For more information, please see full course syllabus of AP Physics C: Mechanics
For more information, please see full course syllabus of AP Physics C: Mechanics
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Ramps & Inclines
- Draw a free body diagram for the object on the incline, utilizing the direction of the object’s motion (or the ramp) as one of the positive axes.
- For any forces not lining up with an axis, break that force up into components parallel to an axis and draw a pseudo free-body diagram.
- Utilize Newton’s 2nd Law equations along each axis direction to solve for unknown quantities.
Ramps & Inclines
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
- Intro
- Objectives
- Drawing Free Body Diagrams for Ramps
- Step 1: Choose the Object & Draw It as a Dot or Box
- Step 2: Draw and Label all the External Forces
- Step 3: Sketch a Coordinate System
- Example: Object on a Ramp
- Pseudo-Free Body Diagrams
- Box on a Ramp
- Example I: Box at Rest
- Example II: Box Held By Force
- Example III: Truck on a Hill
- Example IV: Force Up a Ramp
- Example V: Acceleration Down a Ramp
- Example VI: Able of Repose
- Example VII: Sledding
- Intro 0:00
- Objectives 0:06
- Drawing Free Body Diagrams for Ramps 0:32
- Step 1: Choose the Object & Draw It as a Dot or Box
- Step 2: Draw and Label all the External Forces
- Step 3: Sketch a Coordinate System
- Example: Object on a Ramp
- Pseudo-Free Body Diagrams 2:06
- Pseudo-Free Body Diagrams
- Redraw Diagram with All Forces Parallel to Axes
- Box on a Ramp 4:08
- Free Body Diagram for Box on a Ramp
- Pseudo-Free Body Diagram for Box on a Ramp
- Example I: Box at Rest 6:13
- Example II: Box Held By Force 6:35
- Example III: Truck on a Hill 8:46
- Example IV: Force Up a Ramp 9:29
- Example V: Acceleration Down a Ramp 12:01
- Example VI: Able of Repose 13:59
- Example VII: Sledding 17:03
AP Physics C: Mechanics Online Course
I. Introduction | ||
---|---|---|
What is Physics? | 7:12 | |
Math Review | 1:00:51 | |
II. Kinematics | ||
Describing Motion I | 23:47 | |
Describing Motion II | 36:47 | |
Projectile Motion | 30:34 | |
Circular & Relative Motion | 30:24 | |
III. Dynamics | ||
Newton's First Law & Free Body Diagrams | 23:57 | |
Newton's Second & Third Laws of Motion | 23:57 | |
Friction | 20:41 | |
Retarding & Drag Forces | 32:10 | |
Ramps & Inclines | 20:31 | |
Atwood Machines | 24:58 | |
IV. Work, Energy, & Power | ||
Work | 37:34 | |
Energy & Conservative Forces | 28:04 | |
Conservation of Energy | 54:56 | |
Power | 16:44 | |
V. Momentum | ||
Momentum & Impulse | 13:09 | |
Conservation of Linear Momentum | 46:30 | |
Center of Mass | 28:26 | |
VI. Uniform Circular Motion | ||
Uniform Circular Motion | 21:36 | |
VII. Rotational Motion | ||
Rotational Kinematics | 32:52 | |
Moment of Inertia | 24:00 | |
Torque | 26:09 | |
Rotational Dynamics | 56:58 | |
Angular Momentum | 33:02 | |
VIII. Oscillations | ||
Oscillations | 1:01:12 | |
IX. Gravity & Orbits | ||
Gravity & Orbits | 34:59 | |
X. Sample AP Exam | ||
1998 AP Practice Exam: Multiple Choice | 28:11 | |
1998 AP Practice Exam: Free Response Questions (FRQ) | 28:11 |
Transcription: Ramps & Inclines
Hello, everyone, and welcome back to www.educator.com.0000
In this lesson we are going to talk about ramps and inclines.0003
Our objectives include drawing and labeling free body diagram, showing all forces acting on an object on ramp.0007
Drawing a pseudo free body diagram showing all components of forces acting on the object.0014
Utilizing Newton’s laws of motion to solve problems involving objects on ramps.0019
We have done a few of these already but I think it is worthwhile to really take some time and make sure we get these down because they come up so commonly.0024
Drawing a free body diagrams for ramps.0033
Choose the object of interest and draw these on a dot or a box.0036
Draw and label all the external forces acting on the object and then scatter coordinate system choosing the direction of the objects motion as one of the axis.0039
That means your X axis might be tilted in some direction and that is ok.0048
For the case of an object on the ramp, the direction of the objects motion should most likely be up or down that ramp.0053
If we have something like a ramp with a box on it, that is the same with the hammered with boxes on ramps for some reason.0060
Some angle θ, there is a box what I would do to draw my free body diagram is draw my axis at the same angle as the incline of the ramp.0073
I would call that my x axis then I will do my best to draw a line perpendicular to it for my y.0087
There is x and y.0098
We will draw a dot for our object.0099
We will have our normal force perpendicular to the plane of the ramp.0104
We will have our frictional force.0109
In this case, assuming it is lying down the ramp will oppose that.0111
We will call that our frictional force.0114
There is our normal force straight down.0116
That would be our free body diagram.0122
When the forces do not line up with the axis, we can draw that pseudo free body diagram when we break some of those forces up and their components.0126
We will redraw our diagram of all forces parallel to the axis.0134
If we had something like let us draw it again.0138
Our free body diagram for axis x and y.0144
I'm just going to show how the angle of our plane is the same as the angle of our ramp there.0153
We had a frictional force, we had a normal force, we had weight down.0160
If that angle θ and this angle also with θ are going to break MG up in the components that are parallel to the y axis and parallel to the x axis.0169
That is the way to do that is with pseudo free body diagram on a separate diagram.0183
If we want all of our points in the AP test.0189
We will draw our axis again and XY.0194
We have our dot, we still have our frictional force, our normal force.0199
We put in our components of weight.0208
We have the component that is adjacent to our angle which I call MG perpendicular because it is perpendicular to the plane on the ramp is MG Cos θ.0213
MG parallel the component of the weight that is parallel with the ramp was MG sin θ.0226
Θ is the angle of the ramp itself or it is also that angle right there.0238
Alright, so boxes on ramps.0247
There are other forces the free body diagram in the pseudo free body diagram for box sitting on a ramp of angle θ.0250
Then write Newton’s second law of equation for the X and Y directions.0256
As we take a look here, let us start by drawing the forces on our box itself.0261
Normal force will have a frictional force and MG.0268
Cannot ever get too much practice doing this.0275
We will draw our free body diagram over here.0282
We would have normal force N, frictional force opposing motion and MG.0286
Separate diagram can emphasize that enough our pseudo free body diagram.0295
xy, we got our frictional force.0309
Our normal force.0316
Breaking up weight into a component parallel MG sin θ and perpendicular MG cos θ.0318
If I were to write Newton’s second law of equations there in the x direction net force in the x direction.0330
We are going to call down positive here we can write that as MG sin θ - frictional force = Ma X in the y direction0338
net force in the y = N - MG cos θ is equal the Ma Y but the boxes sitting on the ramp0353
it is not going to accelerate off the ramp nor is it going to fall through the ramp anytime soon.0362
That is all equal to 0 and that would work.0367
Let us do an example here.0372
3 Forces act on the box and inclined plane as shown in the diagram below, weight, friction, and normal force.0374
If the box is at rest the net force acting on it is equal to, if it is at rest the net force must be 0 equilibrium condition.0381
Our answer there must be 0 because the box is at rest.0390
Let us do another one.0395
A 5kg mass is held at rest on the frictionless 30° incline by force F.0397
What is the magnitude of force F?0403
Let us start with their free body diagram.0407
Put it up over here.0410
There is X perpendicular to it.0414
We will draw our Y so XY.0416
We have our normal force, we have our applied force F and MG straight down.0425
I'm going to draw our pseudo free body diagram.0436
We put that over here on the left where we got some room.0439
Our pseudo free body diagram, there is our x and y.0443
We still have our normal force along the y, we have our applied force, but MG we break up in the components.0451
We got the component parallel to the ramp MG sin θ and perpendicular to the ramp into the ramp MG cos θ.0459
Let us call this our positive X and another Y direction when we write our Newton’s second law equation.0477
Let us start in the x, net force in the x direction.0483
It is going to be F - MG sin θ which is going to be equal to MA x0488
Which, since acceleration is 0 because it is at rest it is all equal to 0.0497
Therefore, F = MG sin θ which is going to be 5 kg × G 10 m / s² × the sin of 30° which is ½.0502
The half of 50 is going to be 25 N.0517
Take an example where we look at a truck on a hill.0526
The diagram shows 100,000 N truck at rest on the hill that makes an angle of 8° with a horizontal.0529
What is the component of the trucks weight parallel to the hill?0536
MG parallel we just remember that equation.0540
The component of the weight down the hill is going to be MG sin θ.0544
It gives us the weight MG already which is 1 × 10⁵ × sin 8° which is right about 1.4 × 10⁴ N.0549
Forces up the ramp.0571
The block weighing 10 N is on a ramp inclined to 30° to the horizontal.0572
A 3 N force of friction ff acts on the block as it is pulled up the ramp at constant velocity.0577
It is always important to know constant velocity.0583
The net force must be 0, acceleration is 0 as some force F parallel to the ramp.0586
Find the magnitude of force F.0592
Let us start with their free body diagram.0595
I will draw over here on the left to begin with.0598
The x and y and our object.0609
A dot we have the normal force.0613
We got some force F up the ramp, we have our frictional force.0617
We have the weight of the box which it gives us is 10 N always down.0623
We can draw our pseudo free body diagram.0629
We will do that over here on the right YX.0632
F still acting up the ramp and perpendicular to it for some friction down the plane on the ramp.0647
We are breaking up MG into its components.0655
We have the one perpendicular to the top of the ramp it is going to be 10 N cos 30° is 8.66.0658
We will have 10 sin 30 the plane of the ramp parallel which is going to be 5 N.0668
What is the magnitude of force F?0676
Let us use Newton’s second law in the x direction.0679
F net x = f - the force of friction -10 sin 30 as the equal 0 because it is not accelerating.0682
It is moving at a constant velocity.0695
Therefore, F equals force of friction + 10 sin 30 which is 5,0698
Which is our force of friction that says is 3 N.0705
That is 3 N + 5 N therefore F must be equal to 8 N.0707
Let us take a look at some acceleration down a ramp.0720
A 100 kg block slides down to frictionless 30° incline as shown, find the acceleration of the block.0723
Just like we have been doing, let us draw our axis for free body diagram.0731
Here is our X, there is our Y, there is our box, it is on the frictionless surface.0739
We have normal force and we have its weight down.0750
Friction and applied forces.0759
Let us do the pseudo free body diagram.0761
It looks like we have room beside it to do so.0763
We will draw our pseudo free body diagram over here.0766
We still have our normal force up the ramp, we are going to break MG in the components.0773
We have MG sin θ parallel with the ramp and MG perpendicular which is MG cos θ perpendicular or in the plane ramp.0779
There is our y, there is our x.0793
If we want the acceleration of the block, it is going to accelerate in the x direction I am going to write that Newton’s equation first.0797
F and x equals all the only thing we have there is MG sin θ which must equal M × acceleration in the x direction.0804
Therefore, pretty straightforward acceleration in the x direction is just G sin θ is going to be 10 m / s² sin 30°.0816
Sin 30 is half so that is just 5 m / s².0827
Let us talk about what happens when we place the block on a ramp with an unknown coefficient of friction.0839
The angle of the ramp is slowly increased until the block just begins to slide.0845
To find the coefficient of static friction is a function of the ramps angle of elevation0851
because the angle of the object just begins to slide is known as the angle of repose.0856
Keep lifting it upward to starts to slide, you measure that angle and you can find up the coefficient of friction.0862
Let us see how we can do all that.0868
Here is a free body diagram of our block on our ramp.0871
We have the force of static friction holding it in place.0875
The normal force out of the plane in the ramp and its weight down and we are going to lift it up0879
until that angle is right at the point where the box just begins to move.0884
Right there and then we can draw our pseudo free body diagram just breaking up MG0889
into components parallel and perpendicular to the ramp as shown.0894
If we want to find the coefficient of friction we will start by writing Newton’s second law in the x direction.0899
Net force in the x direction is going to be all we have that force of static friction - MG sin θ and all of that must be equal to MA x.0906
Right at the point where it is just barely beginning to move acceleration is 00923
So therefore, the force of static friction = MG sin θ.0930
Let us do the same basic analysis in the y direction.0939
The net force in the y direction is going to be our normal force - MG cos θ all of the equals Ma Y.0944
But ay= 0 the box is going to spontaneously fly up off the ramp or go through it.0960
Therefore, we can write that the normal force is equal to MG cos θ.0966
Putting all of this together then remember friction is fun.0974
Force of friction equals μ × the normal force that means that μ are coefficient of friction must be the frictional force over the normal force0977
which is MG sin θ / MG Cos θ.0987
M /MG /G sin θ /cos θ is tan θ.0996
If you want the coefficient of friction lifted up until the object just barely begins to slide measure that angle1005
and the coefficient of friction is the tan of the angle.1011
How slick is that?1017
Let us take a look at one more simple problem.1021
Jane rides a sled down the slope of angle θ at constant speed V.1026
Constant speed right away a= 0.1030
Net force =0.1035
The term in the coefficient of kinetic friction between the sled and the slope neglect air resistance.1038
Let us draw our free body diagram.1045
There it is we will call this + Y in that + X direction and there is our object Jane and sled.1058
We have the normal force and the plane of the hill.1067
We have MG force of gravity down and some force of friction opposing motion.1071
From then we can draw our pseudo free body diagram.1083
Let us put that over here on the right.1088
We have our positive y, positive x, we have our normal force.1098
We still have our frictional force and MG we are going to break it in the components.1107
We have MG sin θ parallel to the ramp and MG Cos θ perpendicular to it.1115
We will go to Newton’s second law.1127
Let us look in the x direction first F net x is equal MG sin θ - Force of friction and1129
all of that is equal to Ma X which is equal to 0 because it is a constant speed.1147
Therefore, force of friction equals MG sin θ.1154
If we looked in the y direction net force in the y direction equals N - MG Cos θ equals 0.1161
Therefore, N equals MG Cos θ.1172
The coefficient of friction that the frictional force over the normal force or MG sin θ over MG Cos θ which is just tan θ.1177
Almost the same thing you just get did, why is that?1191
Before when it is just barely beginning to move on that previous problem define the coefficient of friction1197
you have a net force of 0 that was right at the point where everything still balanced.1203
Same idea here, you are a constant speed net forces 0 it is the same analysis just from a slightly different perspective.1208
This is really an angle of repose question because it is moving constant speed.1215
Alright, hopefully, that gets you a little more confident and feeling good about these ramps and incline problems.1221
Thank you so much for watching www.educator.com.1226
We will see you soon and make it a great day everybody.1229
1 answer
Sun Aug 14, 2016 12:53 PM
Post by Cathy Zhao on August 14, 2016
On Example 5, why the acceleration of the block is in the x direction not y direction?
2 answers
Sun Aug 14, 2016 12:45 PM
Post by Cathy Zhao on August 14, 2016
At 4:54, why Fnet y=0? Is Fnet x also=O?
0 answers
Post by Professor Dan Fullerton on March 3, 2016
Assuming you're looking at roughly 5:45, you could write it that way as well. I defined down the ramp as my positive x-direction, so I have mgsin(theta)-F=0. You could just as easily write -mgsintheta+F=0. Regardless, you'll come up with F=mgsin(theta).
1 answer
Thu Mar 3, 2016 5:48 AM
Post by Joy Ojukwu on March 2, 2016
why is it not -mgsintheta + F on x axis