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Lecture Comments (49)

0 answers

Post by Elman Ahmed on July 7 at 07:32:09 PM

Thanks for the sample no. 9 and 10. I got something very similar regarding pulley in the final exam. It was a great review! Glad that i listened to it!

1 answer

Last reply by: Professor Dan Fullerton
Mon Nov 23, 2015 7:35 AM

Post by Jimmy Jones on November 3, 2015

Hey Professor Fullerton,

I got confused on example 3, box held by force.

Wouldn't you do 50cos30 rather than 50sin30, because the question asks for the magnitude of friction which is usually horizontal, not vertical?


2 answers

Last reply by: Jim Tang
Fri Jul 24, 2015 7:33 PM

Post by Jim Tang on July 24, 2015

Hi Dan!

I feel like I know how to do it, but I don't know what I'm doing. Can you elaborate a little further about how massless (ideal) pulley equates to equal tension everywhere in the string and constant acceleration. I can't seem to intuitively understand this. Thanks!

3 answers

Last reply by: Professor Dan Fullerton
Mon Jun 15, 2015 6:03 PM

Post by David Saver on May 8, 2015

in example 8 why is m1 formula T - m1g
and formula for net force for m2 is m2g -T
Why are they the opposite?

1 answer

Last reply by: Professor Dan Fullerton
Sat Nov 29, 2014 7:18 AM

Post by MOGIN Daniloff on November 28, 2014

Hi, I don't understand why in the Atwood machine system the Ts are equal. If this is so then why are the masses different? Is it possible to have different masses yet at the same time have them both pull on one another with equal force?
Thank you.

1 answer

Last reply by: Professor Dan Fullerton
Thu Oct 16, 2014 1:41 PM

Post by luis laosfarfan on October 16, 2014

hello professor, why is it that the downward direction you call it positive would it be negative. and how come when we were dealing with kinematics in 1 2 & 3 dimensions we were using for the x component of the net vector cos and for the y component of the vector sin and now with forces we happen to change it now we use cos for the why components and sin for the x components you in the pseudo diagram  

1 answer

Last reply by: Professor Dan Fullerton
Fri Oct 10, 2014 5:04 AM

Post by Jinwei Wang on October 9, 2014

Hi Professor, I have a problem that I have no ideas how to solve it...
Can you explain it specifically, so I can understand the processes?
A champion archer hits a bullseye in a target mounted on a wall a distance L away and situated at a height h above his bow.  Deduce the equation between the speed (at which the arrow left his bow), the arrow’s initial angle θ with the horizontal, the height, and the distance of the target (whose solution the archer evidently knew.  Neglect air resistance).

1 answer

Last reply by: Professor Dan Fullerton
Thu Jul 24, 2014 5:27 AM

Post by Him Tam on July 23, 2014

In example 5, how does the component parallel to the hill involve sine? Shouldn't it be cosine so it's parallel to the ground?

1 answer

Last reply by: Professor Dan Fullerton
Thu Jul 24, 2014 5:27 AM

Post by Him Tam on July 23, 2014

For the atwood machines, how do you know that m1g-T1 = m1a or that T2 - m2g = m2a?

1 answer

Last reply by: Professor Dan Fullerton
Thu Jul 10, 2014 8:40 AM

Post by Jamal Tischler on July 10, 2014

At example 10. If we had friction, we didnt know what direction it had because we didnt know the masses. How we solve this ?

2 answers

Last reply by: Tom Glow
Mon Jun 30, 2014 10:11 AM

Post by Tom Glow on June 29, 2014

Hey Professor Dan, I have a question about the Atwood Machine.

Say that one of the masses is on the ground and the other is in the air, would it be possible to calculate how much force the mass on the ground exerts as the mass that is in the air changes?  Or would that require complex mathematics?

I would assume that the tension would play into the force exerted by the object on the ground, where the greater the tension the less force applied.

2 answers

Last reply by: Thivikka Sachithananthan
Mon Jun 2, 2014 10:47 AM

Post by Thivikka Sachithananthan on June 2, 2014

for example 4, why does g is not equal to -10? we denoted plus to be in the upward direction. so why is g not negative? thanks.  

1 answer

Last reply by: Professor Dan Fullerton
Fri Jan 10, 2014 7:02 AM

Post by Hyun Cho on January 9, 2014

hey could you help me with ex4? when you said normal force=mg+may, why isnt the acceleration in the y direction -7m/s2? i thought that thr gravity acceleration is -10 and the elevator is accelerating 3 so overall acceleration in the y is -7

1 answer

Last reply by: Professor Dan Fullerton
Wed Sep 18, 2013 9:25 AM

Post by Gaurav Kumar on September 18, 2013

I understand all the math, but I have one conceptual question. In atwood machines, how can we assume the tensions are equal when different masses hang from them?

3 answers

Last reply by: Professor Dan Fullerton
Tue Jul 23, 2013 2:13 PM

Post by Gaurav Kumar on July 23, 2013

In example 10, why are we assuming that the acceleration of m1 is equal to the acceleration of m2?

1 answer

Last reply by: Professor Dan Fullerton
Thu Jun 6, 2013 9:10 AM

Post by Jay Gill on June 5, 2013


1 answer

Last reply by: Professor Dan Fullerton
Tue May 14, 2013 5:35 PM

Post by Jamie Ward on May 14, 2013

These lectures are great! I was wondering where I could find out more about the "math trick" you mentioned where you add the two equations together as in the Atwood machine example. What are the conditions that allow for these equations to be added together like this? Thanks!

2 answers

Last reply by: Nawaphan Jedjomnongkit
Fri May 10, 2013 2:17 PM

Post by Nawaphan Jedjomnongkit on May 10, 2013

From Ex9 why a = g(m2/m1+m2) not a = m2g/ (m1+m2) ??? Thank you

1 answer

Last reply by: Professor Dan Fullerton
Sat Apr 27, 2013 5:46 PM

Post by Edward Xavier on April 27, 2013

concepts were clearly explained with great examples :D

3 answers

Last reply by: Professor Dan Fullerton
Sat May 4, 2013 5:07 PM

Post by Nikki CONSTANT on April 10, 2013

Your lectures are AWESOME!!! Very organized and you hit all the main concepts. Thank you!!!

Dynamics Applications

  • FBDs are tools for visualizing forces on a single object and writing equations to represent a physical situation.
  • The x- and y-axes may be set in such a manner that the object's motion lines up with one of the axes, and is perpendicular to the second axis.
  • Pseudo-FBDs show all vectors or components of vectors parallel to one of the designated axes.
  • The tension is constant in a light string passing over a massless pulley.
  • A scale doesn't read an object's weight, it reads the normal force it exerts back on an object.

Dynamics Applications

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 0:08
  • Free Body Diagrams 0:49
  • Drawing FBDs 1:09
    • Draw Object of Interest as a Dot
    • Sketch a Coordinate System
  • Example 1: FBD of Block on Ramp 1:39
  • Pseudo-FBDs 1:59
    • Draw Object of Interest as a Dot
    • Break Up the Forces
  • Box on a Ramp 2:12
  • Example 2: Box at Rest 4:28
  • Example 3: Box Held by Force 5:00
  • What is an Atwood Machine? 6:46
    • Two Objects are Connected by a Light String Over a Mass-less Pulley
  • Properties of Atwood Machines 7:13
    • Ideal Pulleys are Frictionless and Mass-less
    • Tension is Constant in a Light String Passing Over an Ideal Pulley
  • Solving Atwood Machine Problems 8:02
  • Alternate Solution 12:07
    • Analyze the System as a Whole
  • Elevators 14:24
    • Scales Read the Force They Exert on an Object Placed Upon Them
    • Can be Used to Analyze Using Newton's 2nd Law and Free body Diagrams
  • Example 4: Elevator Accelerates Upward 15:36
  • Example 5: Truck on a Hill 18:30
  • Example 6: Force Up a Ramp 19:28
  • Example 7: Acceleration Down a Ramp 21:56
  • Example 8: Basic Atwood Machine 24:05
  • Example 9: Masses and Pulley on a Table 26:47
  • Example 10: Mass and Pulley on a Ramp 29:15
  • Example 11: Elevator Accelerating Downward 33:00

Transcription: Dynamics Applications

Hi everyone and welcome back to

We are going to take a look at dynamics applications and problem solving in this lesson.0003

Our objectives are going to be to draw and label a free-body diagram (FBD), showing all the forces acting on an object on a ramp.0008

We will also draw a pseudo free-body diagram (P-FBD)showing all components of forces acting on the object -- some overlap with what we have done previously in reinforcement.0016

We will utilize Newton's Laws of Motion to solve problems of objects on ramps.0025

Gain an understanding that tension is constant in a light string passing over a massless, or ideal pulley. 0030

We will analyze systems of two objects connected by a light string over a massless pulley, and finally, we will determine the reading on a scale in an accelerating elevator.0036

So, with that, let us go back to FBDs again -- a quick review. 0047

FBDs are tools used to analyze physical situations and they show all the forces acting on a single object. 0052

Then, we draw all the forces on that object and we draw the object as either a box or as a dot. 0060

When we are drawing FBDs -- what we are going to do is we are going to choose the object of interest and draw it. 0069

Then label all the external forces and draw them. 0075

And then sketch the coordinate system choosing the direction of the objects motion as one of the positive axis.0078

When we do this for the case of an object on a ramp, that is going to be up or down the ramp, which means typically we are going to have an off-set or a tilted set of axis. 0084

Quick review -- we have a block sitting on a ramp -- What do we do about the forces acting on it?0100

We already said we have the normal force, we have the weight, and the force of friction and we draw them just like they are on the ramp so the answer here would be 4. 0105

Once we have that down, we are going to complicate matters a little bit. 0116

With the P-FBDs -- when the forces do not line up with the axis, we draw a new separate FBD and break up those forces into their components that do line up on the axis. 0120

So here is our box on a ramp. Let us draw the forces -- the FBD, and the P-FBD -- for it sitting on the ramp. 0132

Then we are going to write Newton's Second Law equations for the x and the y directions. 0141

So for this box we have its weight down, normal force, and the force of friction, since it wants to slide down the ramp. 0146

Our FBD -- we will draw our axis -- We have mg down. We have the force of friction and the normal force. 0157

And as we said -- this weight does not line up with an axis. 0172

So P-FBD (y,x) -- we have mg perpendicular to the ramp, mg parallel to the ramp, and of course normal force and frictional force do not need to be adjusted.0178

A couple of formulas that went with this -- mg parallel -- the component of weight down the ramp or parallel to the objects motion is mg sin θ, mg perpendicular -- the component of weight into the ramp, was mg cos θ. 0199

With that we could write Newton's Second Law equations. 0216

In the x direction, the net force in the x direction just means look at the x axis and draw all the forces acting in that direction. 0220

In this case if I call to the right up the ramp positive, that is going to be the force of friction minus mg parallel or mg sin θ and that is equal to ma. 0228

In this case since it is just sitting there, there is no acceleration -- that is equal to 0.0242

Or in the y direction -- net force in the y direction is the normal force minus mg perpendicular or mg cos θ and in the y direction it is not accelerating either. 0247

So that is all equal to 0. There is our setup. 0261

For the box at rest here we have three forces acting on our box on an inclined plane, as shown in the diagram and the vectors are not drawn to scale. 0268

If the box is at rest, the net force acting on it is equal to...0277

Well before you get too involved in a problem like this -- it is at rest. 0281

At rest means acceleration, 0. It is going to stay at rest. No net force, therefore, answer 4 must be correct. 0287

Now we have our box held by a force. 5 kg mass is held at rest on a frictionless 30 degree incline by force F. 0300

What is the magnitude of F? 0309

Well let us start with our FBD. We have F acting up the ramp; we have the normal force perpendicular to our surface, and we have mg. 0312

So now I am going to do my P-FBD over here. 0329

We still have F up the ramp, and we still have our normal force, but now we have mg sin θ or mg parallel down the ramp and mg cos θ. 0334

So now I can go write my Newton's Second Law equation for the x direction. 0351

Net force in the x direction is going to be equal to -- if I call this direction positive, that is going to be F minus mg sin θ -- that has to be equal to 0 because it is held at rest. 0359

Therefore, F must be equal to mg sin θ, which is 5 kg × g to approximate 10 m/s2 × the sin of the angle θ sin 30 degrees. 0376

We know that sin 30 degrees is half, so that is 50 × 0.5 or 25N.0392

Great. Let us take a look at what we call Atwood machines.0404

Two objects masses m1 and m2 are connected by a light string over a massless pulley. 0409

M1, m2 -- pulley of sum radius r and a string -- all connected. 0416

That is a basic Atwood machine, an experimental or theoretical device designed to help students understand how forces interact, especially when we are talking about Newton's Laws of Motion.0420

So, properties of Atwood machines -- they have ideal pulleys. 0433

If the ideal pulleys are frictionless and massless -- meaning they do not add any inertia to the system -- then you can say that the tension on either side has to be the same. 0437

That only works because this is a massless pulley but it is constant in the light string since it is an ideal pulley -- it has no mass. 0448

So tension 1 here must equal tension 2. 0455

Now as we set these up -- first we are going to adopt the sin convention for positive and negative motion because as one goes up and one goes down it could be a little confusing which way is positive. 0461

So I like to go draw a direction around the pulley and call that the positive direction.0471

Then what we are going to do is analyze each mass separately using Newton's Second Law. 0476

Here we have our system m1 and m2 -- we have called this way around the pulley, positive y and now we want to know what its acceleration is.0483

So the first thing I am going to do is I am going to come in here and I am going to label this tension 1 and that tension 2, just so I do not mix these up later.0493

And as I look at mass 1 to draw its FBD -- there is mass 1 and going down we have m1 g, its weight, and we have t1 tension -- a rope can only pull, so that must be up -- there is t1.0503

And for this mass, because of our axis over here -- down is the positive y direction. 0521

Lets do the same thing for the second mass over here for mass 2, we have m2 g down and we have t2 up. 0529

In this case though, up is going to be the positive direction because of our arrow, the direction that we indicated here. 0540

So over here positive y is that direction. 0546

Now what I am going to do is start writing Newton's Second Law equations to see if I cannot solve for the acceleration of the system. 0551

If I start with mass 1, the net force in the y direction, well m1g in the positive direction minus t1 in the negative y direction must equal m1a.0557

Let us write a Newton's Second Law equation for m2.0577

We have t2 in the positive direction minus m2g and since m2g is in the negative direction over here, then that must equal m2a. 0581

Finally, we know because it is an ideal pulley, that t1 must equal t2. 0593

So what I am going to do now is I am going to see if I cannot combine these equations because I have a couple of unknowns. 0601

I do not know t1, I do not know (a), and I do not know t2. 0606

So with three equations and three unknowns I should be able to solve this. 0610

I will start with m1g - t1 = m1a. Then I am going to add to it our second equation t2 - m2g = m2a. 0615

Now if the left and right sides are equal and the left and right sides are equal, if I add both left sides and both right sides I should still be equal.0628

A little math trick we can pull. 0637

So if I add the left hand sides here I end up with m1g - t1 + t2 - m2g all equal to... 0640

And the right hand sides if I add them up m1a + m2a, but I also know that t1 = t2. 0652

So I am going to replace t1 with t2 in the equation minus t1 + t2, and if those are equal those add up to 0.0666

So my new equation m1g - m2g = m1a + m2a. 0675

And I am trying to solve for a, so I am going to write this as gm1 - m2 on the left hand side equals am1 + m2.0687

And if I divide both sides be m1 + m2 I get that (a) is equal to g × m1- m2/m1 + m2.0700

I solve for the acceleration of the system by using two separate FBDs. 0715

As an alternate solution we could look at this as an entire single system. 0727

I am going to define my system now as m1, the rope, and m2. 0732

So everything inside that dotted line is part of my single system, my single more complex object.0741

And I am defining this direction to be positive y again, so if I re-draw this a little bit I could re-draw this as m1 over here attached to a string, m2 and I am just taking those pieces of the Atwood machine and flattening them out for the purposes of looking at this as a system.0750

On m2 I am going to have a force of m2g that passes through that barrier. So we have m2g this way. 0774

Over here I have m1g passing in that direction. 0783

So over here I have m1g, again because positive y is pointing this way, that is my definition of the positive y direction.0790

Well, if I write Newton's Second Law now for this system, where again I have defined the system as basically what is inside that dotted container, what I get is to the left in the positive direction I have the force m1g, to the right I have the force m2g... 0801 minus m2g because it is in the opposite direction of the positive y.0823

And that must equal the mass of my system. The mass of my system is m1 + m2 times the acceleration of the system. 0828

Now look how slick this is. All I have to do now is divide both sides by m1 + m2 and I end up then with a = gm1 - m2/m1 + m2.0837

The same thing I had before, but just an alternate approach -- analyzing the solution as a whole.0855

Lets talk a little bit about elevators. 0865

For some reason physicists seem to love the concept of putting a scale that measures an objects weight in an elevator. 0868

I do not really know why, they just seem to love it.0875

So let us talk about it because you may see a problem or two come up like this.0878

To begin with, we need to talk about scales. 0882

Scales do not really tell you the weight of an object and you should know that because you can go jump on your scale and for a minute it gets a really, really big reading and then it is a light reading and it levels out a little bit.0884

So it is not reading your weight the entire time, it is reading something else. 0895

What it is really reading is the normal force that exerts on you.0899

If you put a scale down, you stand on it and once it comes to an equilibrium position as you are standing on the scale, what the scale actually reads on the reading is the normal force that it is exerting, the force it is exerting back on you.0904

Scales read the normal force and if we put scales in things like accelerating elevators, we can get some interesting results. 0922

But we can analyze all of them with the stuff we already know using Newton's Second Law and FBDs. 0929

So let us take a look here. 0935

Buddy the dog with the mass of 25 kg is standing on a scale in an elevator when the elevator accelerates upward at 3 m/s 2. 0938

Probably scared Buddy the dog -- there might have been a little barking there.0947

What does the scale read while it is accelerating and what does it read once the elevator has come to a complete stop?0950

Well lets draw a FBD of our situation.0955

Here we go -- There is Buddy. We have Buddy's weight down mg and the force of the scale, the normal force back on Buddy.0960

And let us call up the positive y direction. 0973

Now Newton's Second Law in the y direction -- Fnety = MAy. 0977

So in this case net force in the y direction is just going to be the normal force minus mg and that must be equal to MAy.0986

And if we want to know what does the scale read -- what we are really looking for is the normal force.0996

Therefore the normal force, the scale reading is going to be equal to mg plus (m) times(a) acceleration in the y direction. 1004

Therefore, the scale reading Fn is equal to Buddy's mass, 25 kg times the acceleration due to gravity g (10) plus Buddy's mass again, still 25 kg times the acceleration of the elevator, 3 m/s 2 up, so that is positive.1017

So we have 25 × 10 = 250 + 25 × 3 = 75, the scale is going to read 320N. 1040

Considerably more than Buddy's 250N weight while it is accelerating upward. 1051

What happens when it comes to rest, when it is stopped? 1059

Well, when it is at rest we can use the same equation -- his Fn = mg + MAy, but as we do that now, when it is at rest Ay = 0. 1062

Therefore the normal force, the reading on the scale is just mg or 25 kg, Buddy's mass, times his acceleration due to gravity, 10 m/s 2, therefore the scale reads 250N.1083

Scales in elevators, very popular problems. So let us try and put all of this together for a few minutes. 1106

We have a truck on a hill here showing a 1 × 105 Newton truck, at rest, on a hill that makes an angle of 8 degrees with the horizontal. 1112

What is the component of the truck's weight parallel to the hill?1121

Oh, we can go through and do all the FBDs and P-FBDs, or you could recognize that the weight parallel to the hill, it is just asking for mg parallel. 1126

That is going to be mg sin θ.1136

In this case it tells us mg, the trucks weight, is 1 × 105N × the sin of 8 degrees or about 1.4 × 104N. 1140

The answer is number 3. 1161

How about a force upper ramp? 1167

We have a block here weighing 10N on a ramp inclined at 30 degrees to the horizontal. 1170

A 3N force of friction (ff) acts on the block as it is pulled up the ramp at constant velocity -- that is important, with force f, which is parallel to the ramp as shown. 1175

What is the magnitude of force f?1186

Right away, I start thinking FBDs, let us get ourselves some help here. 1190

So there we have our axis, x and y and our forces -- we have the normal force perpendicular to the surface, we have the force f up the ramp, we have a force of friction down the ramp, and we have the 10N force, the weight.1198

And that does not line up with our axis, so we have to do something about that -- P-FBD time.1221

All right, x y -- F up the ramp again, normal force perpendicular to the ramp, force of friction down the ramp -- now 10N, we have a parallel and a perpendicular component. 1231

The parallel component is going to be 10 sin θ, mg sin θ, which is going to be 10 sin 30 degrees and the perpendicular component 10 cos 30 degrees. 1249

So to find the magnitude of force f, all I am going to do is write my Fnet equation Fnetx =... 1263

Well, what I have is f, up the ramp minus force of friction minus 10 sin 30 degrees and that is equal to mass times acceleration.1276

But we are at constant velocity a = 0, so that is all equal to 0.1288

If I want the force f then, f equals the force of friction plus 10 sin 30 degrees. 1292

Force of friction is 3N, so that is 3 + 10 sin 30 = 5 for a grand total of 8N. 1303

How about acceleration down a ramp?1316

100 kg block sides down a frictionless 30 degree incline as shown. Find the acceleration of the block.1319

Lets start with our FBD. 1326

We know it is going to go down, so I will call that the x direction. There is our y. 1335

Now we have a normal force perpendicular to the ramp and we have the block's weight (mg).1341

Mg does not line up with an axis, so just like we have been doing -- time to come back to the P-FBD. 1349

Our axis again, (x, y), normal force. Now we have our components of mg -- we have mg parallel, mg sin θ, down the ramp, and mg perpendicular, mg cos θ end of the ramp.1358

So if we want to acceleration of the block, I am going to start with the net force in the x direction. 1387 So net force in the x direction is equal to -- we have mg sin θ, the only thing acting in the x direction, and that must equal MA in the x direction.1380

Therefore the acceleration in the x direction must be mg sin θ divided by m, or g sin θ.1403

How cool, we did not even need mass to solve this problem. 1416

All we need to know is acceleration due to gravity, a constant here on the surface of the earth, and the angle. 1419

The mass does not make a difference. 1424

So that the acceleration in the x direction is just going to be 10 sin 30 degrees or 5m/s2. Very slick. 1427

Let us take a look at another Atwood machine problem. 1443

Find the acceleration of the 20 kg mass, given that the masses are connected by a light string over an ideal massless pulley.1446

And the moment you see "ideal massless pulley" right away you can go and make the assumption that the tensions we have on these are going to be equal. 1454

Let us call that t, let us call that (t) right there and we will set them as equal now. 1463

And since that is pretty easy to see the 20 kg mass is going to win here, I am going to define that direction as my positive y. 1467

Let us call this m1 and we will call this m2. 1475

So FBD for m1 -- we have m1g, down, and we have tension, up, and for m1, up is the positive y direction. 1480

Lets do the same for m2 again. For m2 we have m2g, down, we have (t), up, and we will call down the positive y direction.1496

So when I write my Newton's Second Law equations for m1, I end up with t - m1g must equal m1a. 1509

For mass 2, m2g - t must equal m2a.1521

We do this the same way we did before. 1533

We can solve these lots of different ways, but this seems to be working for us right now. 1535

So when I add these up, I am going to have t and -t that will make 0, so I end up with m2g - m1g = m1a + m2a.1540

Or solving for (a), we have (g) on the left hand side, m2 - m1 = a, m1 + m2 or a = g times the quantity m2 - m1/m1 + m2.1556

Now I just substitute in my values, a = g (10) × m2 - m1, 20 - 15/m1 + m2, 20 + 15 -- so that is 10 × 5/35 or 1.43 m/s 2.1578

All right, what happens if we switch up our system a little bit? 1605

Now we have two masses, m1 and m2, connected by a light string over a massless pulley. 1610

So again, the tensions can be equal, but now one of them is on a table on a frictionless surface. 1615

Find the acceleration of m2. 1621

Let us see what we can do here. 1624

Right away I can tell that this thing is going to accelerate in that direction. 1626

So I am going to call that the positive y direction. 1631

And if we start by our FBD for m1 -- I have down m1g -- I have the normal force on m1, and let us call that t in both places -- can call it the same thing since it is equal -- T to the right. 1634

And we also have m2 here, where we have m2g down, and t up. 1656

And here that is the positive y direction and for m1 that is our positive y direction. 1665

So let us write our equations -- Newton's Second Law over here for m1. 1673

I am going to look in the x direction and just say that Fnet is t, which equals m1a. 1676

For m2, same idea -- Net force is going to be m2g - t = m2a.1685

Let us add those together like we did before.1697

Our first equation, t = m1a, and our second equation, m2g - t = m2a. 1700

When I put them all together and I end up with m2g on the left-hand side equals m1a + m2a or m2g = a × the quantity, m1 + m2.1713

Or if I want the acceleration of the system, which will be the acceleration of m2, a = g × m2/m1 + m2.1730

Slightly different problem, but we solved it the same way, using those same skills, those same tools.1745

What happens if we put our masses and pulleys on a ramp?1754

We are getting a little bit more involved every time.1758

Well, in this case, it is kind of tough to tell exactly which one is going to win but I am just going to pick a direction to begin with and I am going to call that direction around the pulley my positive y direction.1762

So once I have done that, I notice it is a massless pulley again so we can call both of those tensions, the tension is going to be equal, and we are looking for the acceleration of mass2 which is the same as the acceleration of mass1, and it is the same as the acceleration of the system.1774

So let us start by drawing our free body diagram for m1.1789

It is on a ramp, so let us call that our positive direction.1794

We also have the y axis and for our object, we have m1g, always down, we have the normal force on it, Fn, and we have force of tension up the ramp.1799

Right away again, we should be thinking P-FBD because m1g does not line up with the axis.1819

So let us do that right here. There we go.1826

We have tension up the ramp. We have normal force, -- now, m1g, we have got to break up into components -- the component parallel to the ramp is going to be m1g sin 30 degrees and perpendicular to the ramp, m1g, cos sin 30 degrees.1833

If we go and we also draw the FBD now for mass2 -- let us do that over here -- we have tension up, m2g down, and we are defining down as the positive y direction.1857

So this, Newton's Second Law equation is easy. Fnety is going to be equal to m2g - t which is m2a.1871

Over here, we have a little bit more work to do.1885

If we wanted to write the equation here, let us look in the x direction since that is the direction it is going to be moving -- I have the t - m1g sin 30 degrees = m1a.1888

If I rearrange this a little bit, t = m1g sin 30 degrees + m1a.1904

All right. Well, I am going to do this one a little bit differently.1916

I am going to replace t in this equation with all of that so when I do that, I get the m2g - m1g sin 30 - m1a = m2a.1919

We are solving for a again, so let us get all the a's on the same side, m2g - m1g sin 30 degrees = a × m1 + m2. 1939

Or a = g × the quantity m2 - m1 sin 30 degrees/m1 + m2.1954

We are just extending what we have been doing to slightly more complicated situations.1970

Let us try one last more to round all this out.1977

Let us go back to our elevators problem.1980

Darryl the Duck, who has a weight of 230N is standing on a scale in an elevator when the elevator accelerates downwards at 3 m/s2. What does the scale read?1983

Remember what we are really looking for here is the normal force at scale.1994

Well, FBD for Darryl the Duck, we have mg down, which is 230N -- we know his weight.1999

We have the normal force or the force of the scale up on him.2008

Let us call down our positive y direction.2012

Net force in the y direction then is going to be mg - the normal force and that must equal ma.2018

Therefore, the normal force must equal mg - ma or normal force = mg (230N) - ma in this case -- well, we do not know his mass.2029

But we know mg = 230N, o if mg = 230, then m must be 230/g or 230/10 which is 23 kg.2050

So mass, 23 kg × the acceleration and since it is down and we call down positive -- that is a positive 3 m/s2.2063

So the normal force then, 230 - 23 × 3 or about 161N.2073

So his typical weight is 230N, but as the elevator accelerates down underneath him, he feels lighter for a second, the scale reads less.2082

That goofy feeling you have when the elevator drops out from underneath you and you feel like you are lighter for a second, well you are not lighter, the normal force is actually less on you.2090

Imagine you are on the bottom floor and the elevator jolts up with you in it. 2100

Don't you feel heavy for a second, like you are being compressed into the bottom of the elevator?2103

That is when the scale reads more than your typical weight.2107

Hopefully, that gets you a good start on some applications of Newton's Second Law and all these different dynamics problems ranging from boxes on ramps to Atwood machines to elevator problems.2111

Hope you have gotten something great out of it.2122

Thank you for watching and make it a great day!2124