This course is designed for high school students preparing to take the AP* Physics 1 Exam. * AP Physics 1 is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product.

Loading...

Из курса от партнера University of Houston System

Preparing for the AP Physics 1 Exam

36 оценки

This course is designed for high school students preparing to take the AP* Physics 1 Exam. * AP Physics 1 is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product.

Из урока

Forces

Topics include forces, Newton’s Laws of Motion, and applications of each. You will watch 3 videos, complete 3 sets of practice problems and take 2 quizzes. Please click on the resources tab for each video to view the corresponding student handout.

- Dr. Paige K. EvansClinical Associate Professor

Math - Mariam ManuelInstructional Assistant Professor

University of Houston

In this third dynamics video,

we will analyze the motion of objects with primarily vertical acceleration.

Topics such as elevators, Atwood pulleys, and

surfaces combined with pulleys are all discussed in this section.

Note that for these problems, we will always set the direction

of the system's acceleration as the positive direction.

Elevator problems are a classic physics question in which the tension and

the cable, and the force of gravity cause the elevator to accelerate vertically.

For each of the scenarios listed below,

calculate the tension in the elevator's cable if it has a mass of 1,000 kg.

For elevator problems such as these, I, it's important to recognize that there

are really only two forces affecting the system as a whole for this elevator.

The tension up of the cable and mg of gravity pulling it down.

So, for part A, for instance, I'm going to draw a quick sketch.

Here's my elevator.

There's a tension acting upward and energy acting downward.

As usual for these vertical motion questions,

the direction of positive is an important question here

because some things could be accelerating up while others accelerate down.

And so our rule of thumb, in problems like this,

is to always set the direction of its acceleration positive.

Now, I notice in part A the elevator is at rest, so there is no acceleration.

So I just going to go with our default by making up positive.

It wants me to solve for

the tension, so I'm going to start with Newton's second law.

Sum of forces on this elevator should equal the elevator's mass

times the elevator's acceleration.

And like we said, there are two forces acting on this elevator,

which should be balanced as I drew in my diagram here.

So, tension was in the positive direction, so it's positive.

Minus mg, which was in the negative direction.

Those are all the forces acting on this thing in the y direction.

No forces to worry about in the x direction.

Should equal the mass of this elevator times its acceleration.

Solving for tension then, I get mg plus ma.

Now like we said a second ago, well this elevator wasn't accelerating.

So the a in the y direction here would be zero, if the elevator is at rest.

That means this term will be zero, and this term will be zero.

So our tension just equals the mass of the elevator

times acceleration due to gravity.

Or in this case, our elevator had a mass of 1,000 kilograms.

Acceleration of gravity was ten.

Notice again, I've already taken care of the negative signs in our setup.

When we set up F equals m a at this point.

Which means that when I finally get to the substitution step,

I'm subbing all my values in as a positive.

I do not want to double count here with my negative signs.

So the tension in the cable would be 10,000 Newtons

pulling that elevator up, equal to the weight.

That's why it's at rest, the two forces are balanced.

Let's try B now.

B says that the elevator's moving upward at a constant speed

of 3 meters per second.

This will look very similar to the scenario we just drew.

My elevator still has a tension upward and an mg down.

Again, if it's going at a constant speed, in this case,

three meters per second upward, it's not accelerating either.

So when I set up my sum of forces in the y direction, equaling the mass

times acceleration, for this elevator, the acceleration will again be zero.

I will have a tension up, positive.

A minus mg down, negative, equaling 0.

So again, we will have a tension equaling mg, or

in this case, 10,000 Newtons.

You cannot see a difference in the math, in the equations,

between an object at rest or moving in a straight line, and

a c, as a con, at a constant velocity like part B.

Let's try part C, because that does change the scenario a little bit.

In part C, the elevator's accelerating upward at a constant rate of three meters

per second squared.

Moving back down.

Part C gives me an elevator.

Now, thinking about the scenario, if it's accelerating up,

there needs to be a larger force upward on this elevator than there is down, so

I'm drawing my vectors to represent that information.

We said that we're going to make our positive direction,

the direction it accelerates.

So this elevator accelerates up, and that's going to allow for

our F equal ma equations to always have a positive acceleration on the right hand

side of the equation, which will become a lot more important when we have two or

more objects like an Atwood pulley later in this module.

So, working through this, sum of forces equals ma,

always where I start these problems.

Again I have two forces, a positive tension because it points in our

vertically upward direction, or positive direction, minus mg will equal ma.

And this time, we do have an acceleration.

It accelerates upward at, notice above, 3 meters per second squared.

So, solving for tension like before, we're going to get an mg plus an ma.

And I'm going to sub in my numbers for

this, and notice we will get it's weight plus some information,

some number here, to get a larger tension than we got earlier.

We're going to have 1,000 for the mass, acceleration due to gravity of ten,

plus 1,000, and three meters per second squared was our acceleration.

And so, solving this by multiplying this out, I get 10,000 plus 3,000.

13,000 Newtons is the tension on this elevator,

which makes sense to me because if you have 13,000 newtons up and

only 10,000 newtons down for a weight,

that's why this elevator accelerates upward in the first place.

You have a net force on this object upward.

The last part of this problem asks, the elevator is accelerating downward at

a constant rate of three meters per second squared, calculate the tension.

And so, for the last portion, D, again we have our elevator.

This time the tension will be less than the weight of the elevator, and

now it accelerates down.

This is where I need to make downward my positive direction so

that when I set up my sum of forces equals ma, for

this elevator in the Y direction, my tension is now a negative tension.

It points at our negative direction, which may seem odd, but it will allow our

acceleration on the right hand side of this equation to only be positive.

Plus an mg, points in our positive direction, equals a positive ma,

the mass of this elevator times its acceleration.

Solving for tension, I'm going to move tension over to the right,

ma over to the left.

I get mg minus ma.

And now I will sub in with my numbers for that as well.

1,000, g is ten meters per second squared from the center of the Earth.

Again, subbing in everything as a positive value at this point.

1,000, and the acceleration, 3.

Again subbing in as a positive value,

we've already taken care of that negative direction, equaling my tension.

Notice that our tension is now 10,000 minus 3,000.

It means that our tension will be less than the weight of the elevator and

it will accelerate down.

In this case, 7,000 Newtons is the tension in the cable.

Meaning that, like we said earlier, we have a net force down.

That's why it accelerates down.

>> An Atwood machine consists of two masses connected by a string

over a pulley.

When released, the larger mass will accelerate downwards,

while the smaller mass while accelerate upwards.

The magnitude of these accelerations will be equal when the string is inextensible,

meaning it does not stretch.

In this example, two masses are connected by an inextensible string over a massless,

frictionless pulley.

The mass on the left is eight kilograms while the mass on

the right is five kilograms.

Calculate the following if the system is released from rest.

The acceleration of the system.

The tension in the string.

The speed of the eight kilogram mass after it has fallen 1.5 meters

to the ground below.

So in this problem over here, let's go ahead and look at the blocks for

the masses that are attached to this pulley over here.

I've got the eight kilograms, I'll label that as my mass one, and

five kilograms mass two.

The force is acting on this.

Well, I've got m1g, and m2g.

They are the downward forces center of the Earth, since this is force of gravity.

I also have tension over here,

from the string.

However, because this pulley is massless, this is the equivalent tension over here,

tension one on each side so it can cancel out and

I'm not going to use this in any of my calculations.

Let's go ahead and solve for the question.

First part asks for acceleration.

So again, I'm going to look at the whole system.

[SOUND] Set the sum of my forces equal

to mass times acceleration.

Now let's consider which block is heavier.

So I've got eight kilograms versus five,

which means that eight kilograms has greater weight, and

it's going to accelerate in the direction of that one, being pulled downwards.

So here is the direction of my acceleration, five kilograms

will be moved up, whereas the eight kilograms is going to be moved downwards.

I'm going to consider the direction of my acceleration as the positive,

so I'm going to label that as my positive direction.

So, in this case, I have m1g.

Notice how m1g is in the same direction as acceleration

which means that m1g is positive.

How about m2g?

If we notice, acceleration over here is pointed upwards.

This, lighter block wants to move upwards, however, m2g pointing

towards the center of the Earth is downwards, it's in the opposite direction.

So I've got negative m2g.

And now I can go ahead and set this equal to m1 plus m2,

multiply it by acceleration.

And I can solve for

the acceleration,

m1g minus m2g,

over m1 plus m2.

Plugging in values, 8 multiplied by 10 for

gravitational acceleration, minus 5, multiplied by 10.

That's again the g and mg, divided by the masses added up together, 8 plus 5.

This gives me an acceleration of 2.31 meters per second squared.

Let's go in and see what part b asks us for.

So the next part here asks us for the tension in the string.

Remember earlier I did not use the tension in the string, however, I can solve for

this now.

I'm going to use

mass one as I solve for this tension.

[SOUND] So with my forces in the vertical direction equal ma.

So if I'm looking at mass one, let's consider if tension is positive or

negative.

Notice how this tension is pointing upwards.

It's pointing in the opposite direction of the acceleration, which is downwards.

The direction of acceleration is positive, so I have negative

tension plus

m1g equals ma.

So now I can go ahead and solve for this tension [SOUND].

Tension equals ma minus m1g.

Let's keep in mind this tension is still negative over here.

So I'm going to go ahead and plug in my numbers now, m1 is 8.

The acceleration there is 2.31 minus, again, 8.

You could have factored out the mass if you would've wanted to.

Multiplied it by g, which is 10.

Solve for this.

And I end up with negative 61.52

Newtons which means my tension

equals 61.52 Newtons.

Okay?

Let's go ahead and see what they want us to solve for next.

The next part asks for the speed of this eight kilogram mass

after it has fallen 1.5 meters to the ground below.

Again, I'm going to use my kinematics equations.

So in this case I don't have time and I don't need time.

I can go ahead and use the equation Vf squared

equals V initial squared plus 2ax.

I'm looking at everything over here with the acceleration that I've solved for and

received.

I can solve for my final velocity.

However, what about initial velocity?

Well, this object started from rest.

That tells you my initial velocity is zero.

So I have square root, 2ax is all I'm left with.

[SOUND] Plug in numbers now.

Root 2, multiplied by 2.31,

multiplied by 1.5.

Solve for this, and

I end up with a final velocity of 2.63 meters per second.

[SOUND] Now let's combine what we have learned about planes and pulleys.

A 10 kilogram mass is placed on a horizontal table, as shown here.

A 15 kilogram mass is connected to the left side of this block by a string and

suspended over a pulley.

A 20 kilogram mass is connected to the right side of this block by a string, and

is also suspended over a pulley.

All pulleys are frictionless and massless, and

the string is massless and inextensible.

Calculate the acceleration of the system.

Calculate the tension in the strings.

Calculate how far each mass will travel

in 0.8 seconds if the system is released from rest.

So, as I usually do,

I will start this problem off by drawing a diagram of the scenario.

It tells me that there's a mass on a horizontal table,

and that there are pulleys at either end,

each one with a hanging mass attached on either end of this block.

It then tells me, calculate the acceleration of the system, the tension in

each string, and how far each mass will travel in a specific amount of time.

So, before I start anything else, I always do the same thing.

I'm going to draw my forces and my free body diagrams.

I'm going to also figure out what direction is the positive direction.

So, let's see.

Let's call this m1, m2, and m3.

Also note that it tells me that the one on the table is 10 kilograms.

The one hanging off the left hand side, 15 kilograms,

and the one hanging off the right hand side, 20 kilograms.

So through your experience looking at a scenario like this,

you could probably agree that that mass three will fall downward,

while mass one will be pulled up because of the imbalance of the masses there.

Next issue is on positive direction.

So since we set that, that means that the mass that was on the left-hand side,

upward will be its positive direction, because it will accelerate up.

It's the lighter mass.

The one on the table will accelerate right, and

the one on the right hand side will accelerate down.

Next, I'm going to draw all of the forces on each of our masses.

So notice on mass one, on the left hand side, we'll start there.

M1g, gravity pulls that object straight down, more to the center of the earth.

The string, tension 1, will pull up, because that's what strings do.

They can only pull.

They can't push, and they always pull in the direction that string is oriented.

That's why when I get over here to mass two,

there will be a T1 pulling left on that mass.

On this mass two now, there is another string.

It's a completely different string.

T2 will pull this right, but it will pull mass three up.

Other forces on mass 2 will be m2g down,

and a normal force up.

M3g down on this third mass, will pull that down.

So I have all of these different forces.

Also notice in this problem it doesn't tell me that the table is frictionless.

Let's go ahead and assume in this problem that the table is frictionless,

to solve for the acceleration of the system.

We could, of course, include a friction which would point to the left on m2, okay?

So, to find the acceleration of the system,

you could set up an f equals ma for each individual mass, and

have three equations that you will then solve as a system of equations.

There is, of course, a shortcut by setting up your equation for

the whole system, which is what I will do now.

For the whole system,

as you've seen us do before, I'm going to set up sum of forces equals ma.

And I'm going to include all of the forces that act on the whole system, that move

it in the direction of motion, and then I will include on the right-hand side,

the total mass of the system times the acceleration.

And notice that if all three parts are connected, they all have to move together,

they all have to move with the same acceleration.

That's what that inextensible part of the strings meant.

And so, I'm going to start adding these together.

Let's start with m1, I notice that I have an m1g, and

that's a negative force because it points in our negative direction for that mass.

Then I have a positive T1 because it pulls upward,

which is our positive direction for that mass.

Those are the only forces impacting the motion for that mass.

So let's move over to mass two.

On mass two I have a tension to the left, that's the negative direction,

T1, for mass two.

And the I have a positive T2 pulling to the right,

which is our positive direction.

M2g in the normal force, balance out in the y direction and do not speed up or

slow down this object at all in the x direction.

So I don't need to include that here.

Going over to m3, I will have a negative T2, it pulls upward on mass three,

which is our negative direction for that mass, and then a positive m3g on that

right-hand side because it pulls it down, the positive direction for m3.

All of those forces added together.

Now I'm using the total mass of the system, because I've added all the forces

on the system, times the acceleration that they all share.

Notice like several problems before,

the tensions cancel because they are within the system.

Once these forces are within the system and

we've now looked at a bigger picture here, the whole system,

those tensions cancel out, and they will do so every time.

You do not need to write those down when solving a problem like that.

Feel free to skip that step.

So I'm left with something like this now.

I'm going to rearrange it just slightly, m3g minus m1g.

What would equal m1 plus m2 plus m3.

Find the acceleration.

If I wanted to solve this in variables for the acceleration,

m3g minus m1g over m1 plus m2 plus m3.

And I can my sub in values for each of these.

For instance, I know that mass one, looking up at my diagram to

remind ourselves, mass one was 15, mass three was 20.

So 20 kilograms times my acceleration of gravity,

again all positive values, minus 15 times 10.

And then adding all my masses together.

My mass one plus 15, my mass two

was 10, my mass three was 20.

And when I go ahead and divide my numerator by my denominator,

I get an acceleration of the system of 1.11 meters per second squared.

It's hard to say direction here because we'd have to talk about

each individual mass.

The acceleration of M1 would be vertically upward.

The acceleration of M2 would be 1.11 meters per second squared to the right.

While the acceleration of m3 would be 1.11 meters per second squared

down, vertically.

That's what the first part of the question asked.

Now it says, calculate the tension in each string.

Well, to do so I need to break up my system

because our tensions cancelled when we did the whole system.

So I'm going to set up an equation for mass one.

That will help me solve for T1.

I'm going to set up my equation for mass three.

That will help me to solve for T2.

Again, labeling my work so I don't get lost.

This is for Mass 1.

Setting up sum of forces equals ma, for just Mass 1.

Well, on Mass 1 I have a positive T1 because it pulls up a minus

m1g because it pulls down, that's our negative direction, equals m1a.

We only added the forces from m1, on this equation, here.

So, T1 will equal m1g plus m1a.

Notice that this looks a lot like the equation from earlier, for

an elevator accelerating upward.

Which it should, because it's just an object with a tension and

a weight accelerating up.

So, the mass of our number one, 15 kilograms,

an acceleration of ten, plus the mass

of this object again, 15, times the acceleration which we said was 1.11.

My tension 1, that is pulling up

on Mass 1, is 167 Newtons.

One last step then, to find the other tension.

I'm now going to set up my

equation for Mass 3.

Again, I will start with sum of forces equals ma.

Don't forget about the positive negative sign difference.

So my T2 will be negative, it points upward, plus m3g.

Those are the two forces on m3, times deceleration.

Solving for T2,

I get m3g minus m3a equals T3.

And remembering that the Mass 3 was 20 kilograms.

20 times 10 minus 20 times 1.11 meters per second squared.

Again, something all that is positives,

we've already taken care of everything to do with direction and negative signs.

That will give me T3.

In this case, T3 equals 178 Newtons.

The two tensions were different.

In fact, we can actually have a quick check here, because notice that the block

that was on the table had two tensions, T2 to the right and T1 to the left.

But that block needs to accelerate to the right.

By comparing our two tensions, I can see that T2, I'm sorry, this was called T2,

let's go ahead and change that real quick, to be consistent throughout the problem.

T2, T2, T2.

But that T2 pulls right, with 178 Newtons, T1 pulls left with 167.

That's why this object accelerates to the right.

Last part of the problem.

Calculate how far each mass will travel in .8 seconds

if the system is released from rest.

So this is a good kinematic equation question.

It is a constant acceleration,

nothing about the system is changing other than its movement.

The acceleration stays, this 1.11 meters per second squared the whole time.

So to find the displacement, I just need to pick a kinematic equation.

That has the displacement and the time.

So why don't we go with this version?

X naught plus V naught t plus one half a t squared.

It tells me it starts from rest.

It's pretty customary to call the position you start at as 0.

And so, let's say we're talking about the object on the left hand side moving up.

They all have the same displacement.

But setting that equation up for the object on the left,

one half the acceleration for that object.

It's not in free fall.

It's not a negative 9.8.

We just said it accelerates upward, so a positive 1.11 meters per second squared.

It wants to know, where's the object located .8 seconds later?

And so I need to square that value.

When I solve for this, I notice that the displacement of the object on the left,

is 0.355 meters upward.

While the displacement of the block on the table would be 0.355 meters to the right.

And the block on the right hand side, 0.355 meters downward.

One reason why this question wants to know how far, far implies to me distance,

not displacement.

So I would just give the number 0.355 meters for my distance.

Coursera делает лучшее в мире образование доступным каждому, предлагая онлайн-курсы от ведущих университетов и организаций.