WEBVTT physics/ap-physics-1-2/fullerton
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Hi everyone, and welcome back to educator.com. This lesson is on vectors and scalars.
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Our objectives are going to be to differentiate between scalar and vector quantities, to use scaled diagrams to represent and manipulate vectors, being able to break up a vector into x and y components, finding the angle of a vector when we are given it's components...
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...and finally, performing basic mathematical operations on vectors such as addition and subtraction.
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When we talk about vectors, what we are really talking about are different types of measurements in physics, different quantities, and there are really two types. Scalars and vectors.
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Scalars are physical quantities that have a magnitude or a size only. They do not need a direction. Things like temperature, mass, and time. I know what you are thinking. Time has a direction, right? forward or backward. Well, not really.
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When we are talking about direction, we are talking about things like north, south, east, west, up, down, left, right, over yonder, over yander.
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That sort of direction. Forward or backward when we are talking about just a positive or negative value is not what we are talking about here.
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On the other hand, vectors are quantities that have a magnitude and a direction. They need a direction to describe them fully, things like a velocity. You have a velocity of 10 m/s in a direction.
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A force is applied in that direction. Or a momentum, you have a momentum in a specific direction.
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Vectors, we typically represent by arrows. The direction of the arrow tells you the direction of the vector, obviously, and the length of the arrow represents the magnitude or size. The longer the arrow, the bigger the vector.
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Let's take a look at a couple of vector representations. Let's call this nice happy blue one A, and this red one down here B.
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Notice they both have the same direction but A is much smaller than B. A has a smaller magnitude than B. B is the longer arrow, with larger magnitude.
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Now the other thing that is nice about vectors is as long as you keep their magnitude, their size and their direction the same, you can slide them around anywhere you want. You can move a vector as long as you do not change it's direction or its magnitude.
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So if we want to, we could take vector A and instead of having it there, we can slide it somewhere over here, for example, give it the same direction and magnitude, make this one go away, and now there is A.
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With the same value, same direction, same magnitude, we are allowed to move them like that.
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Let's talk about how we would add up two vectors. A vector such as A and B. The little line over that means it's a vector. If we want to try and put together, A and add it to vector B, to get sum vector, C. The sum of those two vectors.
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Well, graphically, here is the trick. Take any vectors you want to add, however many there are and if we slide them around so they are lined up tip to tail, we can then find the resultant, the sum of the vectors.
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So here we have A and B but they are not lined up tip to tail. So, what I am going to do is I am going to redraw these so I am going to put A over here and B, I am going to line up so that it is now tip to tail with A. Hopefully something roughly like that.
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So now we have A and B lined up so that they are tip to tail. To find the sum of the two vectors, all we have to do is draw a line from the starting point of the first to the ending point of our last vector, that must be the sum of the vectors, C.
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Alright, does it make a difference what order you add things? Well if you think back to math, B plus A should be the same thing, and it is.
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Lets prove it. We are going to redraw this now but we're going to do B first.
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So what I am going to do is I am going to draw B down here, there's roughly B and now I am going to put A on it but I am going to line them up tip to tail, in this direction this time so B comes first and then A.
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Once again when I go to draw the resultant, I go from the starting point of the first to the ending point of the last. Notice that I have the same thing. Same magnitude, same direction, same vector, same result.
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Alright, how about graphical vector subtraction? Here we have A again and B. Put the line over them to indicate they are vectors. What do we do for A minus B?
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The trick here is realizing that A minus B is the same as A plus the opposite of B.
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What is the opposite of B? Well it is as simple as you might guess.
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If we have B pointing in this direction with this magnitude, all I have to do is switch it's direction and there is negative b.
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So if I want A plus negative B, let's just redraw them again, tip to tail. We will start A down here. There is A. Now negative B goes something like this.
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So A plus negative B, or A - B, we go from the starting point of our first again to the ending point of our last. A plus negative B equals C. Basic vector manipulation.
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Now when we have these vectors and they are lined up at angles, often times we can simplify our lives from a math perspective if we break them up into component vectors or pieces that add up to the sum.
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If we pick those pieces carefully, so they line up with an axis, the math gets a whole lot easier and I am a huge fan of easier math.
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Let's assume that we have some vector, A right here at some angle θ from the horizontal. We could replace this with a vector along the X axis and a vector along the Y axis.
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Notice that the blue vector plus the green vector, if we add them together, gives us that A vector, the vector we started with.
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So we are going to take this A vector and we are going to replace it with this blue one and this green one. Two vectors that are a little simpler to deal with mathematically.
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Let's call this the X component of A and let's call this the Y component of A. How do we figure out what those are?
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If you notice, here, we have made a right triangle. Here is our hypotenuse, this is the opposite side because it is the opposite the angle and AX must be the adjacent side, it's beside the angle.
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Now I can use trig to figure out AY is. AY, since it's the opposite side is going to be equal to A, the hypotenuse times the sine of that angle.
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On the same note, AX is going to be A times the cosine of θ because this is the adjacent side. remember SOHCAHTOA?
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Sine of θ equals the opposite side divided by the hypotenuse, cosine of θ equals the adjacent side divided by the hypotenuse and tangent of θ equals the opposite side divided by the adjacent.
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All we are doing is we are finding out what this opposite and this adjacent side happens to be. So we can break up this vector A into components AX and AY that are going to be much easier to deal with mathematically.
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We could also go back to finding the angle of the vector. If we know two of the three sides of these triangles, if we know both of the components, we can find the angle, if we know the hypotenuse and the opposite, we can find the angle.
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How do we do that though? Well we have to go back to our trig functions.
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Tangent of θ equals the opposite over the adjacent side. Therefore θ must be the inverse tangent of the opposite side divided by the adjacent side.
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But what if we do not know opposite or adjacent? Well sine θ is equal to the opposite over the hypotenuse. So if we know opposite over hypotenuse, we can find θ by taking the inverse sine of the opposite side divided by the hypotenuse.
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So if you know any two sides of this right triangle you are making with components, you can find the angle using basic trigonometry.
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Let's talk for a few minutes about vector notation. You can express vectors in many different ways.
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You can just draw it on a sheet of paper, you can express it mathematically, but want to do this as efficiently as possible. so I am going to show you some examples in 3 dimensions but you can always scale those back to just two dimensions
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Let's start off by making an axis. We've got YX and lets have a ZX coming out towards us. If we have some vector A, we could express it as having an X component, a Y component, and a Z component.
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On the other hand though, we could also look at in terms of what are known as unit vectors.
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If we take a vector of length 1 along the X axis, magnitude of 1 along the X axis, we are going to call that specific vector ihap, length one along the X axis.
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In the Y axis, we will do the same thing. A vector of unit length, of length 1 in the Y direction, we will call jhap.
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In the Z direction, same idea. A vector of length 1 in the Z direction we'll call khap. Specific vector constants. So we could write A now, as some value, X value times ihap plus its Y value times jhap plus its Z value times khap.
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So whatever the X value is, you multiply it by a vector unit length 1 in the X direction.
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Y value times the unit vector of length one in the Y direction and the Z value times the unit vector of length one in the Z direction.
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Another way to express vectors. That can be very useful when we get to the point of doing vector addition. Let's assume we have our axis here again.
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Y, X and Z. And here let's put a vector that is 4 units in the X, 3 in the Y and out toward us, 1.
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So let's call this point P, 4,3,1, which is defined by some vector P which is 4 units in the X, 3 in the Y and 1 in the Z.
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Let's also define another vector Q. Let's go 2 units in the X, we won't go any in the Y, zero in the Y and let's come out toward us in the Z direction 1,2,3,4.
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Let's call that point Q which is 2, 0, 4 and we'll label the vector from that origin to that point vector Q.
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How do we add these vectors in multiple directions? Well, what we could say is that vector R is going to be equal to Vector P plus vector Q.
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Therefore, let's write P as equal to 4,3,1 in this bracket notation for vectors and vector Q is equal to 2,0,4 in vector bracket notation.
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Well if the left hand side is equal to the right here and the left hand side is equal to the right here, then we can add the left hand sides and add the right hand sides, they should still be equal.
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What we can say then, if we add those two, and add those two. Therefore P plus Q which is equal to our R must be equal to, well, in vector bracket notation, we add up the X components 4 plus 2 is 6.
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We add up the Y components, 3 plus 0 is 3, and we add up the z components, 1 plus 4 is 5.
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So the resultant, R, would be 6 units in the X, 3 in the Y and then 5 towards us. Something like that in 3 dimensions.
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Adding up vectors using that vector notation can make things a lot simpler especially when you don't want to go drawing all of the time.
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Let's take a look at a vector component problem. A soccer player kicks a ball with the velocity of 10m/s with an angle of 30 degrees above the horizontal. Find the magnitude of the horizontal component, and vertical component of the ball's velocity.
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I am going to start off with a diagram here. A Y axis, and an X axis and realize that the soccer player is kicking the ball with an initial velocity of 10m/s, so there is our vector, 10m/s at an angle of 30 degrees above the horizontal.
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We want to know the horizontal component and the vertical component. As you recall, if we want the vertical component, if this is our initial velocity, P, then the Y component of that velocity is going to be V 10m/s times the sine of 30 degrees.
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10m/s sine 30 should be 5m/s.
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In similar fashion, the X component of velocity V is going to be V cosine θ again or 10m/s times the cosine of 30 degrees.
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Cosine 30 is 0.866 so 10 times that is going to be 8.66m/s. We have broken up V into it's X and Y components.
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Alright, another one. An airplane flies with the velocity of 750 kmph 30 degrees south of east. What is the magnitude of the plane's eastward velocity?
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Well let's draw a picture again. North, south, east and west. The airplane flies with a velocity of 750kmph 30 degrees south of east. That means start at east and go 30 degrees south.
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So I am going to draw it's velocity as roughly that. 750kmph at an angle of 30 degrees south of east. If we want it's eastward velocity, the eastward component, that means we want the X component here.
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X component of it's velocity, the X is going to be V cosine θ or 750kmph times the cosine of 30 degrees 0.866 should give us something right around 650kmph.
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Let's take a look at another one where we have to deal with vector magnitudes. A dog walks a lady 8 meters due north then 6 meters due east, I'm sure you've all seen that before. A big dog, a little person trying to walk it but really the dog is in charge?
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Determine the magnitude of the dog's total displacement. Well if the dog walks the lady 8m due north, we'll have a vector 8m north and then 6m east. Determine the magnitude of the dog's total displacement.
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Well if we start it down here and line these up tip to tail so that the total displacement is a straight line from where you start to where you finish, is going to go from here right to there. That's the displacement.
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How do we find the magnitude of that? Well if we look, that's a right triangle. We can use the pythagorean theorem. A² plus B² equals C² where A is our 8m, B is our 6m, C is going to be our hypotenuse or the displacement.
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Therefore, this is going to have a magnitude of the square root of 8m² plus 6² or the square root of 64 plus 36, square root of 100 is going to be 10 meters.
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Say we wanted to know what this angle is. If we wanted to know that, we could take a look and say, you know, the X component of that green vector is going to be 6m, the Y component must be 8m.
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Therefore if we wanted that angle, θ is going to be the inverse tangent of the opposite side over the adjacent which is the inverse tangent of 8m over 6m which comes out to be about 53.1 degrees.
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That would be our angle, θ there too.
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If it had asked us for the angle, it only asked us for the magnitude of the dog's total displacement which we found to be 10 meters.
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Let's take a look at some more vector addition. A frog hops 4m at an angle 30 degrees north of east.
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He then hops 6m at an angle of 60 degrees north of west. What is the frog's total displacement from his starting position?
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This just screams for us to draw a picture here first. So let's draw our axis here. I have a Y axis, and an X and as we look at this, There's our X, here is our Y, The frog starts out 4 meters at an angle of 30 degrees. There is 4 meters at an angle of 30 degrees.
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Then he is going to go and hop 6m at an angle of 60 degrees north of west. So 6m at an angle of 60 degrees north of west is probably something kind of like that.
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That angle is 60 degrees north from west and that is 6m long, thats 4m long. What is the frog's total displacement from the starting position.
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Well, I could find that out graphically, by drawing a line from the starting point of the first to the ending point of the last.
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Or, if I wanted to do this analytically, or a little bit more exactly, I could take a look if our blue vector is A, A is equal to it's X component is going to be 4m cosine 30 degrees, and it's Y component is going to be 4m sine 30 degrees.
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Our B vector, there in red, is going to be, well we have got 6m cosine 60 degrees for it's X component, but it is to the left, so let's make sure that's negative and it's Y component is 6m sine 60 degrees.
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So if I wanted to find the resultant, the sum, vector C. C is just going to be equal to A plus B, so that's going to be 4m cosine 30 degrees. The X component of A, plus the X component B, negative 6m cosine 60 degrees. So that will give us the X component of C.
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For the Y component, we add their Y components together. 4m sine 30 degrees from A plus 6m sine 60 degrees from B.
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When I do the math here, I find out that C equals 4 cosine 30 plus negative 6 cosine 60, that is going to be about 0.46m and the Y component 4m sine 30 degrees plus 6m sine 60 degrees comes out to be 7.2 meters.
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So there is our C vector. 0.46, so not much in the X, 7.2 in the Y. While we are here, let's find out it's magnitude and angle.
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The magnitude of C, I take the C vector and take it's absolute value, I can find out by using the pythagorean theorem again since I know it's components.
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That is going to be the square root of 0.46² plus 7.2² it comes out to be about 7.21m.
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If we wanted it's angle as well, I am expecting a big angle here just by looking at the picture. θ is going to be equal to the inverse tangent of the opposite side over the adjacent side.
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The opposite side is the Y, 7.2 over the adjacent 0.46 for an angle of 86.3 degrees which is over here 86.3 degrees north of east.
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So we could express the vector with magnitude and a direction or we could express it just by leaving it in the vector bracket notation. If we wanted to we could have even written it as 4 6m ihap plus 7.2m jhap. They are all equivalent.
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Let's take a look at one more sample problem, the angle of a vector. Find the angle θ depicted by the blue vector below given the X and Y components.
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Since I am given the opposite side, opposite the angle θ and the adjacent side, the side next to the angle, but not the hypotenuse, I am going to use the tangent function since tangent of θ equals opposite over adjacent.
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Therefore θ is going to be the inverse tangent of the opposite side over the adjacent side. Or θ equals the inverse tangent of the opposite side 10 divided by the adjacent, 5.77 or 60 degrees.
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Hopefully this gets you a good start on vectors and scalars. We will be using them throughout the entire course. They are very important.
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Thanks for watching educator.com. Make it a great day.