WEBVTT physics/ap-physics-1-2/fullerton
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Hi, I am Dan Fullerton. Welcome back to Educator.com.
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Let us talk about Newton's Second Law of Motion.
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Now our objectives are going to be to draw and label a free-body diagram showing all the forces acting on an object and also draw a pseudo-free body diagram showing all components of forces acting on an object.
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We will explain the relationship between acceleration, net force, and mass of an object, and use Newton's Second Law to solve a variety of problems.
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Finally, we want to make sure we understand the difference between mass and weight, and the conditions required for equilibrium.
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Free-body diagrams or FBD's -- these are tools that we use to analyze physical situations.
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What they do is they show all the forces acting on a single object.
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The object itself may be drawn as a dot.
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Some folks like to draw it as a box. It does not matter, either one.
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Now when you draw a FBD, choose the object of interest and draw it as either a dot or a box.
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Then you are going to label all the external forces acting on the object and only forces go on that diagram.
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Finally, sketch a coordinate system showing the direction of the object's motion as one of the positive axis.
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For example, let us take a look at a circus elephant falling off a tight rope -- sad story -- it is just pretend, do not worry.
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Neglecting air resistance -- draw a free-body diagram for the falling elephant.
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I am going to use my amazing physics artistic skills to draw an elephant.
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There it is in FBD terms and I am going to label all the forces acting on it.
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The weight of the elephant -- the force of gravity -- which I typically write on FBD as mg -- the force of gravity on an object on a FBD can save yourself a little bit of work if you write weight as mg.
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Or, we could draw it as a box -- there is our elephant and the one force acting on it is the weight of the elephant pulling it down.
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How about if we had the falling elephant with air resistance?
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Well, there is my elephant again.
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We still have the elephant's weight -- the force of gravity on it -- mg, and we have some force of air resistance.
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The force of air resistance is going to oppose the force of gravity.
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The faster something falls, the force of air resistance resists that motion -- pushes up.
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When eventually the force of air resistance and the object's weight balance out -- they are equal -- the object does not accelerate anymore.
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It is in equilibrium, like we learned with Newton's First Law. It maintains a constant velocity.
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We would call that terminal velocity -- when the force of air resistance equals the force of gravity on an object.
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It maintains a constant velocity; it does not speed up; it does not fall any further.
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Let's draw a FBD for a glass of soda sitting on a table.
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Here is our glass of soda.
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We have the weight pulling it down, but the glass of soda is not moving, it is at rest and it is remaining at rest; it is not accelerating.
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There must be another force on that glass. What is that force?
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It is the force of the table on the soda and it must oppose -- absolutely balance that weight.
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We are going to call this force, the normal force.
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When we say normal, we are not talking about the opposite of weird, we are talking about the geometric interpretation of normal.
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It means perpendicular. In this case, the force is perpendicular to the surface of the table.
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There is our table. If we have our glass sitting on it -- perpendicular to the surface of the table -- the normal force comes out of that surface.
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Perpendicular to this tablet, a normal force would be coming out this direction.
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Normal perpendicular -- we will label that fn.
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Now, in this case it is pretty easy to see that the net force again, must be zero -- the object remains at rest.
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Example problem: Which diagram here represents a box in equilibrium?
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Again, the key here is to recognize that equilibrium means that the net force equals 0 -- all the forces are balanced.
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Well, these are not balanced; it cannot be one.
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Those are not balanced; it cannot be that one.
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So, 2 up, 2 down, 5 right, 5 left -- that has to be equilibrium -- Net force = 0.
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All the forces balance. It will continue in it's current state of motion.
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All right, example five: Now we have a block sitting on a ramp. Which diagram below best represents the forces acting on the block?
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If we think about it, let's draw them here first.
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We have the normal force, which must be perpendicular to the surface -- so there is our normal force this time.
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The weight -- the force of gravity on all objects -- is down toward the center of the Earth: mg.
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Or, in this case it is labeled as weight, as fw in these diagrams.
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Which way does the block want to go?
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It wants to slide down the incline right?
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So the force of friction must be up the ramp. Which one of these four choices nears what forces we see on that?
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Has to be number four here.
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All right. When forces do not line up with axis, you can draw a pseudo free-body diagram (P-FBD)
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We are going to break up the forces that do not line up with the axis into components that do.
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When you do this on an AP exam, your FBD only shows the forces, not components -- then draw a separate P-FBD.
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If you show components on a FBD, often times they will take off points.
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You need to have two separate diagrams -- one showing just the forces and then a separate P-FBD where you have broken forces that are not lined up with an axis into their components.
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Let us draw the FBD for a box sitting on a ramp.
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We have the applied force -- some force pulling it up the ramp.
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Of course, we are going to have a normal force perpendicular to our ramp.
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We have the weight of the box -- straight the force down, the force of gravity.
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And in this case, if we have a force wanting to pull it up, we can draw this as having a force of friction down that way to balance it -- depending on what all our forces are.
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If we were to draw a FBD -- let me draw it over here on the right, our FBD.
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There is one axis -- I am just going to tilt my x and my y.
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Now when I draw this, I have f pointing up the ramp. I have a normal force that is perpendicular.
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If we are pulling it up the ramp, we could have a force of friction -- the opposite direction and weight is straight down.
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That is our FBD. That we do not touch now.
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We have drawn our diagram showing all the forces.
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To do the P-FBD, though, we are going to make a separate diagram.
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We are going to keep our axis y -- x. F already lines up with an axis, so that is fine.
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Normal force lines are up already. Friction lines up with an axis, but the force of gravity -- the weight of the object -- does not.
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What are we going to do with that?
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That one we are going to have to break up into components.
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If that is weight, we are going to have to do a little bit of Geometry here to see what is going on.
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If I were to extend the ramp back here -- this is our angle θ -- that is 90 - θ.
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That angle must be θ again.
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I am going to break up the weight into an x component -- a component that is parallel with the x axis.
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Let's call that mg.
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It is parallel because it is in the same direction as the object's motion, or the direction it wants to move -- mg parallel and we have a component perpendicular to that axis -- mg perpendicular.
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Now notice mg parallel -- this side is opposite our angle θ, so mg parallel is going to be equal to mg sin θ -- mg perpendicular is the adjacent side -- mg cos θ.
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My P-FBD, when I go to draw it -- let me label it here -- is going to show the components of the object's weight instead of the weight itself.
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Now all my forces with line up with the axis. There is my x. There is my y.
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We have f. We have our normal force -- force of friction.
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Now this mg parallel and mg cos, or mg cosine θ -- easy enough. Draw it right there, mg cosine θ.
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Mg parallel right here -- I am just going to shift so it is on the axis; it is already lined up with an axis, but I am going to redraw it over here - mg sine θ.
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You will see in some cases where teachers prefer to see that all at the same point in force of friction, mg sine θ are right beside each other, almost in the same direction.
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I prefer to draw mine end-to-end so it is easy to see that they add up.
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Here is our FBD and here is our P-FBD.
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You have to show those separately on the AP questions in order to get the full credit for a problem.
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Let us take the example of a car on a bank turn.
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Roadways are often angled around steep turns to assist the cars in making the turn.
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Let us draw the free-body diagram and the pseudo free-body diagram for a car on a banked roadway.
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I will start off by drawing my curve and we will put our car on it -- pretend we are looking at the car from the back.
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There are the wheels and there is the license plate.
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This is at some angle θ. So, the forces acting on it -- we can draw on our FBD.
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I will draw my x axis and I will draw my y axis -- (y,x).
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In this case we have a normal force that is coming up out of the ramp -- so our normal force fn -- or capital N if you prefer points that way.
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We have the weight -- the force of gravity, down.There is our FBD.
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And we can throw friction in here if we needed to, depending on the situation, but let's keep this one simple for now.
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If we wanted then to draw the P-FBD for this -- I am first going to redraw this -- y, x, mg -- pointing down -- and just to illustrate this, I am going to draw the normal force up here again.
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Now as we look at the geometry of the problem, our angle θ is going to be -- Let's draw our components-- there is the x component of the normal force. There is the y.
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That is going to be our angle θ by geometry, therefore, this side is going to be fN sine θ, the opposite, and this one is going to be fN cos θ.
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When I do my nice, pretty P-FBD, there is x, there is y, my object -- of course, mg down -- I have this way, fN sine θ and I have fN cos θ.
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There is my P-FBD. All the forces shown as components.
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So now, Newton's Second Law of Motion, perhaps the most important formula or relationship in all of physics.
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We will use this again, and again, and again.
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The acceleration of the object is in the direction of and directly proportional to the net force applied.
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It is inversely proportional to the object's mass.
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If we want to write this in formula form -- acceleration of vector is equal to the net force applied on an object divided by the object's mass -- it's inertial mass.
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The way that it is more commonly written: Fnet equals ma -- Force = mass x acceleration.
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You can apply this in many, many different ways.
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As we look at these two laws, there are some interesting observations to take from them.
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Newton's First Law says an object at rest will remain at rest and an object in motion will remain in motion at constant velocity in a straight line unless acted upon by a net force.
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Basically what it is saying is, if Fnet = 0, then acceleration equals 0.
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Newton's Second Law says the acceleration of the object is in the direction of and directly proportional to the net force applied and inversely proportional to the object's mass.
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Notice how if Fnet over here is 0, A has to be 0.
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Newton's First Law is redundant. It is actually a special case of Newton's Second Law where the net force on an object is 0.
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If you understand Newton's Second Law, through and through, Newton's First Law you do not really need to know.
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It is already embedded in Newton's Second Law. The First Law is redundant.
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Let us look at how we apply Newton's Second Law.
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General strategy -- Draw a FBD -- tremendously helpful tools.
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For any forces that do not line up with the x or y axis, break those up into components that do and then we are going to draw that P-FBD.
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Next, write expressions for the net force in the x and y directions.
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Since the net force equals ma, we can use Newton's Second Law to solve the resulting equations and determine whatever the unknown quantities are that we are after.
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Let us see how this works.
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We have a force of 25N East and a force of 25N West acting concurrently on a 5 kg cart.
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Find the acceleration of the cart.
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You can probably do this in your head, but it is worth walking through the steps to see how this could applied with a simple situation before we complicate matters.
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A FBD -- there is our cart.
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We have a force of 25N to the East and we have a force of 25N to the West.
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Next, I am going to write my Newton's Second Law equation: Fnet = ma.
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Since this is in the x direction, I am going to specify this and say net force in the x direction is = to mass times acceleration in the x direction.
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Now, Fnet(x) -- all this means is that is says looks at your FBD.
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Look for all the forces acting in this x direction and write them down.
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In this case, I have Fnet(x) that I am going to replace with 25 to the left, so that is -25.
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I have 25 to the right, so that is +25.
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That must be equal to ma(x)-25 + 25 = 0, therefore 0 = ma(x), therefore a(x) must be equal to 0.
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No acceleration, which you probably knew before we started the problem, but the steps are what is important.
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Let's take another look here.
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We have a .15 kg baseball moving 20 m/s stopped by a player in .010 s. What is the average force stopping the ball?
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The way I would start these sorts of problems -- it is usually a good idea to write down what kind of information you are given.
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Here I know the mass is equal to .15 kg.
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It is initially moving at 20 m/s -- 0 = 20 m/s.
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I am just going to draw in that that is to the right +x.
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Final velocity is 0. It comes to rest and the time it takes is .01 s.
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As I look at that -- trying to find the average force stopping the ball -- I know the mass -- it sure would be useful to have acceleration.
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I do not know acceleration, but I can use my kinematics -- my kinematic equations to find it.
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Acceleration is change in velocity over time -- Δ anything is it's final value minus it's initial divided by time, or 0 - 20 m/s/.01 s is going to be negative 2,000 m/s ².
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Now that I know acceleration, I can use Newton's Second Law to continue the problem.
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If the net force equals mass times acceleration, that implies then -- since we know acceleration is -2,000 m/s².
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I also know that mass is .15 kg, then the net force must be equal to our mass .15 kg x -2,000 m/s².
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Calculator time -- this implies then that the net force equals -300N.
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Why the negative? What does the negative mean here?
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If you think about it, we called to the right the positive direction -- the initial direction the baseball was moving.
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The negative just implies that this force has to be in the opposite direction in order to stop it.
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That is the opposite direction of the ball's initial velocity.
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Moving on -- let us take a look at a block on a surface.
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Two forces, F1 and F2, are applied to a block on a frictionless, horizontal surface as shown in the diagram.
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If the magnitude of the block's acceleration is 2 m/s², what is the mass of the block?
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We know it is accelerating. F1 is bigger than F2, so it must be in that direction at 2 m/s².
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Well, FBD -- always a great place to start.
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We have 2N to the right, F2, and we have a lot more -- we have 12N to the left.
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I am going to write Newton's Second Law, Fnet = ma.
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Since I am interested in just the x direction, the net force in the x direction equals mass times acceleration in the X direction.
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I am after mass, so net force in the x direction -- 12 to the left, 2 to the right -- let us make this easy and define to the left as positive.
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That means we have 10N in the positive direction, to the left -- must equal our mass times our acceleration, 2 m/s to the left.
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So that is positive, therefore mass is going to equal 10/2 or 5 kg.
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Tremendous. Let us go a little bit further.
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We have a 25N horizontal force northward and a 35N horizontal force southward acting concurrently -- that means at the same place and at the same time -- on a 15 kg object on a frictionless surface.
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What is the magnitude of the object's acceleration? Again, let us start with the FBD.
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There are horizontal forces both North and South, but I am going to draw an overhead view.
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We have a force of 25N North and we have 35N South.
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The net force should be pretty easy to see. It is going to be 10N South.
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The acceleration is going to be the net force divided by the objects mass, which is going to be 10N South, divided by 15 kg, or 0.67 m/s ² South.
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Let us talk about mass versus weight.
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Mass is the amount of stuff that something is made up of. It remains constant.
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Yes, you can change the mass of an object by taking pieces off of it, or adding pieces.
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For the most part, wherever you go for the same object, it has the same mass -- it is made out of the same stuff.
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Weight, however, which we are writing as mg is the force of gravity on the object.
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Weight varies depending on the gravitational field strength, g.
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Here on the surface of the Earth, g is 9.8 m/s².
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On the surface of the moon, g is about 1.6 m/s².
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You would have a different weight if you went to the moon -- 1/6 the weight you would have on Earth.
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However, regardless of how that works, you have the same mass.
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Let us look at an example here.
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An astronaut weighs 1,000N on Earth.
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What is the weight of the astronaut on Planet X, where the gravitational field strength is 6 m/s ².
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On Earth, mg, the object's weight, is 1,000N, therefore, we could say that the mass of the object -- what does not change, is going to be 1,000N/g on Earth -- 10, or about 100 kg.
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If we go over here to Planet X, mg on Planet X must equal the mass, 100 kg -- that does not change, times g on Planet X, 6 m/s² -- 100 x 6 = 600N.
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An alien on Planet X weighs 400N. What is the mass of the alien?
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On Planet X, mg(x) for the alien must be 400N, therefore, the mass of the alien on X is 400N/g on x, 6 m/s ², or about 66.7 kg.
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Take the alien to Earth, it is going to have a different weight.
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It will not be 400N, but the mass will be the same, 66.7 kg.
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Coming back to equilibrium. Translational equilibrium occurs when there is no net force on an object, therefore, acceleration is 0.
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The equilibrant is a name for a single force vector that you add to any unbalanced forces you have on an object in order to bring the object into translational equilibrium.
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For example, if I have a force that is 25N that direction -- if I want its equilibrant, I need a force that is 25N in that direction so that you add them together -- you get 0.
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You get no unbalanced forces. You have 0 acceleration.
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In the diagram here we have a 20N force due North, and a 20N force due East acting concurrently, again at the same place and same time on an object.
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What additional force is required to bring the object into equilibrium? Or we are looking for the equilibrant.
00:25:49.000 --> 00:25:54.000
Now the way I do this is, is if I look here we have 20N and 20N.
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Let us add them together to get the net force.
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I am just going to slide this vector over so that they are lined up tip to tail so I can add them -- 20N -- and my resultant, the sum of the two vectors is going to be a vector with the length square root of 20 squared plus 20 squared.
00:26:12.000 --> 00:26:23.000
That is going to be square root of 20 ^20 + 20 ² = 28.3N.
00:26:23.000 --> 00:26:29.000
I could replace that 20N North and 20N East with one vector, 28.3N to the northeast.
00:26:29.000 --> 00:26:36.000
It's equilibrant, the vector I would have to add to that system to bring it back into equilibrium, must be the exact opposite of that.
00:26:36.000 --> 00:26:51.000
The equilibrant must be that red vector, which would be 28.3N to the southwest.
00:26:51.000 --> 00:26:55.000
That is what an equilibrant is.
00:26:55.000 --> 00:26:58.000
A little bit on translation equilibrium.
00:26:58.000 --> 00:27:01.000
We have a 3N force and a 4N force, so they are acting concurrently on a point.
00:27:01.000 --> 00:27:06.000
Which force could not produce equilibrium with those two forces?
00:27:06.000 --> 00:27:18.000
A 1N force could, because if we have this lined up correctly we could have a 3N, maybe a 4N force and a 1N force that would somehow sum to zero.
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You would get back to where you started.
00:27:20.000 --> 00:27:32.000
A 7N force -- if we had 3N this way and 4N this way, an equilibrant that was 7N in the exact opposite direction would bring that into equilibrium.
00:27:32.000 --> 00:27:47.000
But, we cannot do 9. If we have 3N to the right and 4N to the right, there is no where I can place a 9N force where we end up with no net force when we are all done.
00:27:47.000 --> 00:27:53.000
No matter what I do, I am going to have a remaining force, an unbalanced force of at least 2N
00:27:53.000 --> 00:28:04.000
A 9N force cannot combine with the 3N force and a 4N force to give you 0 net force, or to bring you into equilibrium.
00:28:04.000 --> 00:28:07.000
Determining acceleration.
00:28:07.000 --> 00:28:15.000
We have a 15 kg wagon that is pulled to the right across a surface by a tension of 100N, an angle of 30 degrees above the horizontal.
00:28:15.000 --> 00:28:23.000
A frictional force of 20N to the left act simultaneously. What is the acceleration of the wagon?
00:28:23.000 --> 00:28:24.000
I like to draw a picture first.
00:28:24.000 --> 00:28:38.000
We will make a little red wagon, because those are the cutest kind really.
00:28:38.000 --> 00:28:40.000
We have a force of 100N at an angle of 30 degrees, and we know we have a frictional force over here.
00:28:40.000 --> 00:28:48.000
For my FBD, I am going to have my little red wagon.
00:28:48.000 --> 00:28:55.000
I have the weight of the wagon, mg. I have some amount of normal force.
00:28:55.000 --> 00:29:08.000
I also have this applied force that is going to be 100N in an angle of 30 degrees, and a frictional force.
00:29:08.000 --> 00:29:23.000
There is my FBD, but now I am going to do the P-FBD, where I am going to break up that 100N force that does not line up with the axis into its components: y, x.
00:29:23.000 --> 00:29:33.000
We will start up with the forces that do line up: mg, force of friction, and normal force.
00:29:33.000 --> 00:29:51.000
The x component here is just going to be 100N cosine 30 degrees and its y component is going to be 100N sine 30 degrees.
00:29:51.000 --> 00:29:56.000
If we want the acceleration of the wagon, we are really talking about the acceleration of the wagon in the x direction.
00:29:56.000 --> 00:30:05.000
I am going to write Newton's Second Law: Fnet = ma, and focus on it in the x direction.
00:30:05.000 --> 00:30:12.000
For Fnet, all I do is I go back to my P-FBD diagram and I look for all the forces acting in the x direction.
00:30:12.000 --> 00:30:36.000
I have the 100N cosine 30 degrees -- that is 86.6N -- and in the opposite direction, I have minus the force of friction, 20N and that must be equal to ma(x) -- 86.6 - 20 = 66.6N = ma(x).
00:30:36.000 --> 00:30:59.000
Therefore acceleration in the x must be 66.6N over the mass of our cart, 15 kg -- mass of our wagon, which implies that the acceleration must be 66.6/15, or about 4.44 m/s².
00:30:59.000 --> 00:31:03.000
Let us take a look at one more example problem.
00:31:03.000 --> 00:31:06.000
We will talk about a suspended mass.
00:31:06.000 --> 00:31:12.000
In this case we have a traffic light suspended by two cables as shown and we will label them T1 and T2.
00:31:12.000 --> 00:31:20.000
We have measured the tension and the cables using a spring scale and we found that T1 is 49N and T2 is 85N.
00:31:20.000 --> 00:31:25.000
Can we find the mass of the traffic light? Of course, the answer is going to be yes.
00:31:25.000 --> 00:31:34.000
Let us start by drawing our FBD -- x and y.
00:31:34.000 --> 00:31:43.000
Now, there is our object and of course we have its weight: mg down.
00:31:43.000 --> 00:31:54.000
We have tension one (T1) and if I do just a little bit of geometry over here -- if that is 30 degrees, that must be 30 degrees and over here, if that is 60 degrees, then that is 60 degrees.
00:31:54.000 --> 00:32:06.000
So I am going to draw a T1 in that direction at an angle of 30 degrees, and T2 over here --it is a bigger angle at 60 degrees.
00:32:06.000 --> 00:32:15.000
So my P-FBD, I have to break up T1 and T2 into their components.
00:32:15.000 --> 00:32:25.000
Take our FBD again -- y, x -- mg of course still points down.
00:32:25.000 --> 00:32:40.000
Let us start with T2 here. Its x component is going to be T2 cosine 60, so I will have here T2 cosine 60 degrees and its y component will be T2 sine 60 degrees.
00:32:40.000 --> 00:32:42.000
Now let us deal with T1.
00:32:42.000 --> 00:32:57.000
Its x component will be to the left and that is going to be T1 cosine 30 degrees, and its y component -- T1 sine 30 degrees.
00:32:57.000 --> 00:33:06.000
If we are trying to find the mass, I am going to start with Newton's Second Law in the direction that has the mass in it.
00:33:06.000 --> 00:33:12.000
I am going to write Fnet = ma, but I am going to look in the y direction.
00:33:12.000 --> 00:33:19.000
I am going to replace Fnet(y) with all the forces I see over here acting in the y direction.
00:33:19.000 --> 00:33:40.000
Fnet(y) = T1 sine 30 degrees, pointing up + T2 sine 60 degrees and I have mg down - mg.
00:33:40.000 --> 00:33:50.000
We know that all of that -- since this is just sitting there and the traffic light is not accelerating -- a, it must be 0, so that is all equal to 0.
00:33:50.000 --> 00:34:12.000
When I do that, I can then say since T1 is 49, then 49 sine 30 degrees + T2 is 85N -- 85 sine 60 degrees must be equal to mg.
00:34:12.000 --> 00:34:29.000
Or 49 sine 30 = 24.5 + 85 sine 60 = 73.6 must be equal to mg -- 9.8(m).
00:34:29.000 --> 00:34:38.000
Divide both sides by 9.8 and I come up with a mass of about 10 kg.
00:34:38.000 --> 00:34:46.000
Newton's Second Law: f = ma and free-body diagrams and pseudo free-body diagrams to help us apply those concepts.
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Terrific tool that we are going to use all the time here in Physics.
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