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Dynamics B Subtitled Better
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Preparing for the AP Physics 1 Exam
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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.
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Dr. Paige K. EvansClinical Associate Professor
Math
Mariam ManuelInstructional Assistant Professor
University of Houston
Newton's second law can be applied to various scenarios and systems.
We will start our analysis here by looking at problems in which
an object accelerates across a flat surface.
This will include topics of horizontal surfaces with friction,
inclined planes, and multiple boxes connected with strings.
Here is an example problem using horizontal surfaces and friction.
A 5 kilogram block is placed on a horizontal surface and pulled to the right
with an applied force of 30 newtons, at an angle of 20 degrees above the horizontal.
The coefficient of kinetic friction between the block and
the surface is point two.
Calculate the following, the normal force on the block,
the force of kinetic friction on the block, the acceleration of the block,
and the displacement of the block after 2.5 seconds have passed.
Before I solve for anything on this question, I'm going to go ahead and
draw out my pre-body diagram.
going to start with the weight of this object, or mg.
I'll also draw out my vertical upward force exerted on this object by
the surface, that's the normal force.
I've been told that there is an applied force of 30 newtons at an angle of 20
degrees above the horizontal, and that's directly towards the right.
[SOUND] So here is my applied force at an angle of
30 degrees towards the horizontal.
They give me a coefficient of friction.
Well, I need to draw out my friction vector.
So this object is having a force applied towards the right,
which means that it's going to move towards the right.
Friction, my opposing force would then be directed
towards the left in the opposite direction of the motion.
Let's go ahead and solve for part a.
It's asking for the normal force.
So, the normal force is a vertical force.
I'm going to go ahead and look at the sum of my vertical forces.
[NOISE] That's sum equals ma.
Well, this object's not being lifted up from the surface, which tells me that it's
not accelerating in the vertical direction, that acceleration is 0.
[NOISE] So now, it looks like vertical force is equal 0.
Let's go ahead and place in what all the vertical forces are acting on this object,
and set them equal to 0.
[NOISE] Up is considered positive, and down is considered negative.
Normal force would then be positive [NOISE], mg would be negative.
The vertical component of my F applied is going to be pointing upwards.
[NOISE] Therefore,
I have positive F applied [NOISE] sin theta.
[NOISE] All of this equalling 0.
I can go ahead and plug in numbers.
And solve for my normal force now.
[NOISE] The mass is 5.
Gravitational acceleration is 10.
[NOISE] And the angle is 20 degrees.
[NOISE] My answer for normal
force is 22.6 newtons.
Notice that the answer I got was positive.
Well, this makes sense because normal force is pointed upwards.
[NOISE] Now, let's go ahead and
look at part b.
Part b of this question asks us to solve for
the force of kinetic friction on the block.
I can go ahead and solve for frictional force.
[NOISE] So I'm trying to solve for my frictional force.
I can go ahead and do that by using the equation.
[NOISE] Frictional force equals coefficient of kinetic friction,
which is provided to us.
Multiplied by the normal force, the force being exerted by the surface.
This gives me [NOISE] 0.2 multiplied by
the answer we had just received, 22.6.
[NOISE] My frictional force turns
out to be 4.52 newtons.
And of course, I will include the direction over here.
This force is pointed towards the left.
This is for my pre-body diagram, comes in handy because I can always look back and
see which direction this vec, vector was pointed in.
The next part of this question asks us to solve for the acceleration.
Again, I can look at my pre-body diagram.
And I know that this object is moving horizontally, so
I'm going to look at [NOISE] the sum of my horizontal forces, set them equal to ma.
All the forces that I have acting horizontally,
this includes my negative frictional force.
Remember, because it was towards the left.
My positive [NOISE] force applied cosine of theta.
Recall, I'm using the horizontal component of that applied force.
[NOISE] Set all of this equal to mass times acceleration.
Frictional force negative 4.52 [NOISE] plus 30.
[NOISE] Cosine of 20 degrees.
[NOISE] Set this equal to the mass, which is 5 kilogram times the acceleration.
From here, when I solve for the acceleration,
[NOISE] I end up with 1.54 meters per second squared.
Now, I haven't put a direction down here yet, but let's talk about how I would
figure out whether this object's accelerating towards the left,
towards the right.
I can compare my two horizontal forces.
I know that the force of friction is 4.52, that's the one directed towards the left.
How about this 30 cosine of 20?
That magnitude comes out to [NOISE] 12.24.
What that tells me, is that if I compare my frictional
force to my applied force horizontally, that applied force is greater.
So this object accelerates towards the right.
[NOISE] Finally, as we're looking at part d, it asks us for
the displacement of the block after 2.5 seconds have passed.
I like this question because it's bringing back,
it's borrowing back the information from kinematics.
[NOISE] So if I need to solve for the displacement,
I'm going to use my equation x equals 1/2
at squared plus V initial multiplied by t.
Well, this object starts from rest, which means my initial velocity is 0.
That cancels out this component of the equation.
I've been provided with the time to 0.5 seconds and
the acceleration, be careful not to use gravitational acceleration.
This object's being accelerated horizontally towards
the right with the 1.54 that we just solved for.
[NOISE] So, go ahead and plug those numbers in.
[NOISE] And my displacement is 4.8 meters.
[NOISE] And that would be towards the right.
>> Now, let's try an example in which an object accelerates on an inclined plane.
Jeff pushes a 115 kilogram refrigerator up an inclined plane that has
been placed over a set of stairs.
With a constant force of 1,000 Newtons up the ramp.
The plane is 4 meters long and
is set at an angle of 32 degrees above the horizontal.
The coefficient of friction, of kinetic friction between the refrigerator and
the ramp is 0.4.
It then asks a series of questions about the forces and
acceleration of the refrigerator.
In this problem, we have a 115 kilogram refrigerator be,
being pushed up an incline plane.
Remember, a plane is a two-dimensional object.
So when it tells me that that plane is four meters long is giving me
this side of the plane.
The plane is not the same thing as a wedge.
Even though this might be a ramp you might draw as a right triangle.
It tells me the angle here is 32 degrees.
This refrigerator here.
Is being pushed up the ramp with an applied force.
So that's the first force that I'm going to draw on my free body diagram.
Remember that it's good form and good practice to draw if you're,
especially if you're asked on the exam to draw a free body diagram.
Draw your vector starting from the center of the object.
And don't break them up into components.
Other forces would be mg, for gravity, acting straight down.
There will also be a normal force.
Well, normal force is always acting perpendicular to the surface.
And so, the normal force is this direction.
And there's a frictional force.
Because it tells me about this coefficient of kinetic friction.
Friction always acts parallel to the surface, but
against the direction of motion.
Since the refrigerator is being pushed up the ramp,
the friction will act down the ramp.
And so, those are my four forces.
To analyze a problem like this which we are going to do in a second,
I need to decide how to set up our axes.
If I were to set up my equations perfectly, horizontal and
vertical, our regular Cartesian coordinate system.
Well, I would have to break apart the applied force, the normal force, and
the frictional force into its components.
To get a vertical or horizontal component.
Instead what I will do, as a trick.
And there's a way to simplify the problem which we always look for,
I'm going to tilt my x and y directions to look something like this.
So my positive x will be going up the ramp.
My positive y will be going out of the surface of the ramp.
That means that Fa, the applied force, will be positive, and
I will not have to break it up.
The only force then, that I will have to break up, is mg.
Now of course if this were a free body diagram you'd stop here.
But to solve the problem, going to break up mg into its horizontal and
vertical components.
Or at least tilted on my x and y directions.
This angle here, you can see that label, is theta.
So, if I want the horizontal, the x component,
that's going to be the sine of mg.
And the adjacent side, the vertical component would be the cosine.
So let's look at our.
Says find the normal force exerted on our refrigerator.
I notice that that is in my y direction in this problem.
So I will start for part A.
By setting up my sum of forces
equals ma equation my fundamental principle in the y direction.
I know that its not accelerating in y direction because its staying on the ramp.
So that whole right side of my equation will equal zero.
Because the acceleration is zero.
There are two forces with vertical y components in this setup.
And will have a normal force which is positive.
It points in my positive y direction.
And I will have a component of mg.
In this case, to get to the side of the triangle that is in the negative
y direction, it will be the adjacent of that angle theta.
So mg cosine theta equals 0.
This may look like a somewhat common result solving this But
you have probably seen before that normal force equals, mg, cosine theta.
Subbing in my numbers to find this normal force on this incline plane.
Well the mass of this refrigerator, let me go up so you can see that,
115 kilograms, g 10 cosine of 32 degrees.
Notice, I didn't plug in 10 with a negative sign.
Because the way we've set up this equation in this first step here,
we have made any forces going left or down negative already.
We don't want to do that twice.
So by this point in the problem, while I'm subbing in my numbers.
Using forces I will always sub in as a positive value.
But we've already taken care of direction earlier in our setup.
So the normal force here, when I put this in my calculator, I got 975 Newtons.
That's the normal force for part A that is experienced by the fridge.
Part B then says the force of friction exerted on the refrigerator.
So this is that force of friction that pulls left back against the ramp.
So scrolling down some here, I'll change the colors.
B again, I'm going to set up my fundamental equation here.
F equals ma.
And I know that it accelerates.
Which is something that we're going to find later on in the problem, so
going this route would give me two unknowns.
I wouldn't know friction, and I wouldn't know the acceleration.
So this is not probably the best route to go for this problem yet.
Instead, I will go with my equation for kinetic friction, which looks like this.
Friction equals mu that coefficient of kinetic friction
times normal force which is a number we already know.
So if you were to do this in variables mu of k,
normal force from earlier was mg cosine of theta.
Subbing in our values then, mu was 0.4 in this problem, 0.4.
And mg cosine theta was 975 Newtons.
So the total force of the friction, and we're just talking about magnitudes at
this point for each of these forces, would be 390 Newtons.
That's how much force friction
pulls back down this ramp to resist the motion of this refrigerator up the ramp.
Let's go back up and see what part C asked.
Part C says the net force exerted on the refrigerator.
Well immediately I can tell, by looking my diagram and the motion,
there is no net force.
Perpendicular to that ramp in our y direction,
because it's not being accelerated off the ramp or in to the ramp.
The normal force and that vertical component of mg, that y component,
will cancel out.
So I only need to worry about the net force in the x direction.
I've got what looks like a positive applied force and a negative friction
force, and those two are what I should combine for this portion.
So, [SOUND] changing colors again, part C wants to know the net force on the topic,
which we know is in the x direction.
And when it asks for that force, I'm solving for this portion of the equation.
Or I could solve for the right hand portion of the equation.
If I knew the mass and the acceleration of this object in our x direction.
Since I don't, I'm going to do the left hand side.
I need to add together all the forces to the x direction.
In this case, there's a positive Fa, applied force, to the right,
minus our force of friction that we just found.
And it's minus because it points in our negative x direction.
So they told us that Jeff pushes on this refrigerator
with a force of 1,000 Newtons.
And if you've every felt a Newton, perhaps on a spring scale,
you know a Newton is not very large.
So that's definitely very possible.
Minus the friction force we just found, which is 390 Newtons.
That gives me a total net force in the x direction on this object,
since there is no net force in the y direction.
So this is the total.
610 Newtons in the positive x direction.
Again, it only asks for magnitude here, so I don't necessarily need a direction, but
we are aware of what direction it points.
The last portion of this question asks, let's go back up and
read it, the acceleration of the refrigerator up the ramp.
That means now I can finally use my entire Newton's second law equation.
Part D, again I'm going to set up sum of forces equals ma in the x direction.
We've actually all ready set up the left hand side of this equation.
There was a positive applied force to the right, a negative friction force
to the left, and that'll equal the mass of this refrigerator times its acceleration.
In the x direction because we added together forces in the x direction
to give us an acceleration in the x direction.
So setting this up, the force applied was 1000 minus 390, our friction force.
The mass of this refrigerator was 115 kilograms,
and then solving for the acceleration.
So the acceleration of this refrigerator was 5.3 meters per second squared.
As usual, keep an eye on sig figs, your significant figures.
Try to be within 2 to 4 on the exam, don't go out too far.
And make sure to include units.
>> This next example we will solve includes multiple boxes connected
by massless inextensible strings.
Three boxes of mass, 5 kilogram, 10 kilograms, and
20 kilograms are connected in this order by strings on a horizontal surface.
A 150 Newton force is applied to the 20 kilogram box, and
is directed towards the right.
Calculate the acceleration of the system,
the tension in the strings between each set of boxes, the coefficient of friction.
Calculate the acceleration of the system and
the tension in the strings between each set of boxes.
The coefficient of kinetic friction between the surface and each box is 0.3.
Before I solve for anything that this question asks for
I'm going to go ahead and draw out my free body diagram.
Starting out over here towards the right I've been told that I have an applied
force that's being applied to the 20 kilogram box,
that in this case for me is labeled as m3.
So I'd start from the center and I draw out my applied force.
Let's look at what other forces this object has acting on it.
It's going to exert its weight on the surface.
So I have m3g as the gravitational, or weight force.
The surface is going to exert a force right back.
That will be my normal force.
And then I can go ahead and actually draw out that weight and
normal force vector on all of these.
So over here I've got m2g,
normal force, m1g,
and normal force.
Now let's look at what other forces I have acting on it.
Well, the question mentions that I need to solve for the acceleration, and
the tension in the strings.
So let's go ahead and talk about that.
The tensions in this string that you haven't exerted, your equal and
opposite forces from Newton's third law recall.
So I have tension over here.
And from the same string, the tension here as well.
I'm going to go ahead and label this as Tension 1.
Remember, it's just the same exact tension being exerted over
here because this is still that same string.
Then over here, I've got both boxes pulling on this string,
Tension 2.
And now, let's see, I also have friction,
they mentioned the coefficient of friction.
So, my frictional force is going to oppose the motion.
The motion is guided towards the right with the applied force.
So here I have my Frictional force.
And I can go ahead and draw out my Frictional force on each of these.
So I have my Frictional force 1 for object 1.
I will also have, and I'll draw this down here, Frictional force 2.
And Frictional force 3.
Notice all the Frictional forces are pointing towards the left, which means
when I do my calculations, I'll need to be sure to use them as negatives.
Let's go ahead and do those calculations now.
It asks for acceleration.
Well, this whole acceleration's occurring in my horizontal direction.
So I'm going to look at the entire system, whole system.
And I will look at the sum of my horizontal forces
equaling mass times acceleration.
Let's look at all the horizontal forces I have.
I've got all three Frictional forces on each object.
And then I have my applied force,
towards the right, the only positive that I've listed here.
Setting them equal to mass times acceleration,
notice I have added all of the masses together.
They're all accelerating with that one acceleration because they're
all connected.
So I can do this over here.
Now notice what happens here.
I plug in variables for each.
Frictional force,
which if you recall, is the coefficient of friction multiplied by normal force.
I'm just going to scroll back up over here to show you.
Normal force is equaling the mg of each object.
Normal force here is equaling m1g.
Normal force here is equaling m2g.
And normal force here is equaling m3g.
There are no other vertical forces or components vertically up of force,
that are being implemented here.
So I only have normal and mg equaling each other.
Now I can go ahead a plug in variables.
So, my coefficient stays the same, it's 0.3.
Noticed I've added the negative.
5 multiplied by 10 per g minus again, 0.3,
10 kilograms, multipled by 10 per g minus 0.3,
20 kilograms multiplied by 10 per g, and
then plus my applied force, which is 150.
All of this equaling 5 kilograms plus 10 plus 20 multiplied to the acceleration.
From here I can go ahead and solve for the acceleration.
And I end up with 1.29 meters per second squared.
Well, is this going towards the right or the left?
In one of the previous problems we discussed how you can add up the magnitude
of all of your forces in each direction.
I know towards the right I have 150.
If I combine all of my forces towards the left I get negative 105,
or 105 towards the left, which means that my forces towards the right are greater.
This acceleration is positive, it's directed towards the right.
The next part of this question wants me to solve for
the tension in the strings between each boxes.
So I have two strings, two tensions to solve for, Tension 1 and Tension 2.
I'm going to start with the box at the very end, Box 1.
Sum of the Forces, horizontally equal mass times acceleration.
That acceleration horizontal, I have that now.
I've got negative frictional force,
positive tension.
This is tension 2 that I'm solving for.
Equals mass of box 1 times acceleration.
Just going to go back up and show you that.
My frictional force here is pointed towards the left.
Tension, tension 2 that is pointed towards the right.
This is the easiest one for me to use right now because I already know what
frictional force is, I only have one unknown here which is tension 2.
So I can solve it and then you'll see how Iwill use that to solve for tension 1.
going to plug in variables.
Coefficient. Times m1g which is that
normal force on that object.
Plus tension 2 equals m1 times acceleration.
Now we place in the numbers.
This was the 5 kilograms.
G is 10 plus tension 2.
5 kilograms multiplied by the acceleration which
was the 1.29 meters per second squared.
From here, when I solve for tension 2.
I end up with 21.45 Newtons, that's towards the right.
And I can now go ahead and solve for tension 1.
So I'm going to use this information over here, and
I'm going to use the next box over, I'm going to use block 2.
Similarly, sum of my horizontal courses equals, ma.
You always want to write out your summation equation, or
your before you start putting your variables in.
I have frictional force.
Minus tension 2, plus tension 1, which is the one that I am solving for.
All of this equals m2 times acceleration.
Let me scroll up and show you what I've used here.
I'm solving for m2.
Solving using the forces on that one.
So I have T1, tension 1 that's directed to the right.
It's positive.
Tension 2 notice over here is towards the left which
is why I plugged it in as a negative.
And then of course I have my negative frictional force acting on it as well.
Now we're ready to start plugging numbers in.
[SOUND] First I'm going to go ahead and
just place in these variables.
So coefficient times m2g minus
T2 plus tension 1 equals m2 times
a 0.3.
This mass is 10 kilograms.
G is 10.
Tension 2, I already have that now, which makes this really easy.
21.45.
Plus the tension 1.
All of this equals the mass.
10. And the acceleration, I also have that.
1.29.
From here when I solve for tension 1,
I end up with 64.35 Newtons.
And that is towards the right.
>> This last example illustrates the difference between static and
kinetic friction.
While kinetic friction can always be determined using the equation,
friction equals mu times normal force, you will only determine
the maximum force of static friction when using this similar version.
In this question, a block is placed on a horizontal table and a force of 100
Newtons is applied to the left at an angle of 60 degrees below the horizontal.
Calculate the force of friction on the 40 kilogram block, if the coefficient of
static friction is 0.6 and the coefficient of kinetic friction is 0.3.
In some ways this question's almost a trick question because
it requires a very specific understanding of friction, which is often a,
the cause of many mistakes in calculations for students.
So let's pay careful attention of what's going on here.
There's a block on a horizontal table.
So I'll draw my block.
There is a force.
Now we're going to draw a Free body diagram here of 100 newtons
applied left at an angle of 60 degrees below the horizontal.
So, I'm starting my vector from the center of the object.
There's an applied force here of 100 newtons.
You can label it in numbers if you like, but definitely always label the variables.
And this angle here, 60 degrees.
So below the horizontal.
I also want to label the other forces on this object.
Other forces would include a normal force, vertically upward,
perpendicular to the surface.
It also mentions friction.
Since friction is always parallel to the surface but
resisting the direction of motion.
Friction will be to the right, and
of course in all of these problems there is always an mg
straight down to the center of the earth, and those are my four forces.
Again, I'm setting up my axis, X and Y directions like so.
Positive x, positive y.
Horizontal and Vertical.
Which means there's only one force here that I'm going to have to break up
in my horizontal and vertical direction.
Which if it asks on the, on the AP exam to draw a free body diagram,
I would stop here and draw my components elsewhere.
But since this is just for me.
I'm going to go ahead and draw my components.
This force I'm breaking into a right triangle.
It pushes left and vertically downward on this block.
And so this side of the triangle would be Fa cosine for that angle theta, and
this would be Fa sine of that angle theta.
And now it asks, calculate the force of friction on this block.
You might be very tempted to use this equation, mu times the normal force.
Well, if this worked for kinetic friction, that would always work and
always give us a result that we want.
It would be the value that we need.
But this is talking about static friction, and
to use the equivalent equation for static friction.
Means that that only gives us the maximum frictional force that an object give back,
and an object does not always have its maximum static friction.
It's a misleading and
will give you the wrong answer to use this equation unless you're at a maximum.
So going in that vein, it would be wrong to use that scenario.
Instead I'm going to go by route of Newton's second law to solve this problem,
and you'll see what I mean in a moment.
I'm going to set this up in the x direction.
All the forces in the x direction I'll add together to
equal a mass times the acceleration of this block.
And since it's at rest still, the object's not moving,
that's why I'm using static friction, the acceleration will be 0.
Now I need to add together the two forces that act in a horizontal direction.
In this case, there's a negative Fa.
And I need only that horizontal component.
Cosine theta.
There's also a frictional force acting to the right.
Notice, friction's not always negative, the positive and
negative signs are based on the direction, and what we chose as positive.
In this case, I chose positive to be the direction to the right.
This will all equal 0, because it was not accelerating.
Notice then, to solve this problem,
that friction will equal just the horizontal component of my applied force.
That's all it has to push back with to balance out.
If friction were pushing more than this, perhaps the maximum value our mu times
N equation would give us, that means friction would be moving this object to
the right, which is definitely not the case.
Friction only resists motion.
It doesn't cause the motion.
So in this case, I'd be plugging in the numbers that they gave me.
So, let's see here.
The force applied was 100 Newtons, and
the angle that they gave me was 60 degrees.
And if you remember your unit circle, you know that the cosine of 60 is one-half.
If not,.
That might be good things to kind of keep you,
keep in your mind when you're solving problems like this.
Also your constant sheet underneath the exam would have that information, for
basic angles, for trig functions.
And so when I solve this, I get a frictional force of 50 newtons.
And if I wanted this to be the full vector, I need to give a direction.
In this case, in my positive direction.
Or alternatively, I could have said 50 newtons to the right.
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