WEBVTT physics/ap-physics-1-2/fullerton
00:00:00.000 --> 00:00:03.000
Hi everyone and welcome back to Educator.com.
00:00:03.000 --> 00:00:08.000
We are going to take a look at dynamics applications and problem solving in this lesson.
00:00:08.000 --> 00:00:16.000
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.
00:00:16.000 --> 00:00:25.000
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.
00:00:25.000 --> 00:00:30.000
We will utilize Newton's Laws of Motion to solve problems of objects on ramps.
00:00:30.000 --> 00:00:36.000
Gain an understanding that tension is constant in a light string passing over a massless, or ideal pulley.
00:00:36.000 --> 00:00:47.000
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.
00:00:47.000 --> 00:00:52.000
So, with that, let us go back to FBDs again -- a quick review.
00:00:52.000 --> 00:01:00.000
FBDs are tools used to analyze physical situations and they show all the forces acting on a single object.
00:01:00.000 --> 00:01:09.000
Then, we draw all the forces on that object and we draw the object as either a box or as a dot.
00:01:09.000 --> 00:01:15.000
When we are drawing FBDs -- what we are going to do is we are going to choose the object of interest and draw it.
00:01:15.000 --> 00:01:18.000
Then label all the external forces and draw them.
00:01:18.000 --> 00:01:24.000
And then sketch the coordinate system choosing the direction of the objects motion as one of the positive axis.
00:01:24.000 --> 00:01:40.000
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.
00:01:40.000 --> 00:01:45.000
Quick review -- we have a block sitting on a ramp -- What do we do about the forces acting on it?
00:01:45.000 --> 00:01:56.000
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.
00:01:56.000 --> 00:02:00.000
Once we have that down, we are going to complicate matters a little bit.
00:02:00.000 --> 00:02:12.000
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.
00:02:12.000 --> 00:02:21.000
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.
00:02:21.000 --> 00:02:26.000
Then we are going to write Newton's Second Law equations for the x and the y directions.
00:02:26.000 --> 00:02:37.000
So for this box we have its weight down, normal force, and the force of friction, since it wants to slide down the ramp.
00:02:37.000 --> 00:02:52.000
Our FBD -- we will draw our axis -- We have mg down. We have the force of friction and the normal force.
00:02:52.000 --> 00:02:58.000
And as we said -- this weight does not line up with an axis.
00:02:58.000 --> 00:03:19.000
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.
00:03:19.000 --> 00:03:36.000
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.
00:03:36.000 --> 00:03:40.000
With that we could write Newton's Second Law equations.
00:03:40.000 --> 00:03:48.000
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.
00:03:48.000 --> 00:04:02.000
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.
00:04:02.000 --> 00:04:07.000
In this case since it is just sitting there, there is no acceleration -- that is equal to 0.
00:04:07.000 --> 00:04:21.000
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.
00:04:21.000 --> 00:04:28.000
So that is all equal to 0. There is our setup.
00:04:28.000 --> 00:04:37.000
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.
00:04:37.000 --> 00:04:41.000
If the box is at rest, the net force acting on it is equal to...
00:04:41.000 --> 00:04:47.000
Well before you get too involved in a problem like this -- it is at rest.
00:04:47.000 --> 00:05:00.000
At rest means acceleration, 0. It is going to stay at rest. No net force, therefore, answer 4 must be correct.
00:05:00.000 --> 00:05:09.000
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.
00:05:09.000 --> 00:05:12.000
What is the magnitude of F?
00:05:12.000 --> 00:05:29.000
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.
00:05:29.000 --> 00:05:34.000
So now I am going to do my P-FBD over here.
00:05:34.000 --> 00:05:51.000
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.
00:05:51.000 --> 00:05:59.000
So now I can go write my Newton's Second Law equation for the x direction.
00:05:59.000 --> 00:06:16.000
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.
00:06:16.000 --> 00:06:32.000
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.
00:06:32.000 --> 00:06:44.000
We know that sin 30 degrees is half, so that is 50 × 0.5 or 25N.
00:06:44.000 --> 00:06:49.000
Great. Let us take a look at what we call Atwood machines.
00:06:49.000 --> 00:06:56.000
Two objects masses m1 and m2 are connected by a light string over a massless pulley.
00:06:56.000 --> 00:07:00.000
M1, m2 -- pulley of sum radius r and a string -- all connected.
00:07:00.000 --> 00:07:13.000
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.
00:07:13.000 --> 00:07:17.000
So, properties of Atwood machines -- they have ideal pulleys.
00:07:17.000 --> 00:07:28.000
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.
00:07:28.000 --> 00:07:35.000
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.
00:07:35.000 --> 00:07:41.000
So tension 1 here must equal tension 2.
00:07:41.000 --> 00:07:51.000
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.
00:07:51.000 --> 00:07:56.000
So I like to go draw a direction around the pulley and call that the positive direction.
00:07:56.000 --> 00:08:03.000
Then what we are going to do is analyze each mass separately using Newton's Second Law.
00:08:03.000 --> 00:08:13.000
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.
00:08:13.000 --> 00:08:23.000
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.
00:08:23.000 --> 00:08:41.000
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.
00:08:41.000 --> 00:08:49.000
And for this mass, because of our axis over here -- down is the positive y direction.
00:08:49.000 --> 00:09:00.000
Lets do the same thing for the second mass over here for mass 2, we have m2 g down and we have t2 up.
00:09:00.000 --> 00:09:06.000
In this case though, up is going to be the positive direction because of our arrow, the direction that we indicated here.
00:09:06.000 --> 00:09:11.000
So over here positive y is that direction.
00:09:11.000 --> 00:09:17.000
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.
00:09:17.000 --> 00:09:37.000
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.
00:09:37.000 --> 00:09:41.000
Let us write a Newton's Second Law equation for m2.
00:09:41.000 --> 00:09:53.000
We have t2 in the positive direction minus m2g and since m2g is in the negative direction over here, then that must equal m2a.
00:09:53.000 --> 00:10:01.000
Finally, we know because it is an ideal pulley, that t1 must equal t2.
00:10:01.000 --> 00:10:06.000
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.
00:10:06.000 --> 00:10:10.000
I do not know t1, I do not know (a), and I do not know t2.
00:10:10.000 --> 00:10:15.000
So with three equations and three unknowns I should be able to solve this.
00:10:15.000 --> 00:10:28.000
I will start with m1g - t1 = m1a. Then I am going to add to it our second equation t2 - m2g = m2a.
00:10:28.000 --> 00:10:37.000
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.
00:10:37.000 --> 00:10:40.000
A little math trick we can pull.
00:10:40.000 --> 00:10:52.000
So if I add the left hand sides here I end up with m1g - t1 + t2 - m2g all equal to...
00:10:52.000 --> 00:11:06.000
And the right hand sides if I add them up m1a + m2a, but I also know that t1 = t2.
00:11:06.000 --> 00:11:15.000
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.
00:11:15.000 --> 00:11:27.000
So my new equation m1g - m2g = m1a + m2a.
00:11:27.000 --> 00:11:40.000
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.
00:11:40.000 --> 00:11:55.000
And if I divide both sides be m1 + m2 I get that (a) is equal to g × m1- m2/m1 + m2.
00:11:55.000 --> 00:12:07.000
I solve for the acceleration of the system by using two separate FBDs.
00:12:07.000 --> 00:12:12.000
As an alternate solution we could look at this as an entire single system.
00:12:12.000 --> 00:12:21.000
I am going to define my system now as m1, the rope, and m2.
00:12:21.000 --> 00:12:30.000
So everything inside that dotted line is part of my single system, my single more complex object.
00:12:30.000 --> 00:12:54.000
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.
00:12:54.000 --> 00:13:03.000
On m2 I am going to have a force of m2g that passes through that barrier. So we have m2g this way.
00:13:03.000 --> 00:13:10.000
Over here I have m1g passing in that direction.
00:13:10.000 --> 00:13:21.000
So over here I have m1g, again because positive y is pointing this way, that is my definition of the positive y direction.
00:13:21.000 --> 00:13:43.000
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...
00:13:43.000 --> 00:13:48.000
...so minus m2g because it is in the opposite direction of the positive y.
00:13:48.000 --> 00:13:57.000
And that must equal the mass of my system. The mass of my system is m1 + m2 times the acceleration of the system.
00:13:57.000 --> 00:14:15.000
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.
00:14:15.000 --> 00:14:25.000
The same thing I had before, but just an alternate approach -- analyzing the solution as a whole.
00:14:25.000 --> 00:14:28.000
Lets talk a little bit about elevators.
00:14:28.000 --> 00:14:35.000
For some reason physicists seem to love the concept of putting a scale that measures an objects weight in an elevator.
00:14:35.000 --> 00:14:38.000
I do not really know why, they just seem to love it.
00:14:38.000 --> 00:14:42.000
So let us talk about it because you may see a problem or two come up like this.
00:14:42.000 --> 00:14:44.000
To begin with, we need to talk about scales.
00:14:44.000 --> 00:14:55.000
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.
00:14:55.000 --> 00:14:59.000
So it is not reading your weight the entire time, it is reading something else.
00:14:59.000 --> 00:15:04.000
What it is really reading is the normal force that exerts on you.
00:15:04.000 --> 00:15:22.000
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.
00:15:22.000 --> 00:15:29.000
Scales read the normal force and if we put scales in things like accelerating elevators, we can get some interesting results.
00:15:29.000 --> 00:15:35.000
But we can analyze all of them with the stuff we already know using Newton's Second Law and FBDs.
00:15:35.000 --> 00:15:38.000
So let us take a look here.
00:15:38.000 --> 00:15:47.000
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.
00:15:47.000 --> 00:15:50.000
Probably scared Buddy the dog -- there might have been a little barking there.
00:15:50.000 --> 00:15:55.000
What does the scale read while it is accelerating and what does it read once the elevator has come to a complete stop?
00:15:55.000 --> 00:16:00.000
Well lets draw a FBD of our situation.
00:16:00.000 --> 00:16:13.000
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.
00:16:13.000 --> 00:16:17.000
And let us call up the positive y direction.
00:16:17.000 --> 00:16:26.000
Now Newton's Second Law in the y direction -- F^nety = MAy.
00:16:26.000 --> 00:16:36.000
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.
00:16:36.000 --> 00:16:44.000
And if we want to know what does the scale read -- what we are really looking for is the normal force.
00:16:44.000 --> 00:16:57.000
Therefore the normal force, the scale reading is going to be equal to mg plus (m) times(a) acceleration in the y direction.
00:16:57.000 --> 00:17:20.000
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.
00:17:20.000 --> 00:17:31.000
So we have 25 × 10 = 250 + 25 × 3 = 75, the scale is going to read 320N.
00:17:31.000 --> 00:17:39.000
Considerably more than Buddy's 250N weight while it is accelerating upward.
00:17:39.000 --> 00:17:42.000
What happens when it comes to rest, when it is stopped?
00:17:42.000 --> 00:18:03.000
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.
00:18:03.000 --> 00:18:26.000
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.
00:18:26.000 --> 00:18:32.000
Scales in elevators, very popular problems. So let us try and put all of this together for a few minutes.
00:18:32.000 --> 00:18:41.000
We have a truck on a hill here showing a 1 × 10^5 Newton truck, at rest, on a hill that makes an angle of 8 degrees with the horizontal.
00:18:41.000 --> 00:18:46.000
What is the component of the truck's weight parallel to the hill?
00:18:46.000 --> 00:18:56.000
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.
00:18:56.000 --> 00:19:00.000
That is going to be mg sin θ.
00:19:00.000 --> 00:19:21.000
In this case it tells us mg, the trucks weight, is 1 × 10^5N × the sin of 8 degrees or about 1.4 × 10^4N.
00:19:21.000 --> 00:19:27.000
The answer is number 3.
00:19:27.000 --> 00:19:30.000
How about a force upper ramp?
00:19:30.000 --> 00:19:35.000
We have a block here weighing 10N on a ramp inclined at 30 degrees to the horizontal.
00:19:35.000 --> 00:19:46.000
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.
00:19:46.000 --> 00:19:50.000
What is the magnitude of force f?
00:19:50.000 --> 00:19:58.000
Right away, I start thinking FBDs, let us get ourselves some help here.
00:19:58.000 --> 00:20:21.000
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.
00:20:21.000 --> 00:20:31.000
And that does not line up with our axis, so we have to do something about that -- P-FBD time.
00:20:31.000 --> 00:20:49.000
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.
00:20:49.000 --> 00:21:03.000
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.
00:21:03.000 --> 00:21:16.000
So to find the magnitude of force f, all I am going to do is write my F^net equation F^netx =...
00:21:16.000 --> 00:21:28.000
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.
00:21:28.000 --> 00:21:32.000
But we are at constant velocity a = 0, so that is all equal to 0.
00:21:32.000 --> 00:21:43.000
If I want the force f then, f equals the force of friction plus 10 sin 30 degrees.
00:21:43.000 --> 00:21:56.000
Force of friction is 3N, so that is 3 + 10 sin 30 = 5 for a grand total of 8N.
00:21:56.000 --> 00:21:59.000
How about acceleration down a ramp?
00:21:59.000 --> 00:22:06.000
100 kg block sides down a frictionless 30 degree incline as shown. Find the acceleration of the block.
00:22:06.000 --> 00:22:15.000
Lets start with our FBD.
00:22:15.000 --> 00:22:21.000
We know it is going to go down, so I will call that the x direction. There is our y.
00:22:21.000 --> 00:22:29.000
Now we have a normal force perpendicular to the ramp and we have the block's weight (mg).
00:22:29.000 --> 00:22:38.000
Mg does not line up with an axis, so just like we have been doing -- time to come back to the P-FBD.
00:22:38.000 --> 00:23:00.000
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.
00:23:00.000 --> 00:23:23.000
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.
00:23:23.000 --> 00:23:36.000
Therefore the acceleration in the x direction must be mg sin θ divided by m, or g sin θ.
00:23:36.000 --> 00:23:39.000
How cool, we did not even need mass to solve this problem.
00:23:39.000 --> 00:23:44.000
All we need to know is acceleration due to gravity, a constant here on the surface of the earth, and the angle.
00:23:44.000 --> 00:23:47.000
The mass does not make a difference.
00:23:47.000 --> 00:24:03.000
So that the acceleration in the x direction is just going to be 10 sin 30 degrees or 5m/s². Very slick.
00:24:03.000 --> 00:24:06.000
Let us take a look at another Atwood machine problem.
00:24:06.000 --> 00:24:14.000
Find the acceleration of the 20 kg mass, given that the masses are connected by a light string over an ideal massless pulley.
00:24:14.000 --> 00:24:23.000
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.
00:24:23.000 --> 00:24:27.000
Let us call that t, let us call that (t) right there and we will set them as equal now.
00:24:27.000 --> 00:24:35.000
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.
00:24:35.000 --> 00:24:40.000
Let us call this m1 and we will call this m2.
00:24:40.000 --> 00:24:56.000
So FBD for m1 -- we have m1g, down, and we have tension, up, and for m1, up is the positive y direction.
00:24:56.000 --> 00:25:09.000
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.
00:25:09.000 --> 00:25:21.000
So when I write my Newton's Second Law equations for m1, I end up with t - m1g must equal m1a.
00:25:21.000 --> 00:25:33.000
For mass 2, m2g - t must equal m2a.
00:25:33.000 --> 00:25:35.000
We do this the same way we did before.
00:25:35.000 --> 00:25:40.000
We can solve these lots of different ways, but this seems to be working for us right now.
00:25:40.000 --> 00:25:56.000
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.
00:25:56.000 --> 00:26:18.000
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.
00:26:18.000 --> 00:26:45.000
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.
00:26:45.000 --> 00:26:50.000
All right, what happens if we switch up our system a little bit?
00:26:50.000 --> 00:26:55.000
Now we have two masses, m1 and m2, connected by a light string over a massless pulley.
00:26:55.000 --> 00:27:01.000
So again, the tensions can be equal, but now one of them is on a table on a frictionless surface.
00:27:01.000 --> 00:27:04.000
Find the acceleration of m2.
00:27:04.000 --> 00:27:06.000
Let us see what we can do here.
00:27:06.000 --> 00:27:11.000
Right away I can tell that this thing is going to accelerate in that direction.
00:27:11.000 --> 00:27:14.000
So I am going to call that the positive y direction.
00:27:14.000 --> 00:27:36.000
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.
00:27:36.000 --> 00:27:45.000
And we also have m2 here, where we have m2g down, and t up.
00:27:45.000 --> 00:27:53.000
And here that is the positive y direction and for m1 that is our positive y direction.
00:27:53.000 --> 00:27:56.000
So let us write our equations -- Newton's Second Law over here for m1.
00:27:56.000 --> 00:28:05.000
I am going to look in the x direction and just say that F^net is t, which equals m1a.
00:28:05.000 --> 00:28:17.000
For m2, same idea -- Net force is going to be m2g - t = m2a.
00:28:17.000 --> 00:28:20.000
Let us add those together like we did before.
00:28:20.000 --> 00:28:33.000
Our first equation, t = m1a, and our second equation, m2g - t = m2a.
00:28:33.000 --> 00:28:50.000
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.
00:28:50.000 --> 00:29:05.000
Or if I want the acceleration of the system, which will be the acceleration of m2, a = g × m2/m1 + m2.
00:29:05.000 --> 00:29:14.000
Slightly different problem, but we solved it the same way, using those same skills, those same tools.
00:29:14.000 --> 00:29:18.000
What happens if we put our masses and pulleys on a ramp?
00:29:18.000 --> 00:29:22.000
We are getting a little bit more involved every time.
00:29:22.000 --> 00:29:34.000
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.
00:29:34.000 --> 00:29:49.000
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.
00:29:49.000 --> 00:29:54.000
So let us start by drawing our free body diagram for m1.
00:29:54.000 --> 00:29:59.000
It is on a ramp, so let us call that our positive direction.
00:29:59.000 --> 00:30:19.000
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.
00:30:19.000 --> 00:30:26.000
Right away again, we should be thinking P-FBD because m1g does not line up with the axis.
00:30:26.000 --> 00:30:33.000
So let us do that right here. There we go.
00:30:33.000 --> 00:30:57.000
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.
00:30:57.000 --> 00:31:11.000
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.
00:31:11.000 --> 00:31:25.000
So this, Newton's Second Law equation is easy. F^nety is going to be equal to m2g - t which is m2a.
00:31:25.000 --> 00:31:28.000
Over here, we have a little bit more work to do.
00:31:28.000 --> 00:31:44.000
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.
00:31:44.000 --> 00:31:56.000
If I rearrange this a little bit, t = m1g sin 30 degrees + m1a.
00:31:56.000 --> 00:31:59.000
All right. Well, I am going to do this one a little bit differently.
00:31:59.000 --> 00:32:19.000
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.
00:32:19.000 --> 00:32:34.000
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.
00:32:34.000 --> 00:32:50.000
Or a = g × the quantity m2 - m1 sin 30 degrees/m1 + m2.
00:32:50.000 --> 00:32:57.000
We are just extending what we have been doing to slightly more complicated situations.
00:32:57.000 --> 00:33:00.000
Let us try one last more to round all this out.
00:33:00.000 --> 00:33:03.000
Let us go back to our elevators problem.
00:33:03.000 --> 00:33:14.000
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?
00:33:14.000 --> 00:33:19.000
Remember what we are really looking for here is the normal force at scale.
00:33:19.000 --> 00:33:28.000
Well, FBD for Darryl the Duck, we have mg down, which is 230N -- we know his weight.
00:33:28.000 --> 00:33:32.000
We have the normal force or the force of the scale up on him.
00:33:32.000 --> 00:33:38.000
Let us call down our positive y direction.
00:33:38.000 --> 00:33:49.000
Net force in the y direction then is going to be mg - the normal force and that must equal ma.
00:33:49.000 --> 00:34:10.000
Therefore, the normal force must equal mg - ma or normal force = mg (230N) - ma in this case -- well, we do not know his mass.
00:34:10.000 --> 00:34:23.000
But we know mg = 230N, o if mg = 230, then m must be 230/g or 230/10 which is 23 kg.
00:34:23.000 --> 00:34:33.000
So mass, 23 kg × the acceleration and since it is down and we call down positive -- that is a positive 3 m/s².
00:34:33.000 --> 00:34:42.000
So the normal force then, 230 - 23 × 3 or about 161N.
00:34:42.000 --> 00:34:50.000
So his typical weight is 230N, but as the elevator accelerates down underneath him, he feels lighter for a second, the scale reads less.
00:34:50.000 --> 00:35:00.000
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.
00:35:00.000 --> 00:35:03.000
Imagine you are on the bottom floor and the elevator jolts up with you in it.
00:35:03.000 --> 00:35:07.000
Don't you feel heavy for a second, like you are being compressed into the bottom of the elevator?
00:35:07.000 --> 00:35:11.000
That is when the scale reads more than your typical weight.
00:35:11.000 --> 00:35:22.000
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.
00:35:22.000 --> 00:35:24.000
Hope you have gotten something great out of it.
00:35:24.000 --> 00:35:27.000
Thank you for watching Educator.com and make it a great day!