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.

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来自 University of Houston System 的课程

Preparing for the AP Physics 1 Exam

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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.

从本节课中

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 the last module, we discussed the topic of kinematics,

the study of how objects move.

In this module, we're going to take a look at dynamics.

That's the study of why objects move.

Much like the gravitational force found in freefall, we will see that the forces on

an object have a direct impact on the acceleration that object experiences.

The relationship between force and

motion can be understood through Newtons three laws.

Newtons first law is called the law of inertia and it's often stated like this.

An object with no net force acting upon it will remain at rest or

move in a straight line motion with a constant velocity.

This tendency for an object to maintain a constant velocity, or

stay at rest, is what we call inertia.

This statement may conflict with your daily life experience.

A common misconception is that a net force is required to

keep an object in motion at constant speed in a straight line.

Let's be clear here what we mean by a net force.

Yes, many forces can be present on an object when it moves at constant velocity.

For instance, a coffee cup being pushed across a table in

a straight line at a constant speed.

But the net force, that being the total force on the coffee cup is zero,

as you can see in this diagram.

The push from the person is balanced by the friction in the x direction.

The force of gravity is balanced by the normal force from the table in

the y direction.

Because the net force is zero in each direction,

the object is not accelerating in either.

While Newton's first law describes the motion of an object when no

force acts upon it.

Newton's second law, instead,

describes scenarios in which there is a net force present on the object.

This is usually summarized by the equation,

sum of forces equals mass times acceleration.

From this equation we can see that a net force on

an object will cause it to accelerate.

And therefore, change the magnitude or direction of its velocity.

Suppose for instance, that a box sits at rest on a horizontal frictionless surface.

While at rest, all the forces acting upon the box are balanced,

like the coffee cup example from earlier.

Now a rope is attached to a two kilogram box and

pulls with a tension of 20 Newtons to the right.

The box will begin to move towards the right with an acceleration of 10

meters per second squared.

The net force acting on the box is what caused the increase in

the objects velocity from zero.

In fact if we doubled the tension in the string from this example,

the box would change its velocity more quickly.

A tension of 40 Newtons to the right would have caused the box to

experience an acceleration of 20 meters per second squared to the right instead.

In other words, doubling the force would double the acceleration.

Notice that the direction of the acceleration of the box is

the same as the net force.

>> Newton's third law describes how objects interact through forces and

can be stated as follows.

When object A exerts a force on object B, an equal force is

exerted from object B back on object A in the opposite direction.

Notice the two objects are necessary for a force to exist.

The two equal but opposite forces exerted during this

interaction are called a Newton action-reaction force pair.

Suppose that a coffee cup is at rest on a table.

Since it is not acceleration, we know all the forces on this object are balanced.

By this, we mean that all the forces vertically upward on the object,

equal all of the forces vertically downward on the object.

Similarly, all the forces left equal all the forces right on the object.

This is true along any axis.

Let's draw the force vectors in this scenario for

each of the Newton force pairs acting on the cup.

Gravity pulls downward on the coffee cup with a force of mg.

The coffee cup will exert an equal and

opposite force of mg up on the Earth itself.

Notice that through their mutual attraction, due to the force of gravity,

the coffee cup does actually pull upward on the Earth with a force of

equal magnitude.

The coffee cup also experiences a force upward from the table

called a normal force.

This means that the coffee cup must exert an equal and

opposite force downward on the table.

We have now drawn two separate force pairs.

Careful not to make the mistake of assuming that the Newton reaction force

to mg on the cup, is actually the normal force from the table.

Yes, these forces are equal in magnitude and oriented in opposite directions.

But the normal force acts between the table and

the cup, not the Earth and the cup.

Since these forces interact between different objects,

different pairs, they cannot be a Newton action, reaction force pair.

Let's look at a quick example which an understanding of

Newton's Third Law is necessary.

A small car of mass 1500 kilograms is traveling to the right when it

crashes head on into a large truck of mass 3000 kilograms, traveling left.

If the small car experiences an average acceleration of 200 meters

per second squared to the left while they are in contact, calculate the average

acceleration experienced by the large truck during the collision.

In this problem,

we have a small car travelling with an initial velocity to the right when

it collides head on with a larger truck with initial velocity to the left.

When they collide, however, they interact through a pair of forces.

Labelling those forces, this is a good case of the Newton force pair of equal and

opposite reaction action force pair.

Where in this case,

the smaller car will push to the right with a force on the large truck.

At an equal but

opposite force the truck will push on the car with an equal force to the left.

So it gives us information about the car and

wants to know more information about the truck.

So let's look at the car a little more closely.

I'm going to write out my equation.

My fundamental principle here.

The sum of forces equals mass times acceleration for that car.

We know that there is a force on that car from the truck and now will equal then.

And that's the only force in the horizontal direction on the car.

We ignore for the moment all the forces on the truck.

The mass of the car times its acceleration will be on the right-hand side of

our equation.

So subbing in my numbers, we know the mass of that car to be 1500 kilograms.

And the acceleration to be 200 meters per second squared to the left and

I made it negative because that acceleration is directed to the left.

Solving for force, 3 times 10 to the 5th Newtons will

be the magnitude of that force on the small car.

And I will make that negative because it is a force directed to the left.

Now that I know that information that will help me solve with

some information about the truck as well.

I can starting with my fundamental principle F equals ma.

The sum of all the forces on the truck will equal its

mass times its acceleration.

Again, the only force on it would be the one from the small car and

that should equal its mass, the mass of the truck, times its acceleration,

which is what we are looking for.

Understanding that the forces are equal means that I can sub in 3 times 10

to the 5th newtons, and I am subbing that in as

a positive because like we noticed up here the force on that truck is to the right.

It's in the opposite direction as it was on the car.

Equals the mass of that truck in this problem,

3000 kilograms and I'm solving for the acceleration.

Moving down just a hair so that I have a little bit more room.

The acceleration that that truck then experiences is going to be

a positive 100 meters per second squared.

It accelerates to the right because of the force on the truck was to the right.

Notice also, that the acceleration of the truck,

here, is less than the acceleration of the car.

Even though the forces were equal, they can exhibit and

experience different accelerations because the masses of the objects were different.

>> So far, we have drawn forces at various points in the lesson.

But let's take a moment to discuss best practices when drawing free body

diagrams in physics.

Here are some of the forces that we will encounter in this module.

Other variable names are acceptable, but

these are the representations that we will use throughout the course.

On the AP exam, there are very specific rules when drawing a free body diagram.

One, draw a vector starting from the center of the object and

pointing in the direction of the force.

Two, the length of the vector should roughly correspond to the magnitude of

the force.

A force that is twice as large as another should have a vector arrow that is

twice as long.

Three, do not break up forces in a Free Body Diagram into components.

Leave only the total vector in your diagram.

Four, label all forces with appropriate variable names,

such as the ones listed previously.

Free body diagrams can be very helpful in solving problems involving

multiple forces.

Here's an example.

A 10 kilogram box is at rest on a horizontal surface when a 30 Newton

tension applied to the left at an angle of 40 degrees above the horizontal.

Calculate the net force exerted on the box.

All right.

So this question wants me to solve for the net force that is exerted on this box.

Before I solve for that, I'm going to go ahead and draw out my free body diagram.

Start from the center of the object.

And the first vector here, the first force that I have is my gravitational force

which is equivalent to the weight, and the weight, of course, equals mg.

The next one that I'm going to draw is my upward force, normal force.

Which is the force exerted on this object by the surface.

The question tells me I have an applied force towards the left at an angle,

so I have my F applied here.

It's at an angle of 40 degrees.

And now we're ready to solve this question.

So, for net force, let's look at this in components.

I'm going to first look at the vertical component.

Now, unless the question tells you otherwise,

you're going to assume that this object is not being lifted up from the surface.

What that means is that when I am looking the sum of my vertical courses,

I know that it's going to equal mass times acceleration, F equals ma.

But because it is not moving vertically my acceleration equals 0.

This means that my net forces in the vertical direction equals 0.

They are balanced, there's no movement occurring vertically.

Horizontally, however, let's go ahead and look at what our net force is there.

Some of my horizontal forces equals ma.

Now, if this question were asking me to solve for acceleration, I would figure out

what my net force is horizontally, come back and solve for the acceleration.

Right now though, I'm only concerned with net force.

So let's go ahead and solve for that.

[SOUND] The only force that I have acting is

the horizontal component of my applied force.

So notice the applied force was acting at an angle.

The applied force here would be 30 newtons.

And in order for me to figure out the horizontal component I

need to do basically the component method that we had discussed the kinematics unit.

I'm solving for the adjacent vector so I'm going to use cosine.

So this would be F applied cosine of theta.

However, notice it's a vector.

Direction is important.

This is point towards the left.

So I have negative F applied cosine theta,

which gives me negative 30 cosine

40 degrees, and when I solve for

this I end up with negative 20 Newtons.

So my net force, being in the horizontal direction,

is negative 20 Newtons or 20 Newton directed left.

One thing I want to point out is