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For more information, please see full course syllabus of AP Physics C: Mechanics
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Lecture Comments (6)

1 answer

Last reply by: Professor Dan Fullerton
Wed Sep 2, 2015 5:11 AM

Post by Parth Shorey on September 1, 2015

In the figure below How much force would we need to exert on the free end of the cord in order to lift the plank (mass M= 300 kg) with constant velocity ? (Ignore the masses of the pulleys)?

*So basically there are 6 pulleys, but what I don't understand why is it Mg/6? Why did they use 6 and what did the 6 represent? I understand there are 6 pulleys but why was it added to determine force?

1 answer

Last reply by: Professor Dan Fullerton
Tue Sep 1, 2015 4:52 AM

Post by Parth Shorey on August 31, 2015

I just have a really long question, just making sure if your still online to reply. Considering the last comment was in 2014!

1 answer

Last reply by: Professor Dan Fullerton
Sun Dec 7, 2014 3:31 PM

Post by Dawud Muhammad on December 6, 2014

hey professor.?? will the addition prcoess work when u add mass of the pulley and and the radius(m1g+m2g=Ia)....??

Atwood Machines

  • The tension is constant in a light string passing over a massless, frictionless (ideal) pulley.
  • To analyze an Atwood Machine, first adopt a sign convention for positive and negative motion, then analyze each mass separately using Newton’s 2nd Law equations.

Atwood Machines

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 0:07
  • What is an Atwood Machine? 0:25
    • What is an Atwood Machine?
  • Properties of Atwood Machines 1:03
    • Ideal Pulleys are Frictionless and Massless
    • Tension is Constant
  • Setup for Atwood Machines 1:26
    • Setup for Atwood Machines
  • Solving Atwood Machine Problems 1:52
    • Solving Atwood Machine Problems
  • Alternate Solution 5:24
    • Analyze the System as a Whole
  • Example I: Basic Atwood Machine 7:31
  • Example II: Moving Masses 9:59
  • Example III: Masses and Pulley on a Table 13:32
  • Example IV: Mass and Pulley on a Ramp 15:47
  • Example V: Ranking Atwood Machines 19:50

Transcription: Atwood Machines

Hello, everyone, and welcome back to

I am Dan Fullerton and in this lesson we are going to talk about Atwood machines.0003

Our objectives include drawing and labeling a free body diagram, showing all forces acting on an object.0008

Understanding the tension is constant and a light string passing over a mass-less pulley.0014

Analyzing systems of 2 objects connected by a light string over a mass-less pulley.0019

Atwood machines, what is an Atwood machine?0024

Basically what it is, it is a couple of masses on by light string over a pulley there on a ramp or straight up or on the table.0028

The whole goal is to help students practice and learn about Dynamics Newton’s laws and how to apply these different concepts.0037

Here is an example.0045

We have 2 objects mass m1 and m2 connected by a light string over a mass less pulley over radius r.0046

Pretty obvious to see if you look at the M1 if that is a bigger mass than m2 is going to be accelerating down as M2 comes up.0052

We are going to analyze some of these.0061

Now, properties of that Atwood machines.0064

Ideal pulleys are frictionless and mass less, they had no inertia to the system.0067

We will get into real pulleys here pretty soon.0072

Tension is constant the light string passing over an ideal pulling.0075

The tension here is = the tension there.0079

How do we set up these problems?0086

First, adopt a sign convention for positive and negative motion.0088

In this case I see that we are going to be going down on the M 1 side so0092

I would probably draw an arrow like this and I am going to call that my positive y direction.0096

That means if M1 is moving down in that direction I'm going to call that positive.0101

If M2 is moving up on the side I am going to call that positive.0105

Let us get into these.0112

How do we solve them?0113

We are going to draw free body diagram for each mass.0115

Write Newton’s law equations for each mass and solve for the unknowns.0118

We will also talk about an alternate way to solve these after we have the basics down.0122

First, let us take a look will call this the positive y direction or call this T1, the tension 1 on the M1 side.0126

This T2 on the M2 side.0133

I am going to draw free body diagram for M1 first.0135

There is a dot from my object.0140

We have T1 pulling it up.0141

Down is our positive y direction because we are here and we called down the positive y on this side.0145

And we have M1 G.0151

Doing the same thing for M2 to draw free body diagram.0156

We have our dot for M2.0159

We have M2g pulling down.0162

We have T2 up.0166

But now on this side up is the positive y direction.0167

We can write our Newton’s second law of equations.0172

For M1, let us write that down here that force in the y direction is = M1 G - T1 which is all equal to M1 a.0176

We can do the same thing for out second mass.0192

Net force in the y direction is = T2 - M2 G which is = M2 a.0195

We can try to put these together to eliminate some variables to see if we can solve for the acceleration of the system.0210

The way I am going to do that there are bunch of different procedure to do so.0216

In this problem, let us start by taking this equation here.0220

You are going to write it up here M1 G - T1 = M1 a.0223

Underneath that I'm going to write the corresponding equation for M2.0230

T2 - M2 G = M2 a.0234

I'm just going to add those together I am going to add the left hand sides in the right hand side.0239

They are equalities as some of the left hand side has to equal some other right hand side.0243

On the left hand side, I'm going to come up with M1 G - T1 + T2 - M2 G is all equal to.0249

On the right hand side, we have a × M1 + M2.0260

If I pull the a out of those is I add them together.0265

Now, because tension 1 = tension 2, tension on each side and we have an ideal pulley - T1 + T20 that is going to be 0.0268

I now have M1 G - M2 G = a M1 + M2.0280

Let us factor in G out the left hand side.0290

GM1 - M2 = a M1 + m2 and I want the acceleration of the system A therefore = G × M1 - M2 / M1 + m2.0292

I know the acceleration of the system by using free body diagrams applying Newton’s second law0312

and also the fact that we know that the tensions are equal in an ideal pulley.0318

There is an alternate way we could analyze this too.0324

If we look at the whole thing is a system calling this the positive y direction let us define our system here as we got an extended system.0327

I'm just going to draw it inside the dashes there.0339

We are just going to look at all of the forces that crossed the boundary of our system that are not internal forces.0342

I'm going to redraw the system up here just to make it even more apparent what is going on.0350

If I were to take those and spread them out this way we are going to have is something that looks kind of like this.0355

We will have M1 on the left it is connected by a string to M2.0362

There is our system we are calling this way the positive y direction and if I look at the forces0372

that are crossing that boundary of our system on the M1 side we have M1 G.0383

On the m2 side we have m2g.0390

If I were to look at this in applying Newton’s second law to something like this or call the net force in the y direction0396

I'm going to have M1 G and a positive y direction - M2 G.0405

The net has to equal the mass of our system M1 + M2 × acceleration a or just rearranging for a, A = G × M1 - M2 / M1 + m2.0414

Same answer just taking it and looking at it from a systems approach.0435

You can use a system like that as well as when you go to solve these problems.0441

It is fun to do it both ways and check and make sure you get same answer.0446

Let us take an example of the basic Atwood machine.0452

Find the acceleration of the 20 kg mass given the masses are connected by light string over an ideal mass less pulley.0455

It looks to me like the 20 kg mass is the 1 that is going to win here.0464

Let us call this our positive y direction and we will do this when the old fashioned way.0468

Let us first take a look we will call this M1 and M2.0475

We will draw our free body diagram for M1.0479

I got my object I have T1 and T2.0483

We have T1 or T because they are the same on both sides and M1 G down that would be a positive y direction.0490

For our second mass, M2 down will be our positive y direction and we have tension upwards and M2 G down.0500

Now, drawing or writing Newton’s second law equations for M1 we can see they were going to have T - M1 G = M1 a.0515

For our second mass, we are going to have M2 G - T = M2a.0529

We will combine these together T - T that is going to be 0.0538

We will end up with on the left hand side M2 G - M1 G = M1 a + M2 a or G × M2 - M1 = A × M1 + M2.0543

Or a =G M2 - M 1 / M1 + M2.0563

And substituting our values is going to be =10 m / s² M2 20 kg – M1 15 kg a/ 20 + 15 or 35 kg.0572

Gives us an acceleration of about 1.43 m / s².0585

Let us take a look at another one.0595

Masses are hung on a light string attached to an ideal mass less pulley has shown a diagram here below left.0600

The total mass hanging from the left string is = that on the right to their equilibrium here, 1 kg 1 kg.0606

At time T = 0 the 0.2 kg mass is moved to the right side.0613

This one gets bigger that one gets a little bit smaller.0620

How far does each mass move in 1 second?0623

I think we need a little bit more room to solve this one.0626

Let us move to the next slide we got a little bit more room and we will do our analysis.0629

Let us call this a M 1, we will call this T1 and that will be a positive y direction.0636

Here we will have M2 and T2.0648

Looking at our first mass M 1, we have T1 up and M1 G down.0655

For M2 we have T2 up and M2 G down.0666

If I put these together with Newton’s second law equations make sure we define our positive and negative directions that is positive Y and that is positive y.0678

We have T1 - M1 G = M1a and we have M2 G - T2 = m2a.0691

T1 and – T2 can get to 0 because the tension is equal on both sides their T 1 = T2.0706

We come up with M2 G - M1 G = M1 + M20713

or A = G M 2 - M1 / M2 + M1 which is again 10 × the difference in the masses which is going to be a 0.4 kg ÷0725

the sum of the masses 2 kg or 2 m / s².0741

Hopefully, you are starting an idea how these repeat in that general pattern.0750

That has just how far does it move?0754

Now, that is a kinematics problem are going to look in the y direction and the vertical direction0756

We call down the positive y direction the initial equal 0 we do not know the final we do not know Δ y.0762

But we know the acceleration in the y is 2 m / s² I want to know how far it goes in 1second.0770

Using kinematics Δ y = the initial T + 1/2 to 80²0777

or that is going to be since the initial is 0 that term goes away.0786

Δ y = ½ × are acceleration 2 m / s² × our time 1/2⁺or Δ y = 1 meter.0791

We will we can not adjust our Atwood machines a little bit to make a touch more interesting.0810

Here we have 2 masses and M1 and M2 connected by a light string / masses pulley.0815

Assuming a frictional surface find the acceleration of M2.0820

We are going to define that is our positive y direction tensions are the same T and T and let draw our free body diagrams.0824

For M1, we have normal forced up, we have M1 G down and we have T to the right which for M1 is going to be what we call the + 1 direction.0839

For M2, we have tension up, we have M2 G down, we are calling down the positive y direction over here.0860

Finding a 2nd Newton’s law equation in the direction of motion for N1, the net force.0873

What we are calling the line that is really horizontal for M1 is going to be = just T which must be = M1a.0879

Doing the same thing for mass 2 net force = M2 G - T = M2a.0891

Let us see if we can put this together we have T = M1a and we are going to combine that with M2G - T = M2a and what do we get?0904

M2 G = M1 + M2 × a or a = G × the quantity M2 / M1 + M2.0923

Just slight variation on the same theme.0941

We can even do this with their ramp.0945

2 masses M1 and M2 are connected by light string over masses pulley.0948

A sum of frictional surfaces finding acceleration of M2.0952

Let us start with our free body diagrams again will do M1 cassette looks like it is a more complicated0957

We are pretty good doing free body diagrams for objects on ramps.0963

Let us do our x over here there is our y.0968

So x y and their object M1 we have attention of pulling it up the ramp we have normal force and we have M1 G.0978

M1 G does not lineup with the an x again you know the drill let us turn it into a pseudo free body diagram.0994

It will be a little bit more useful to us for problem solving.1000

There is our x, there is our y, and let us define that direction as + y.1005

We are not certain it is actually going to go that way.1021

It could be the other direction but let us define it that way and if we come up with the negative acceleration we know we chose.1023

Tension up the ramp, normal force perpendicular to it and M1 G we got to breakup the components1033

and that is our angle θ that means that is also angle θ.1041

This would be the adjacent side M1 G cos θ.1046

This would be the opposite side M1 G sin θ.1053

I am going to write our Newton’s second law of equation here calling up the x axis for calling + y T - M1 G sin 30° = M1 a.1061

Or solving for T, t = M1 G sin 30° + M1 a.1079

Let us do our free body diagram from as 2 here.1091

For M2 down as our positive y direction we have tension up and M2 G down.1094

M2 G - T must equal M2 a in the Newton’s second law M2 G - T is = M1 G sin 30° + M1 a.1105

All of that must be = M2 a or a little bit of math here.1125

M2G - M1 G sin 30° - M1 a = M2 a.1134

Factoring G × M2 - M1 sin 30° must equal a × M1 + M2.1149

Or finally solving for acceleration that is going to be G × M2 - M1 sin 30° ÷ M1 + M2.1162

We can keep getting trickier here.1184

Let us finish off with one last question our ranking test.1186

Rate from least the greatest the acceleration in net force of these 6 different Atwood machines.1192

The first thing I think I'm going to do here with all of that data is I'm just going to solve a generally without plugging in the values.1199

What I'm going to do is I'm going to do this one from a systems approach.1207

Let us draw our systems.1211

We will call this on the left M1 that one on the right is M2 and we will define our system M1 is that way.1215

We will call them + y direction.1225

We will define our system like we did with our ultimate solution earlier just to change things up as looking like that.1227

When I do that the net force in the y direction are going to have m1 G and the positive y direction - M2 G1237

and all of that has to be equaled Ma where our total mass is M1 + M2 A.1251

Or acceleration is going to be net force in the y ÷ M1 + m2.1259

The net force is just M1 G - m2 G or net force will be = G × M1 - M2.1273

Now, all of this different data seems to be the easiest way to handle it might be to make a table.1285

Let us do that.1289

We will have a row for ABCDE and F.1291

Our data will have a value from s1, a value from S2.1298

Once we have those we can calculate the net force by multiplying G × the difference of those.1303

Let us make a column for net force and then we will know the acceleration.1308

Let us make a column for acceleration which is the net force ÷ the sum of m1 and m2.1313

We can start filling in our data.1319

For situation A, M1 is 5, M2 is 1.1322

The net force is going to be M1 - M2 4 × G10 or 40.1328

The acceleration was going to be the net force ÷ the total mass 40 / 6 which is 6.67.1336

I can just keep going to fill it in there is my table in that manner.1344

For B, we have M1 is 3, M2 is 1, the net force is going to be the difference of those × 10 or 20 in the acceleration 20 ÷ 4 or 5 m / s².1348

For C, mass 1 is 4, mass 2 is 2.1363

Again, we will have a net force of 20 that our acceleration is going to be 20 ÷ the total mass 6 or in this case 3.33.1366

For D, we have initial mass m 1 is 1, M2 is 4, so the magnitude of the net force 4 -1 is 3 × 10 is going to invest 30 N.1378

30 ÷ the total mass 5 will be 6 m / s².1390

For part E, 1 and 2 for masses.1396

Our net force is just going to 10 and accelerations 10 ÷ total mass of 3 is just 3.1400

For F, we have 8 as our first mass, 2 as our second mass, the difference is 6 × 10 =60 N for force.1410

Our acceleration is 60 ÷ 10 or 6 m / s².1422

We are asked to write these from least to greatest for acceleration and for net force.1428

I just use my table to rank them from least to greatest for acceleration.1437

It looks like we have C and E together those are the smallest.1440

Then we go up to B than it looks like D and F for both 6 those are together.1446

And finally, our highest acceleration A.1455

Same thing for the net force that looks like our starting point is E our lowest then B and C are both 20 then we come to D 30 then we come to A 40.1460

Our highest net force F with 60 N.1476

Hopefully, that gets you a good start with Atwood machines.1482

Get you feel comfortable with them and lots more practice with Newton’s second law and how you apply that with free body diagrams and pseudo free body diagrams.1485

Thank you so much for watching

We will see you again soon and make it a great day everyone.1496