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Lecture Comments (8)

1 answer

Last reply by: Professor Dan Fullerton
Sun Aug 14, 2016 12:53 PM

Post by Cathy Zhao on August 14 at 12:51:41 PM

On Example 5, why the acceleration of the block is in the x direction not y direction?

2 answers

Last reply by: Professor Dan Fullerton
Sun Aug 14, 2016 12:45 PM

Post by Cathy Zhao on August 14 at 12:38:22 PM

At 4:54, why Fnet y=0? Is Fnet x also=O?

0 answers

Post by Professor Dan Fullerton on March 3 at 05:48:24 AM

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

Last reply by: Professor Dan Fullerton
Thu Mar 3, 2016 5:48 AM

Post by Joy Ojukwu on March 2 at 08:42:49 PM

why is it not -mgsintheta + F on x axis

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 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

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 direction 0338

net force in the y = N - MG cos θ is equal the Ma Y but the boxes sitting on the ramp 0353

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 x 0488

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 elevation 0851

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 up 0879

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 MG 0889

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 0 0923

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 force 0977

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 angle 1005

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 and 1129

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 friction 1197

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