High School Physics Newton's 2nd Law: Advanced Examples

Hi, welcome back to educator.com, today we are going to be talking about Newton's second law in some really advanced examples.0000

Single dimension is just an introduction to talking multi-dimensionally, now we are going to get the chance to really structure our muscles to see how much we can do with Newton's second law.0006

There are going to be no new specific concepts or major ideas, but it is just going to be the chance to see really advanced complicated problems.0016

We will just be exploring a bunch of examples in this lesson.0025

But first I have got an interesting thing to talk to you about.0029

This is going to be the block and tackle pulley system.0031

You might have heard of block and tackle before.0035

It is used on sailing ships, sometimes on non-sailing ships like motor driven ships.0037

It is a way to be able to use tension to give you mechanical advantage over what you are lifting.0042

You can put less force into your lifting, than it is the actual weight.0048

We will talk about how that is possible, it is pretty interesting.0052

With clever construction, we actually cause the tension in our rope to give us more strength than we first expect.0055

First consider a really basic single pulley lifting system.0061

You have got some weight pulling down, some force of gravity pulling down on our object.0065

If we want to lift this, we going to need a tension pulling here, because this rope is just right here, so it is going to be whatever tension is in that rope.0070

So we are going to have to pull with a force, tension left be equivalent to the force of gravity, for us to be able to lift that block.0078

There is only one direct connection and it is that rope.0086

But, there is a clever thing that we can do.0090

Let us consider a double pulley lifting system like this one right here.0092

So, in this one, we do not have just one rope, we have two ropes.0098

This seems kind of weird, because there is this one connection here, but really what it is, it is the two sides of the pulley.0103

We have got the two different sides of the pulley.0111

So, whatever tension we put in here, is actually going to be translated all the way through, so we are going to have a tension here, and a tension here, so we are going to get double the lift of that tension that we put in.0113

We might have the force of gravity pulling down here, but we can put in a tension equal to one half the force of gravity, and be able to counter out the force of gravity.0124

We can put it in a static equilibrium, with only half the force of gravity.0133

This is really shocking, it seems sort of like magic.0138

The reason is why you will have to work with the fact that, if you want to lift that, you are going to have to pull double the rope as we would in the single pulley lifting system, to this rope has to go a certain amount, and this rope to go up a certain amount.0141

In the other system, we only had to pull up, if we want pull it up a metre, we only had to pull the rope a metre.0157

In this one, if we want to pull it a metre, we have to pull the rope a metre here, and a metre here, which means 2 m .0161

We will talk about this a little bit later, we will get the chance to revisit this really quickly when we are talking about work and energy, but you can certainly understand that there is a bit of trade off here.0167

Let us consider an even more complicated system.0179

Consider a quadruple pulley lifting system.0182

This time, instead of having to just double up once, we are going to have tension here, tension here, tension here, tension here.0184

So we are actually going to get the 'T' to show up 4 times.0193

For one force of gravity, whatever the force of gravity (the weight is of our object), we can actually put in, T = (1/4)F_g, and we will be able to offset that, put everything into static equilibrium.0197

So, by using these 4 pulleys, we are able to spread that tension in the rope and cause it to occur multiple times.0211

It seems sort of fake, -(and this pulley should actually be connected up here, otherwise it will just fall to the ground)-, and unbelievable, but it does actually work, and you will possibly get the chance to see it if you are in a Science museum, or if you are on an old sailing ship, if you get the chance to play with block and tackle, that is what this set up is called.0221

You can actually see mechanical advantage in motion.0239

Whatever the advantage you are looking at, you have to commensurate more amount of rope, but it could be potentially really useful.0241

You would be able to move really heavy things, which you otherwise would not be able to move.0249

The point of all this, is mot to give us a new formula or something new to do in our problems, it is just to make us realize how important it is to be able to carefully think through what we are doing.0252

If we carefully consider all the forces being involved, with a really good free body diagram, we can see really interesting things happen that we might not notice at first.0261

So, drawing really good diagrams is really important and it is something you got to make sure you do, because otherwise you might miss out on really key piece of information.0271

On to the examples.0279

We are going to have a bunch of examples today, and this is the first one.0282

Say we have got a free hanging mass-less string, and it is holding up three objects of unknown mass.0285

We know the tensions in each section of the string though.0291

First off, let us name these masses.0294

We will call them mass-1 (M₁), mass-2 (M₂) and mass-3 (M₃), and we will say that going up is the positive direction.0296

What are the tensions?0307

The tension in the top section, T₁ = 98 N , tension in the middle section, T₂ = 49 N, tension in the third one, T₃ = 9.8 N .0309

What are the masses of the objects?0334

We would not be able to figure it out, but luckily we have got gravity on our side, so we can use gravity to get the information we need.0335

We do a free body diagram for each one of these.0342

This one is being pulled up with a tension of 98 N, but it is also being pulled down by T₂ = 49 N, and also being pulled down by the force of gravity on it.0345

This one is being pulled up by 49 N (because that is what is above it), pulled down by 9.8 N (because that is what is below it), and also being pulled down by its own force of gravity, so F_g2.0358

Finally we have got this one, pulled up by 9.8 N, pulled down by F_g3, but it does not have any tension pulling it down.0372

Let us start figuring out what these masses are.0386

Everything is in static equilibrium, so we know that the acceleration across the board is going to be zero for everything.0388

Let us look at M₃ first.0394

We know that F_net,3, (if it has no acceleration, we know that F_net is equal to zero.), so F_net,3 = 9.8 N - F_g3 = 0 (everything inside of it), so, F_g3 = 9.8 N = M₃g , M₃ = 1 kg .0396

That is basically the method we will use for all of these.0439

F_net,2 = 49 N + 9.8 N - F_g2 = 39.2 N - F_g2 , since this is in static equilibrium, F_net,2 = 0, F_g2 = 39.2 N = M₂g , M₂ = 39.2/9.8 = 4 kg .0443

For the last one, we have that F_net,1 = 98 N - 49 N - F_g = 49 - F_g,1 = 0, F_g1 = 49 N = M₁g = 49 N, M₁ = 5 kg .0504

That is how we do it.0550

Also, one thing we could have done from the very beginning is, we could have figured it out that there is 10 kg total, because if we consider it as a one big system, then we would have, the tension on the system (in static equilibrium) = 98 N - F_g,system = 0.0551

So, 98 N = F_g,system = M_systemg , M_system = 98/g = 10 kg, because if we were to add up (5+1+4) = 10.0580

So there are two different ways of doing it, we could have figured out that the total is going to be 10 kg, and if we have two weights, figure out the third, they all work, they are all different ways of doing it.0611

Figure out which one works best for you, and also the specific kind of question you are looking at.0618

On to the next example.0622

Example 2: An object rests on a frictionless surface (we are looking top-down), and is being acted upon by 3 forces.0625

Since we are looking down on it, it is not going to be affected by gravity, so we do not have to worry about gravity for this one.0634

It is being affected by these 3 forces, but it is in static equilibrium, it is not moving.0640

F₁ = 250 N (at an angle 147 degrees), F₂ = 150 N and F₃ is pulling directly down.0645

We want to find out what is F₃.0655

First, let us start working on figuring out what the components to F₁ are.0659

We know that this is static equilibrium.0663

So, we know that F_net = (0,0) as a vector, both x and y components are going to be nothing, they are going to cancel out to nothing.0667

What does F₁ break down into?0679

What is this angle right here?0681

The whole angle is 147 degrees, and that is a right angle, that means that this is going to be 57 degrees right here.0683

So, we use trigonometry, and we get that (F₁ is over here), 250 × sin(57) , and here we have got 250 × cos(57) .0691

We get, F_1,x = 136 N , F_1,y = 210 N .0711

So we know that 136 N pulling this way, 136 N pulling up as well.0724

Now, what about F₂, we know its magnitude = 150 N, but we do not know what its angle is yet, that is an interesting thing we were not given.0730

That means that there is a possibility, but we do know for sure in static equilibrium, the x's have to cancel out.0740

F₃ do not have any x component, it is pointing directly down.0745

That means that whatever F₂ is, whatever its angle turns out to be, we know that F_2,x = F_1,x = 136 N, otherwise they will not cancel out, it will have an acceleration, it could not be in static equilibrium.0751

So, F_2,x must be 136 N, and that is an interesting thing.0768

But, there is two possibilities, either F₂ could point up, or F₂ could point down.0773

So if we have got two possibilities, that is going to give us two possible answers.0779

We do not know which one it is going to be yet, we can work on that in just a moment.0783

Let us finish the whole problem out.0786

Finishing this out, we know from what we did before, that we have got that, F_1,x = 136 N, and F_1,y = 210 N .0792

So, we know F_2,x has to be 136 N as well.0818

So, 136 N is what we have got, possibility 1, possibility 2.0831

What has to be the other side?0836

Now we can use the Pythagorean theorem.0840

We know that, 150² = 136² + F_2,y²0842

We get, sqrt(4004) = F_2,y, we will get +/- when taking the square root if we are doing it as algebra.0863

Sometimes you can get rid of that, but in this case we know that both (+) and (-) are both possibilities.0882

So, F_2,y = +/- 63.3 N .0888

So there is two possibilities.0896

We know that it is either pulling up at 63.3 N and canceling out the x, or itis pulling down at 63.3 N canceling out the x.0898

We know that 210 N,( because we know that F₃ is going to cancel out the rest of the y component), 210 N + F_2,y + F₃ = 0 .0907

You add up all the forces in the y component, we get zero because acceleration of y is zero, just like the acceleration of x is zero, how we got 136 N being pulled both ways.0924

So, because the acceleration of the y is zero, we are able to get this formula right here.0937

Now we are going to split into the two different possibilities.0945

We have got 210 N + 63.3 N + F₃ = 0 and 210 N - 63.3 N+ F₃ = 0.0948

The possibilities of when F₂ is pulling up, or when it is pulling down, it is going to be pulling either a little bit up or little bit down.0963

It is mainly pulling to the side, because of how strong it has to the side, the F₁'s x component, but it has to pull up or down just like we talked about up here.0971

210 + 63.3 + F₃ = 0, 273.3 + F₃ = 0, F₃ = -273.3 N, or F₃ = -146.7 N .0981

They are both answers, both are possibilities, we do not have enough information to be sure of which one of the two possibilities the answer is, but we have the answer for what the question is asking for here, and we could if we were given a little more information, if we knew that F₂ is pointed up or pointed down, we will be able to figure out whether or not, we would able to figure out which of these answers we would go with.1010

But these are the two possibilities for the third force.1031

Third example: We have got a chandelier that is suspended by a cable from the roof of an elevator, in a fancy hotel.1034

There is a bottom for our elevator, right there.1041

Assuming that the cable is mass-less, and the chandelier weighs 10 kg, what is going to be the tension in the cable when the elevator is at rest?1054

Also, what happens when the elevator starts moving around?1061

So the elevator is moving up, do you think the tension is going to increase or decrease?1064

If the elevator is moving up, have you ever stood in an elevator and paid attention what the floor felt like under your feet?1069

As the elevator starts to move up, you feel slightly heavier.1076

That is because the floor of the elevator has to push against you.1079

The floor of the elevator actually pushes against you to be able to give you an acceleration.1083

It then just maintains the standard normal force that you are used to so you do not go through the floor, but to be able to accelerate you, it has to put a force on you.1087

So, it puts an acceleration through the floor.1096

That acceleration comes by putting in extra force.1100

What happens when the elevator wants to go down?1103

If you are trying to go down in an elevator, you will notice that you experience a brief moment of weightlessness, not weightlessness, but of a less weight, as you start to accelerate down towards the Earth.1105

That is because, normally, you got some normal force pulling you down, but as we accelerate down, we need less normal force, but now we are being allowed to move towards the centre of the Earth.1114

The normal force is receding, because it is allowing you to gain a velocity, it is allowing you to get some acceleration for a brief period of time.1127

Finally, what would the acceleration be necessary, if the elevator were to have no tension in the cable?1135

What is the tension when it is at rest?1142

When is the tension when there is an acceleration of 1 m/s/s?1145

What is the acceleration when it is -1 m/s/s?1150

We will have positive as going up.1153

1 m/s/s will increase the tension because we are going to have to pull that chandelier along.1156

When the acceleration is going down, when it is -1 m/s/s, we will need less tension, because we are going to be able to allow it to go in the direction gravity would normally be taking it, so we just need to give it less force.1161

Finally, what would have to be the acceleration for the elevator to have no tension in the cable whatsoever?1176

We are going to start off with slightly simplified diagrams, so we have got some tension pulling up (elevator is not the important thing here), the elevator is attached up here, we have got some mass of 10 kg which is being pulled down by the force of gravity.1182

Force of gravity = 10 × 9.8 = 98 N, so being pulled down by 98 N.1205

If the acceleration = 0, what is the tension going to be?1213

If a = 0, tension is going to have to cancel out gravity.1218

So, T - F_g = mass × acceleration = 0, because we are completely still, T = F_g = 98 N.1226

So, if the acceleration = 0, if things are static, 98 N.1243

What if the elevator was moving up?1249

That is going to require more tension, so we will expect to see the tension grow.1253

Let us look at this again.1257

If acceleration =1, then T - F_g = Ma, T - 98 N = 10 kg × 1 m/s/s, so, T = 98 + 10 = 108 N .1259

So if you want to make that chandelier accelerate at 1 m/s/s, you are going to need to give it another 10 N past what it takes to stay off gravity, because you have to keep holding off gravity with what would normally be a normal force, but in this case is tension, because it is not resting on anything, it is being pulled up by a cable.1287

What if we had an acceleration of -1?1310

Very easy, just the same thing.1315

We have got, T - F_g = Ma, T - 98 N = 10 × (-1), T = 98 - 10 = 88 N, so we need a little less tension, 10 N less tension to be able to allow it to get that bit of acceleration, because gravity will take it the rest of the way, gravity puts in 98 N, and the tension needs to only bring it 10 N of down force, so that it will be accelerated at 1 m/s/s down.1318

Finally, if we want no tension whatsoever, what would the acceleration have to be?1360

If T = 0, T - F_g = Ma, - F_g = Ma, -98 = 10 × a, a = -9.8 m/s/s, same thing as gravity.1368

If we want to have no tension, we need to let the elevator go into complete free-fall.1397

You just drop the elevator, and there is going to be no tension, because the cable is going to be accelerating with gravity, the chandelier is going to be accelerating with gravity, the elevator is going to be accelerating with gravity, gravity is going to be taking over all the work.1401

So, the only way to get rid of the tension completely is to let gravity do all the work, it does not have to be resisting anything, there is no longer any resistance, you are letting gravity take over.1412

Example 4: A 20 kg baboon climbs a mass-less rope attached to a 22 kg crate.1428

The 22 kg crate on the ground is over a frictionless tree limb, so the baboon climbs the rope with an acceleration of 3 m/s/s, what acceleration will the crate have?1446

What is the maximum acceleration that the baboon can achieve without moving the crate?1455

Our baboon is pulling up, he is putting in some tension, just like if you are climbing a rope to a ceiling, you would be putting in a tension into the rope as it pulls you up, as you use your arms, and the rope resists it, you would be able to yourself up the rope.1460

That is exactly what our baboon is doing here, he is climbing the rope just as you might have to in gym class.1476

He is climbing, so he is putting in a tension.1482

At the same time, that tension is going to pull in here, and it is going to start pulling on the crate.1484

So either it is going to lower its effect of weight so much that the tension is actually going to lift it, so the box will be easier for somebody else to come along and lift, or it is going to get so much tension in the rope that it is actually pulled off the ground, as the baboon is climbing.1490

That is our basic idea, let us start working on it.1506

We will just consider our tree limb to a perfect pulley, mass-less and frictionless, and our baboon is going to climb up things, so if the acceleration of the baboon is 3 m/s/s, what will the acceleration of the crate be?1511

He is going to climb up with some acceleration, which is going to cause some tension in the rope.1524

So, over here, that tension gets canceled out, pulled the other way, rope is always pulling off position, that tension is going to be the same here.1533

Free body diagram for the monkey, he is going to be pulled up, by the tension he is putting into the rope, and he is also being pulled down by his weight.1540

So his force of gravity = 20 × g =196 N.1551

What is the tension?1560

We do not know what the tension is yet, but we do know that his acceleration = 3 m/s/s.1562

So, we can figure out what the tension must be.1570

If we make up positive, and keep in mind, that is going to turn out to be that 'down' from the crate's point of view is positive, that is really strange, because we are actually wrapping it, flipping it all the way, we have to keep in mind that positive stays the same around the pulley, so positive 'up' for the baboon means negative 'up' for the crate.1572

Positive goes into the ground from the point of view of the crate.1595

T - F_g = M_baboon × a_baboon , we do not know what the tension is, but we know everything else.1598

T - 196 = 20 × 3, T = 256 N.1609

So, 256 N, what would be the acceleration on the crate?1623

The crate has got two things happening on it.1626

It has got T, and its own force of gravity = 22 × g = 215.6 N .1629

So, it experiences F_g - T = M_crate × a_crate, (Tension is negative from the crate's point of view, positive from the baboon's point of view.)1642

So, F_g - T = 215.6 - 256 = 22 × a_crate , (F_g and T are not equal, which means we are going to have some lift off.)1687

Solving, a_crate = -1.84 m/s/s .1701

So, the baboon is able to achieve an acceleration of 3 m/s/s, the crate is also going to have to achieve 1.84 m/s/s .1717

What about when we want to have the baboon climb, but not cause any movement?1735

If we want to have the baboon climb, but not cause any movement, then we are going to have to look for what is the maximum acceleration for the baboon is going to happen when we get the most tension, because tension is what enables it.1740

Keep in mind, unlike our block problems from before, they are not rigidly connected, the baboon is climbing hand over hand, going up the rope, he is not attached to a single fixed point, whereas the two blocks are attached to a fixed point, when one of them moves, the other one has to move, but in this case the baboon is able to climb up it.1764

It still puts a tension, otherwise nothing would be able to move him around, he would not have an external force to cause him to actually go up.1787

But he does not connect directly to the crate, so they are going to be decoupled, our accelerations are decoupled.1794

The maximum acceleration for the baboon will be when there is the most tension in the rope, without moving crate.1801

What is the maximum tension we can have without moving the crate?1812

We have to cancel out the baboon, otherwise he is going to start falling.1816

So we have got, T = 196 + T_a, (196 is what which cancels weight, T_a('a' for acceleration) is what is able to give more than that.)1824

Most of this tension could be, is equal to the force of gravity on the crate, because otherwise the crate is going to start to lift off the ground.1856

196, that which cancels out the baboons weight, and the acceleration he gets in addition to canceling out his weight, cannot exceed the weight of the crate.1864

What is the weight of the crate equal to?1872

The weight of the crate = 215.6, so we get the acceleration part of the tension, T_a = 215.6 - 196 = 19.61874

What is the maximum acceleration our baboon can have?1903

The maximum acceleration the baboon can have is going to be, we have already canceled out, we have got our T, we have got our force of gravity pulling down on the baboon + Tension, we are going to split that tension into the one that cancels weight + the tension that actually does acceleration.1906

These two things cancel one another out, and that leads to, equal to, the mass of the baboon × acceleration.1933

So we have got, T_a = M_baboona_baboon, 19.6 = 20 × a , a = 19/20 = 0.98 m/s/s, is the maximum acceleration, which is still pretty good, 1 m/s/s, that is a tenth of the acceleration due to gravity, which seems kind of low, but if you try to climb a rope faster than that, that would be pretty impressive.1946

Final example: Two block of masses, M₁ = 5 kg, M₂ = 10 kg, are roped together on an incline, both having different angles.1993

The first box is on an incline of 50 degrees, and second one is on an incline of 20 degrees.2007

They also each have different masses as we said before.2015

Assuming the pulley and rope mass-less, and the pulley is frictionless, what acceleration will they experience?, what will be the tension in the rope?2022

Very first thing to do, we have force of gravity pulling down on both, but force of gravity is not what really matters, what matters is how much they get canceled out, and how much is parallel.2031

We are going to, remember, whatever the force of gravity is perpendicular, F_normal, they are going to cancel each other out.2044

So, while they are still present, for our purposes, we are not going to pay attention to them, because they are not going to have any effect on what we are doing.2055

Force of gravity perpendicular over here, gets canceled out by the normal force, because the normal force is able to cancel out whatever perpendicular.2061

So the normal force cancel the force of gravity perpendicular to the side of the triangle, once again, we are not going to worry about these for the purpose of our problems.2072

The only thing that we are really concerned about, is what is the parallel component.2082

Up here, we have got a 20 degree angle, over here we have got a 50 degree angle.2089

So, sin(50) × F_g, and over here, sin(20) × F_g.2098

We have not figured out F_g yet, but that will take no time at all.2113

F_g = 5 × 9.8 = 49 N, over here, F_g = 10 × 9.8 = 98 N.2116

If you want to figure out what the parallel components are, we get, sin(50) on this one, it is going to be equal to, the force of gravity on the first block, parallel to the triangle (down the side of the triangle) is equal to 37.5 N .2132

Over here, we are going to get that the force of gravity on the second block, parallel, is going to be equal to 33.5 N.2158

So, we have got what the two parts are parallel.2170

Which direction can we expect to move in?, we know we are going to have to move in this direction because there is more force in the system pulling this way, than this way; this is the smaller one, so this is the side that is going to win.2172

Let us finish this up.2190

So, we know that the parallel force, which is the only force we have to worry about, because we know that gravity perpendicular and normal force cancel one another out, we know that this force of gravity, F_g,1(parallel) = 37.5 N, F_g,2(parallel) = 33.5 N.2193

Now we want to find out what is our acceleration, what is the tension in the rope.2220

So, there are going to be tensions pulling both this way.2225

As we have talked about before, we can actually treat this whole thing as one big system.2227

So, we have got one big system.2232

We can figure out the tensions, come up with a bunch of equations, but there is no reason to, because we know that what are the external forces acting on it, the tensions are not external, those are the things that links our system, that makes our system.2234

So, the tensions are not going to cancel out whether we can treat it as a system, they are what enables us to use it as a system.2246

So we only have to pay attention to what is working outside, if we treat it as a one big system.2251

What forces are acting on the system?2255

That is going to be, = M_system × a, unlike the baboon and crate problem, we got a rigid connection here, if one thing moves, the other has to move, so we know that we can use the same acceleration for the two of them.2260

What forces are here?, let us make to the right as positive, we got 33.5 - 37.5 = (10+5)a, solving, -4 = 5a, a = -0.27 m/s/s, which makes sense, because as we have talked about before, we are going to see an acceleration pulling things that way.2274

In this case, it is a very small acceleration, but it is still pulling it along.2312

This is what our acceleration is, it is parallel, it is not just some general acceleration, this is the magnitude, and what we know, we know if we were to consider it as a vector, it is the amount parallel to the triangle and pulling to the left.2318

It is going to be parallel from both of their point of views.2332

They are going to have very different acceleration vectors.2334

This one is only going to come up, at a 20 degree incline, while this one is going to go down at a 50 degree incline, so one is rising , the other is falling, but as we talked about before, the only thing effecting this is the parallel component, because remember, the normal force cancel out the perpendicular component of gravity.2337

The only thing we really have to worry about here is, what the total acceleration is, and it is up to us, if we had a problem where we had to consider which direction it is moving, then we would have to work on it, but in this case, it is enough to say that it is -0.27 m/s/s, or really 0.27 m/s/s, and then talk about which direction it is moving in.2354

Now, what if we were to solve for the tension is?2375

If we want to solve for what the tension is, we can look at a single block on its own.2378

So, let us consider looking at the smaller block.2385

The smaller block is being pulled down with 37.5 N, it is being pulled up with the tension (which we will figure out in a second).2388

Another important thing is, because the way we got this constructed, tension is parallel.2403

As we have talked about before, it is possible for the rope to be pulling at a different angle, and cause things to not be parallel to the surface, but in this case, we have set up our problem in such a way, or the problem is given to us in such a way that the rope is going to run parallel to the surface, so we do not have to worry about this.2409

So, it is just going to be these two things acting directly, without having to worry about putting in any more angles.2423

We already figured out that this one was the parallel component, so we are good to go from here.2427

This is the negative direction, so tension is pulling in the positive direction, T- 37.5 = M_{little block} × a_{little block} = 5 kg × (-0.27 m/s/s), solving, T = 36.15 N, which makes sense, because the tension in the rope is going to be have to be less than the force of gravity parallel, otherwise it is going to go in the direction of the tension.2432

The larger force is the one that will win out, which is going to have the effect, if they are directly opposed.2493

So, the tension is smaller than the force caused by gravity.2497

That gives us everything for this problem, we have got it solved.2505

I hope you learned something, there is a bunch of different possibilities in really advanced examples that you can do with forces, but just by breaking it down into good free body diagram, carefully analyzing the forces involved, putting down all the things that you know, you can work your way through it, and you can figure out what you are going to do.2507

I hope you enjoyed it, and we will see you at educator.com for the next lesson.2520