you realize the acceleration doesn’t depend on mass) \\ a_{∥} & = (9.8 \frac{m}{s^{2}}) \sin 30^{\circ} - (0.150) (9.8 \frac{m}{s^{2}}) \cos 30^{\circ} (plug in numerical values) \\ a_{∥} & = 3.63 \frac{m}{s^{2}} (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

0.75

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.

\begin{aligned} a_{∥} & = \frac{Σ F_{∥}}{m} (use Newton’s second law for the parallel direction) \\ a_{∥} & = \frac{m g \sin θ - F_{s}}{m} (plug in the parallel forces of gravity and static friction) \\ 0 & = \frac{m g \sin θ - F_{s}}{m} (since the car isn’t slipping, the acceleration is zero) \\ 0 & = m g \sin θ - F_{s} (multiply both sides by m) \\ 0 & = m g \sin θ - F_{s max} (assume F_{s} is equal to its maximum value F_{s max}) \end{aligned}

[Why are we assuming static friction is a maximum?]

\begin{aligned} 0 & = m g \sin θ - μ_{s} F_{N} (plug in formula for maximum static friction force) \\ 0 & = m g \sin θ - μ_{s} (m g \cos θ) (plug in expression for normal force on an incline) \\ 0 & = m g \sin θ - μ_{s} (m g \cos θ) (divide both sides by m g) \\ 0 & = \sin θ - μ_{s} (\cos θ) (savor the awe when you realize the angle doesn’t depend on the car’s mass) \\ \sin θ & = μ_{s} (\cos θ) (solve for sin θ) \\ \frac{\sin θ}{\cos θ} & = μ_{s} (divide both sides by cos θ) \\ \tan θ & = μ_{s} (replace \frac{\sin θ}{\cos θ} with tan θ) \\ θ & = \tan^{- 1} (μ_{s}) (take inverse tangent of both sides) \\ θ & = \tan^{- 1} (0.75) (plug in numerical values) \\ θ & = 37^{\circ} (calculate and celebrate) \end{aligned}

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Charlie Auen
Posted 8 years ago. Direct link to Charlie Auen's post “Wouldn't the driveway be ...”
Wouldn't the driveway be a case where we can't round up? At 37 degrees, the car would slip down the driveway, since the maximum was 36.86... degrees.
Button navigates to signup pageComment on Charlie Auen's post “Wouldn't the driveway be ...”
(17 votes)
Answer
- G.Gulzt
  Posted 6 years ago. Direct link to G.Gulzt's post “If I were to build my dri...”
  If I were to build my drive way based on this math I would choose to round down though ;-)
  Button navigates to signup page
  (7 votes)
Mary Mckeigan
Posted 8 years ago. Direct link to Mary Mckeigan's post “I don't understand the an...”
I don't understand the angle part. Is the angle of incline the same as the angle pictured between the mg and mgcos?
Button navigates to signup pageComment on Mary Mckeigan's post “I don't understand the an...”
(6 votes)
Answer
- Asmita Sengupta
  Posted 7 years ago. Direct link to Asmita Sengupta's post “Yes , the angle of inclin...”
  Yes , the angle of incline and the angle between mg and mgcos are the same. Lets take a simple example. Say , let the angle of incline be 30 , then draw the mg force downwards so as it forms a right angled triangle including angle of incline 30 . In that triangle the other two angles will be 90 ( the right angled part ) and 60. Now draw the mgcos force from the same point as mg. We see the mgcos is perpendicular to the surface of incline , where one angle we already found out was 60 , therefore the angle between mg and mgcos has to be 90-60 = 30 , which is equal to the angle of incline.
  draw the forces on the body on the incline , take this example and solve , your doubt will be dispelled. Thank you !
  Button navigates to signup page
  (10 votes)
Alex Pitts
Posted 7 years ago. Direct link to Alex Pitts's post “I am literally so lost on...”
I am literally so lost on the last example. Like how did sine theta get on the other side without being negative? Wouldn't you have had to subtract it from the side it was on with the cosine theta and the friction coefficient? Also why did we do tan negative one all of a sudden? I don't feel like there was a good explanation for that.
Button navigates to signup pageComment on Alex Pitts's post “I am literally so lost on...”
(1 vote)
Answer
- RacheLee
  Posted 7 years ago. Direct link to RacheLee's post “For the first question yo...”
  For the first question you asked, let's take a look.
  0=sin(theta)-coefficient (cos theta) (hopefully this makes sense to you :/ )
  -sin(theta)=-coefficient (cos theta)
  sin(theta)=coefficient (cos theta) Is this makes sense to you? It should! :)
  The next question was that why did (sin theta) / (cos theta) became tan(theta). right?
  Sine theta is equal to opposite/hypotenuse. Cosine theta is equal to adjacent/hypotenuse. If you do (sin theta) / (cos theta), the hypotenuse thing will cancel out and it is left with opposite/adjacent, which is same as tangent(theta)! Yay!
  Hopefully, this makes sense to you. It took a long time to type all these.
  Button navigates to signup page
  (10 votes)
fortunewookie
Posted 5 years ago. Direct link to fortunewookie's post “In example 1, wouldn't al...”
In example 1, wouldn't all the snow piling up in front of the sled change things?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Sajjad Bin Samad
  Posted 5 years ago. Direct link to Sajjad Bin Samad's post “It would but there is no ...”
  It would but there is no indication in the question about snow piling up . So, you don't need to think about that when solving this problem.
  Comment on Sajjad Bin Samad's post “It would but there is no ...”
  (5 votes)
Ory Alony
Posted 3 years ago. Direct link to Ory Alony's post “Why do you cancel both th...”
Why do you cancel both the masses in the numerator for 1 mass in the denominator?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Charles LaCour
  Posted 3 years ago. Direct link to Charles LaCour's post “When you have something o...”
  When you have something of the form (a*x + a*y)/(a*z) this is equivelent to ((a*x)/(a*z)) + ((a*y)/(a*z)) and you can see that the a's in the numerator and denominator to the left of the + sign cancel as do the numerator and denominator to the right of the + sign giving you (x/z) + (y/z)) which is also equivalent to (x + y)/z
  Comment on Charles LaCour's post “When you have something o...”
  (5 votes)
Srishti H
Posted 5 years ago. Direct link to Srishti H's post “Which way does the force ...”
Which way does the force of friction act on an inclined plane if the force acting on the mass is in a horizontal direction and the mass is moving?
Button navigates to signup pageComment on Srishti H's post “Which way does the force ...”
(2 votes)
Answer
- Andrew M
  Posted 5 years ago. Direct link to Andrew M's post “friction always opposes m...”
  friction always opposes motion.
  Button navigates to signup page
  (3 votes)
akiilessh
Posted 7 years ago. Direct link to akiilessh's post “can somebody explain why ...”
can somebody explain why there should be max. steepness for max. static friction!
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Andrew M
  Posted 7 years ago. Direct link to Andrew M's post “Because when you go past ...”
  Because when you go past that point, the object starts to slide, and that's not static friction anymore
  Comment on Andrew M's post “Because when you go past ...”
  (2 votes)
aahmadsabri1613
Posted 4 years ago. Direct link to aahmadsabri1613's post “why is the time taken sli...”
why is the time taken sliding a block up an incline faster than sliding down an incline
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
elan michael
Posted 6 years ago. Direct link to elan michael's post “Ok I am lost on question ...”
Ok I am lost on question 1...so we draw the force diagram...then we have the equation a= F/M ..then you start subsiting cos and sin and im lost at that point...please help
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Mark Zwald
  Posted 6 years ago. Direct link to Mark Zwald's post “You should probably watch...”
  You should probably watch the videos on vectors and splitting up a vector into component parts. If you don't understand that, you won't understand this video.
  Button navigates to signup page
  (3 votes)
Jess Loren Walsh
Posted 6 years ago. Direct link to Jess Loren Walsh's post “In Example 1, is Fk in th...”
In Example 1, is Fk in the equation, mgXsin(o)- Fk, because if you look at the diagram with respect to the box (move your head to the right, then Fk is going to the left which by convention is negative? Just wanted to double check this as when I first looked at it, Fk is upwards, so I was confused as to why it was negative in the example
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(2 votes)
Answer
- Andrew
  Posted 6 years ago. Direct link to Andrew's post “The child is sliding down...”
  The child is sliding down and to the right. the force of friction is working in the opposite direction up and to the left, however you can not forget that to solve force problems on inclines we're not working with conventional x-y coordinate plane but a relative x-y plane with the relative plane it appears to work in the x direction alone.
  Button navigates to signup page
  (1 vote)

Physics library

Course: Physics library > Unit 3

What are inclines?

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

How do we find the ⊥‍ and ∥‍ components of the force of gravity?

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

What do solved examples involving inclines look like?

Example 1: Snowy sliding sled

Example 2: Steep driveway

Want to join the conversation?

How do we find the $⊥$ ‍ and $∥$ ‍ components of the force of gravity?

What is the normal force $F_{N}$ ‍ for an object on an incline?