High School Physics Newton's 2nd Law: Multiple Dimensions

Hi, welcome back to educator.com, today we are going to be talking about Newton's second law and applying to multiple dimensions.0000

Last time we only talked about applying it in one dimension, today we are going to be talking about what happens in more than one dimension.0006

As in kinematics, each dimension works independently.0011

Acceleration in the y direction is completely unrelated to acceleration in x direction.0016

As far as Newtonian mechanics is concerned, we can consider them completely independently.0020

Now we just turn our formula from before into vector based ones.0030

Now we have got, F_net = ma, so we can consider them as the different components as separate pieces, x_net, y_net is going to have acceleration in the x and acceleration in the y, both going to be completely separate components.0033

It is possible to break a force vector into its components just as we did in kinematics to break velocity, acceleration, displacement into separate components.0053

If we had a force of 10 N acting on an object 30 degrees above the horizontal, what will that force be in component form?0063

First, we drop a perpendicular sown.0068

Now, we have a right triangle, we can easily apply trigonometry.0072

If we want to know what this side is, this is the side that is going to be connected to sin(30), we got the hypotenuse = 10, so it is going to be (10 N) × sin(30) , which comes from sin(30) = side opposite / hypotenuse, if you get used to doing stuff in component form, you will get really fast at this, it is just going to be sin(30) times whatever the full magnitude of the vector is.0075

Similarly on the other side, on the bottom part it is going to be, (10 N) × cos(30) .0120

Now we know what each side would be, so we just use basic trigonometry, and we get that from a table/calculator.0132

We get, the length (the width) is going to be 8.66, and the height is going to be 5, both in newtons, 8.66 N and 5 N in horizontal and vertical directions respectively.0139

Before we talked about some special forces, and they are all still there, the force of gravity (weight) is still there, now it is only going to pull down in one direction, it either gets canceled out, or works at the same time as the horizontal forces are working.0159

Normal force, force that holds you up, still there, if something is resting on the top of a surface, that surface is still going to support it against gravity.0187

Tension, force on whatever object is connecting things, still there, and it is going to pull in whatever direction that rope is angled at.0195

If you have a rope connected to something, and you pull like this, then it is going to cause a force of that, and if you pull like this, that is going to cause a force like that, it just have to do with what angle it is acting.0203

But, I have not explained why it is called the 'normal' force!0220

Why do we call it the normal force?0228

That is a great question.0230

The normal force, it is a fancy word for Math, it simply means perpendicular.0233

The word 'normal' comes from the Latin word 'norma', which is the word for a 'carpenter's square', which is a tool used for making right angles.0240

Normal just comes from a right angle, so it just means perpendicular.0249

Two normal segments would be any two that intersect with a right angle between them.0255

If you had a curve of some sort, you could have a number of right angles that is sprouted out of it.0262

Each one of them is a little bit different, but they are all normal even though they are attached to a curve, because what matters is that they are perpendicular at the point of intersection.0279

So normal is just another way of saying that it is perpendicular.0289

Normal force is the force that is perpendicular, the gravity the surface can cancel out, because the surface can only put up a perpendicular force.0293

Consider these two versions:0301

We have got a man standing on a perfectly flat plane, now what happens is, the force of gravity pulls down on him by a certain amount, and because he is standing on a perfectly flat plane, the normal force is going to be able to be exactly opposite, and the two are going to cancel one another out.0304

So that is great.0319

Now, what happens when force of gravity is pulling on a man, who stands on the side of a building.0320

He does not have anything connecting his feet to the building, so we are not going to think about how he managed to get there to his position where his feet are on the building for a brief second before he plummets down.0327

Does he have a normal force on him?0338

No, because the normal force would have to work by going this way, and that does not make sense because there is nothing pushing him against it.0341

So you cannot have that.0349

The normal force would only work if he were pushed against something, but he has nothing to be pushed against to, he is just going to plummet down to the ground.0350

What happens if you were on an incline?0367

So we looked at the two extreme cases, flat surface and vertical surface, either the normal force is going to be completely there, or it is going to have no effect.0369

If you were on an incline, them some of it is going to be able to be perpendicular.0390

We can now create a right triangle, that has this part perpendicular, and this part parallel to the top of the triangle.0394

There will be a parallel section and a perpendicular section.0406

The important part is that, this is a right angle, and this is a right angle, and this is parallel to this.0418

So we know this is a right angle, so we know how much is perpendicular to the surface, and we know how much is parallel to the surface.0428

Let us figure out what would be the angle up here.0434

If we look at this as a slightly larger version, here is angle θ , and here is a right angle, we can figure out from basic trigonometry what that angle is.0441

We know that across two intersections, we know that these have to equal angles.0462

So, this θ must be equal to this other angle over here, which means if we want to figure out what these two components are, we are going to get cos(θ) × F_g over here (because that is the side adjacent), and sin(θ) × F_g is going to be the parallel.0468

In general that is a rule to remember.0506

But it is important to make a diagram and understand yourself than slavishly fall over formula, if you do that you are liable to make mistake, the math can go wrong, it is up to us pay attention to what we are doing, and that is incredibly important.0517

But in general, if you are able to get this angle here in this right angle format, then we are going to have that sin(θ) is going to be the thing connected to the parallel part, and cos(θ) is going to be the thing connected to the perpendicular part, which makes sense,0546

because if θ was zero, if we were on a flat thing, then cos(0) would be 1, i.e. all of the force of gravity would be translated to the perpendicular and if it were 90 degree (perfectly vertical), then sin(90) = 1, which means all of the gravity would be translated into the parallel.0566

So, sin(θ) correlates to the parallel, cos(θ) correlates to the perpendicular, once we draw the diagram this way.0590

Depending on the diagram you use, you might wind up changing it out, or it might be that your problem works differently, so it is important for you to pay attention to what you are doing, and have it make sense, but this is a general structure that you can follow, also you would like somebody else to understand your work later.0596

First example:0623

We have got an object with mass of 5 kg sitting on a perfectly flat frictionless plane.0624

It is pushed by a force of 3 N to the North, and a force of 4 N to the East.0629

What is the acceleration of the object, what is the magnitude of that acceleration, and at what angle does the combined force move the object along at?0634

When we are saying it is on a perfectly flat frictionless plane, we imagine it like the middle of a frozen lake.0647

We are looking down from above.0653

So, we have got North, East, South, West like on a compass.0655

We make East-West the x positive direction.0666

That seems pretty reasonable, we could make North as x positive, but we want to make it something we are used to, not complicated, so we make East x positive and North y positive.0670

Some if we were to go South, that would be negative in the y component of our vector, and if West, that would be negative in the x component of our vector.0685

With that let us look at what our picture is, we have got our block sitting still at the moment, and it is pushed by a force of 3 N to the North, so 3 N 'up', and 4 N to the East, so 4 N to the 'right', 'up' and 'right' from the way we are visualizing it.0692

The important thing is that we have set up a new coordinate system to understand this.0725

So, 3 N positive y, and 4 N positive x.0730

That means now x component of our force is 4 N and y component of our force is 3 N .0733

So, if we want to find out what the acceleration of our object is, we are going to have to find out what the acceleration is for each of the components, in x and y, and together that will give us the acceleration vector, the combined acceleration.0745

We appeal to Newton's second law: F_net = ma, there is no other forces than this one force, actually two forces, but they do not interact since they are in different coordinates axes.0760

So we do not have to worry about any other forces, so F_x = ma_x , 4 N = (5 kg) × a_x , a_x = (4/5) m/s/s .0772

We do that for the y, F_y = ma_y , 3 N = (5 kg) × a_y , a_y = (3/5) m/s/s .0800

Now we have got our two pieces, we can put them together, we get, a = (4/5,3/5) m/s/s .0831

Here is our first answer.0850

If we want to find out what the magnitude of the vector is, that is just going to be the square root of each of the components squared.0852

So, square root of (4/5)² + (3/5)² = square root of (16/25) +(9/25) = square root of (25/25) = 1 .0859

So, the magnitude of our acceleration vector is 1 m/s/s .0883

Finally, to figure out what the angle it is coming at, we want to set up a triangle so we can see the angle more easily.0888

So, here is a picture of what is going on.0897

We have got 4 N to the right, which has now become (4/5) m/s/s to the right, and also (3/5) m/s/s to the up.0899

If we want to figure out what angle that is, now we have got a triangle where we can set things up, we can see where the angle goes.0919

The angle above the horizontal, above the East, the angle North of East, we are going to able to figure that out by saying, tan(θ) = (y component)/(x component) = (3/5)/(4/5) = 3/4 .0929

Taking the arctan of that, θ = arctan(3/4) = 36.87 degrees .0954

Next example:0967

A 20 kg block is on an incline of 50 degrees with a rope holding in its place.0968

Assuming that the incline is frictionless, what would be the tension in the rope?0973

Then if we cut the rope, what acceleration would the block have?0976

First, what forces are at play here?0980

We got tension pulling this way, force of gravity pulling down.0982

Force of gravity = (20 kg) × (9.8 m/s/s) = 196 N .0991

So, how much of this is perpendicular, and how much of this is parallel?1006

Up here, we are able to get this as θ from the geometry, so this is going to be equal to 50 degrees.1013

So, force of gravity perpendicular = cos(50) × (196 N) = 126 N, and the force of gravity parallel = sin(50) × (196 N) = 150 N .1028

Now, there is one more force operating over here, and that is going to be the normal force.1058

So, the normal force, it can only work on what is operating perpendicular to the surface, how much is operating perpendicular to the surface?1066

The force normal = force of gravity perpendicular, which means that this force will cancel out this force, so the net force is only going to be the force of gravity parallel and the tension in the rope.1073

Now, is this thing moving?1087

We know that it is not, it is currently still, the rope is taut and it is staying in place.1089

So we know that the tension, we will make this the positive direction, Tension - force of gravity (parallel) = mass × acceleration = 0, because we got zero acceleration.1094

So, T = F_g(parallel) = 150 N .1112

So we have got the tension.1121

If we cut the rope, how much acceleration will the block have?1124

Now, it is not going to have anything pulling it away, it is only going to be moving parallel, because what we have done is, we have broken down it into the perpendicular and parallel.1128

The perpendicular is canceled out by the normal force, because that is how much force is translated to through the surface perpendicular into the object.1136

So the force of gravity that is perpendicular to the surface gets canceled out, but the force of gravity parallel to the surface will be able to accelerate the block along a parallel vector to the surface of the triangle, it will accelerate this way.1143

We do not know what 'a' is yet, but we can figure it out.1161

We had, T - F_g(parallel), but now there is no more T, so all we have is F_g(parallel) = (20 kg) × acceleration, (150 N)/(20 kg) = acceleration.1164

So our acceleration = 7.5 m/s/s .1198

What direction is that going?1207

It goes in the direction parallel to this.1209

This is just an acceleration, not a vector, we could figure out what the components are to the F_g(parallel), but that is going to take a lot of time, there will be some part x, some part y, and it will take a lot of effort, instead we can just know what it is off of our diagram,1212

because we do not have to say specifically what it is, we can just say it is working parallel to the top of the triangle, it is working parallel to the hypotenuse of the triangle, because that is really the most interesting thing, that gives us the useful information.1227

So, a =7.5 m/s/s, parallel and sloping down along the triangle.1238

Final example:1252

We have got a 10 kg block on an incline of 20 degrees, and it is attached by a rope to a free hanging block of 5 kg .1254

Assuming that it is frictionless, the rope and the pulley are mass-less, what is the acceleration of the blocks, and what is the tension in the rope?1260

As done before, we are going to look at this as a system.1270

It is a lot easier, although we could break it into the tensions and the gravities experienced by the blocks.1274

This one does not have any force normal operating on it, it is just going to have the force of gravity, which in its case, = 5 × 9.8 = 49 N .1282

How much gravity does this one have operating on it?1298

Its force of gravity = 10 × 9.8 = 98 N .1301

If we want to break this into its perpendicular and its parallel component, then here is our angle of 20 degrees.1310

So here we are going to have, cos(20) × F_g, so force of gravity (perpendicular) will be canceled out by the normal force, F_g(perpendicular) = cos(20) × 98 N = 33.5 N, but sorry!!, I gave you the wrong one there (parallel), but there is no need to worry since the normal and F_g (perpendicular) are going to cancel out, so we just have to worry about the parallel here.1320

We know the acceleration perpendicular to the surface of the triangle, because we know that they are going to get canceled out, because of the way the normal force works.1386

So, all we care about is, how much is parallel.1394

How much is parallel will be, sin(20) × F_g (parallel) = sin(20) × 33.5 N .1396

Like before, we are going to consider this as positive direction, so positive goes around the corner, goes down, we are going to consider it the same as we did in the last lesson when we worked out a problem similar to this.1420

The only thing left, we have not talked about these tensions working in the rope.1432

But, if we look at the system as a whole, the only thing pulling in an external fashion, are the two forces of gravity (the one that is going completely down), the tensions are going to be canceled out when we look it as a system, because the rope is going to be taut the entire time, so we can consider it to be a rigidly moving system, where the accelerations are going to be the same.1440

We have done a problem like this before, you should go back a lesson and look at the final example from that one, because that is very similar, we are just throwing in the incline to make the things little bit more difficult.1469

Now we know what our forces are, the only two forces we need to care about, are forces of gravity, for this one, = 49 N, we want to find out the acceleration and the tension in the rope.1478

For this one, force of gravity (parallel) = 33.5 N .1496

So, what is the mass over here?1505

This has a mass of 10 kg, this has a mass of 5 kg .1507

So, what are the forces acting on the system?1513

This is the positive direction, F_net (system) = mass (system) × acceleration, the accelerations being the same (since rigid).1515

So, 49 N - 33.5 N = (10 kg +5 kg) × a , 15.5 N = 15 × a, a = 1.03 m/s/s .1536

So there is going to be an acceleration this way, and an acceleration this way.1563

For this one it is going to be going directly down (for the smaller block), and for the larger one it is going to move parallel and up the surface of the triangle, and they are both going to be 1.03 m/s/s .1568

What is the tension?1580

For this, we just look at the free-body diagram.1582

Let us consider the smaller one.1585

It has some tension T, and it has some force of gravity pulling down.1587

We know what the force of gravity already is, it is 49 N .1593

What is the tension, we do not know that, but we do know that it has an acceleration = 1.03 m/s/s, and we know its mass = 5 kg .1596

We have got that, F_net = ma , F_g - T = 5 × 1.03 , 49 - (5 × 1.03) = T , T = 43.85 N , which makes a lot of sense because if the block is going to be able to fall, there need to be slightly less tension in the rope, because it has to have a net force that is stronger downwards than upwards, otherwise it will not accelerate down, it will accelerate up.1610

That finishes it for this lesson, hope you enjoyed it.1661