Welcome back to another AP physics, one video in our second unit of dynamics.
This video will encapsulate Common Force applications that you will counter in the exam.
The first situation would be inside an elevator.
Let's start by labeling some forces on this diagram.
We have the force of gravity going down and our person is in contact with the elevator floor.
There's a formal exert, a normal force upwards.
But what is the quantity of this normal force?
Well, remember, diagrams and in second life have altogether.
Let's go back to the second law.
And that force or in this case is a net force minus the force of gravity is equal to the mass times
deceleration.
Now let's specify the question.
So let's say the elevator is moving at a constant speed, either upward or downward.
It says elevators at a constant velocity.
That means deceleration doesn't change as does zero.
Therefore, the net force of the normal force minus the force of gravity is equal to zero.
We can then solve for the normal force, which is equal to the point, the gravity or the mass times
acceleration due to grab G.
Now we have the same situation inside an elevator.
Let's say the question specified a different constraint.
This time I see the elevators moving up and is speeding up.
I would mean the separation is positive.
There may be other wordings of positive acceleration.
We'll be using this description, for example.
That means the normal force of gravity is equal to the best times, positive acceleration.
You know, the forces of gravity to the other side and expand the force of gravity to MGE, which ends
up as normal force equaling the mass and the quantity of a plus G.
So then normal fourth is equal to justice.
Keep in mind that the normal force is greater than what we calculated earlier.
Under constant velocity, since the overall acceleration is upwards, I mean, the net force is upwards.
That means the upward era of normal force has to be greater than the downward force, the gravitational
force to look like this in the end.
I have one more scenario for elevators, so let's revisit the previous setup this time, let's say the
elevator is moving up but slowing down.
So in other words, it's experiencing negative acceleration.
So in that fourth, which is the normal course, what is the force of gravity is equal to negative M.A.
We can follow normal force, which ends up as normal force, equal to the mass times, the quantity
of G minus eight.
So this would be the final value for normal force.
Now, notice that normal force is less than when the elevator's under constant velocity.
Since our overall acceleration is downwards, that means the net force is downloads.
So then the downward arrow would be greater upward arrow.
So look, something like this in the end for a proper diagram of the elevator scenario.
Now for a summary of elevator force applications here on the left, again, no celebration, no real
force equal to the force of gravity in the middle under a positive acceleration or upwards of celebration.
Normal force is greater than the force of gravity.
And here to the right, under negative acceleration when it's going downward.
Normal force is less than the force of gravity.
It's not required to memorize these relationships.
But you know why this is the case and how to derive these relationships if needed.
Now, for applications with the pool, there will be two common instances of Kohli's, which is shown
in this image where there will be both systems and they're under a rotating device called a pole.
Another common scenario is this image or one and is on the table and the other end is suspended in and
through a pole here.
Let's take a look at the first scenario, which I'll call a vertical problem, let's say in this scenario.
He is heavier than a.
So it will result with the pulley rotating clockwise, with B, falling downward and a rising.
Essentially unravel this system, a pull you end up looking like this.
So keep this in mind.
But let's first draw the forces acting on the system.
Both are suspended in the air by force and are pulled downward by the force of gravity.
So we will able to measure such and we can transfer these over to our unraveling pulley system.
In doing so, we see that the forces will look like this.
These gravitational force towards the right age, gravitational force towards the left, and the tensions
towards the left and right respectively.
Now, keep in mind, that is towards the right when unravel such as this.
Now let's look at the forces tension and tension, we are the same as they are from the same court.
It's good to know that from the same court of stream, that means the force will have the same tensions.
And here's its and be are in the same court or different directions.
That means they cancel out so we can proceed with new and second law or detentions cancelled out.
That just leaves a two gravitational forces left in the net force.
Since they're in the same direction, we can say that they contribute to the same net force.
So in the end.
B's positive gravitational force is going to the right, minus the negative, a gravitational force
and force left as it goes to the mast and the celebration.
Can you expand this to say MBG and agy is it would be a massive plus Mesabi Times of celebration.
So this equation will represent the forces in a police system.
But what about an object yourself, such as blocking individually?
Let's take a look.
Here's blockade's diagram.
Here's the question we found earlier.
We can solve for acceleration by dividing M.A plus m.v for both sides.
So acceleration is equal to this quantity here.
Let's revisit the forces on the diagram.
So starting with the top one.
Apartheid, divorce, minus the downward force of gravity is equal to the massive eight times the celebration,
no celebration without earlier was part of the system, but the system's acceleration applies to each
individual object within the system.
So the system's acceleration would be acceleration for blocking as well.
The next slide, we add the force of gravity on aid to both sides and we expand as shown here.
Next up, we sent this to a into the equation for air here to get this equation, ashow.
And lastly, we factor out energy from the equation to get engine is equal to EMAG and the quantity
of and B minus, I may over emphasize and B plus one.
Now, it's not required to memorize again, but understand how an equation like this was derived and
understand how a system's acceleration applies to each object in the system as well.
Now, for the other police situation, as shown in this image and alcohol with the horizontal body.
So let's see.
For calculation purposes, Asab pulls the system downward so ends up moving like this.
So as usual, restaurant free body diagrams walk as a normal fast upwards apostrophe downwards.
And the tension from the string pulling towards the right or block B. There's the force of gravity downward
and the tension upward.
So let's unravel the pole, which would make the diagram look like this.
Notice that the normal force and the force of gravity on block eight are shown as they're not in the
same direction as Napoli's acceleration.
So they won't be part of the net force that will use later.
So here, since Tetuan, A and B are part of this, in part, they're equal in magnitude opposite direction,
so they cancel out.
So let's use Nunavik second law with what's like the only remaining force, which is be the force of
gravity on block B.
So that would be equal to the net force, which is equal to the net mass of the system.
Massive eight plus B times acceleration.
And expand this to say, MBG, we go to my boss and be all times acceleration.
And that would be the equation representing the horizontal pulley.
Now for applications on an incline, whether the object slides and it's all right down the front.
So once we draw a first time around, it would look something like this where the force of gravity going
downwards, as always, friction force opposing the motion, which would be up the incline and the normal
force and angle perpendicular to the surface.
Remember, a force is perpendicular to the surface.
No normal force would be an angle to end it and climb like this.
We would also split the force of gravity.
Here we can use trigonometry to say about this component of gravity is equal to F.G. sign theta.
And we can also say the other component of gravity is equal to, oh, sandpaper.
Now we can rotate the diagram so that the incline.
It's actually the horizontal, which we make our diagram looks something like this.
Now we can balance the forces with second law in the Y direction.
As the object doesn't move that way, the net force is equal to zero.
So the normal fourth minus the force of gravity cosine theta are equal to each other.
So normal forces, you would MGE close eye theta.
The X direction at the side of gravity, minus the force of friction, which is equal to the mass times,
the celebration.
Keep in mind this inclined extraction the same direction as the objects sliding down incline.
It's just that we imposed it differently so we can expand this equation as they decide that minus mumu
COSAC data is equal to me and dividing by massive boaties, we can say, gee, Cybernet minus Chico's,
I'd be as equal to the celebration.
So the final wrap up unit to memorize what all of the industry lost and implies, also try to memorize
each type of force and has calculator, but be wary of normal course and tension force being depending
on the scenario.
Practice recognizing the forces, acting on a system and drawing them on a free body diagram.
And lastly, understand the various force applications.
The next video will go our practice questions where I go over questions from the student.