Practice: PRACTICE 1: You move 15 m due east, then 30 m north, then 25 m east. Draw displacement vectors and find the magnitude and direction (use the *absolute* angle) of your total displacement.

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Intro, Units & Conversions | 32 mins | 0 completed | Learn |

Intro to Vectors (Basic Trigonometry) | 58 mins | 0 completed | Learn |

Vectors with More Trigonometry | 52 mins | 0 completed | Learn |

Additional Practice |
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Units & Conversions |

Dimensional Analysis |

Significant Figures |

Uncertainty |

Geometry & Trigonometry |

Vector Composition & Decomposition |

Unit Vectors |

Vector Addition & Subtraction |

Concept #1: Intro to Trig Review

**Transcript**

Hey guys, so now that we've done a lot of work with vectors we've done vector composition, decomposition, vector addition and all kinds of different directions I want to take this one step further and I don't know if you've noticed but all the vectors we've used so far were either flat in the X axis, flat in the Y axis or they were in the first quadrant or in the fourth quadrant and the reason for that is that those are much simpler and we don't have to get into the ugly details of trigonometry to solve those questions but now that you have a good foundation of how vector addition, composition, decomposition works it's time to go one step further so I'm going to do a two video summary review of all the trig you are going to need to know for physics which is actually not a lot there's only a few things from trig that you really need to know and that's what these two videos are for and you need to be crazy good at this it's going to help out a lot throughout the course, so let's jump into it.

Concept #2: Trig Review & Using Arctan

**Transcript**

Hey guy so it turns out that there is actually very few things you need from trig in physics which is awesome news, but you need to be a ninja at them at these things because they are going to show up all the time right so first I am going to show you some unusual cases or uses of the Pythagorean theorem and soh cah toa and the reason I say unusual is because most of the time you're going to be able to use your third vector equations to solve these problems every now and then you are going to get some weird thing thrown at you and you're going to have to you might have to use these guys so before I start I want to point out that every tiangle in physics would be a right triangle so this is 90 degrees here and you're going to have all triangles of four variables the three sides and the angle now there's an angle here but it doesn't count because if you know this angle you know that angle as well so these are the four main variables and as long as we have two of them you can find the other two using a combination of Pythagorean theorem and soh cah toa work. So here I'm giving you A equals 5 A X equals 4 and we're looking for A Y and we're looking for theta the easiest way to do this is to realize that I have 2 sides so I can find the third side using pythagorean theorem. Whenever you have two sides you can find the third one whichever combination of two you have works, once you do that you can then find theta. Now that being said you can actually find theta without finding the A Y first as long as you know any two sides you can find theta using soh cah toa so. So if you are willing to do that just to show you this particular case because you're not going to use it very often. Here you have this is your opposite this is your adjacent and this is your hypotenuse and you have your adjacent and hypotenuse so if you go here adjacent, hypotenuse it's cosine CAH means that the cosine of theta is a adjacent over hypotenuse but you don't want the cosine of theta you want theta so theta is the arc cosine of the adjacent which is 4 over the hypotenuse which is 5 you plug this into a calculator you get your lovely answer but I don't actually want to go that route I just want to show you that real quick I want to find A Y first using pythagorean theorem it's the easiest way to do it, it's probably what you should do. A squared plus B squared equals C squared so A X squared plus A Y squared these are the components equals the whole vector right let's plug in the numbers here 4 squared A Y is what we are looking for so I'm going to do this and this guy here A is 5 squared I'm going t move the 4 over here becomes a negative so I get A Y squared equals 25 minus 4 square 16 this is a 9 so A Y is just the square root of 9 which is 3. So A Y equals 3. Now I can find the angle and now I have all three sides so I can pick whichever one of these three I want soh cah toa I'm going to pick toa because that's the one we usually use so it's good practice. In toa it says that the tangent of theta equals opposite over adjacent but I want theta so theta is the arctangent of opposite which is now we know it's a 3 over the adjacent which is a 4 and if you do this carefully you calculate it you get 37 degrees so that's this right here 37 degrees I'm going to highlight the answers in green, 3 and 37. Now let's look at this other one a little unusual here I got the angle up there usually this is a 30 usually we're going to want the angle to be against the X axis this angle here is against the Y axis. This would've been against the X axis but we want the angle here typically what you would do is move this angle and make it a 60 just get rid of this guy and use 60 instead but let's first ourselves to use that angle to practice soh cah toa and pythagorean theorem, so what else do I know I know A Y equals 8 and I'm looking for A X and I'm looking for A. Now you actually cannot use pythagorean theorem here because you only have one side you don't have both sides so your going to have to use soh cah toa right you have the angle you have the this is the adjacent side that you have but you don't have the opposite side or hypotenuse. So just pick which one you want to solve for first do you want to solve for the opposite or do you want to solve for the hypotenuse you want to basically get out of this lock that you're in I'm going to solve for the opposite angle first that's just arbitrary decision but if you look here opposite is here and adjacent here so tangent is going to be the angle that I'm going to the Toa part of soh cah toa that I'm going to use so it says that tangent of theta is opposite over adjacent now usually what you do with this is you take the arc tangent to find theta but now I actually have theta and I'm looking for something else so this is going to be the tangent of theta of 30 rather sorry equals opposite which is what I'm looking for divided by adjacent which is 8 so the this 8 can multiply and come back come up here I'm going to get 8 tangent of the 30 is the length of my opposite side and if I do this I get that my opposite side which is A X equals put this in the calculate I get 4.62. So this is 4.62 now I have two sides I can find the third one using pythagorean theorem A X squared A Y squared equals A squared A X is 4.62, A Y is 8 and if I take the square root of all this and I plug it in the calculator I get 9.24. So let's keep going now I'm going to show you a bunch of other small quick things here this first piece you already know this vector composition if I'm moving this way I have an A X I move up this is an A Y I can use these two guys to find my A and my theta and the way that works is the magnitude of A is given by the square root of A squared plus A X squared A Y squared this is pythagorean theorem and this angle here is given by the arc tangent of your A Y over A X right now the flipside of that is I could give you an A the length of the vector and ask you what are the components of that which are these two guys and then we're going to find A X and A Y and the way to do that is A X equals A cosine of theta and A Y equals A sine of theta. Remember we want X to go with cosine we want the angle to be against the X axis so you absolutely need to know this for equations you are going to use them all the time right so next thing here you need to circle remember this guy from trig probably not good memories but 0 is right here on the unit circle its the positive X axis and the unit circle goes like this this is the positive direction of the unit circle counter-clockwise should've made that red but whatever this is the blue is going to be the negative direction which is clockwise so what you need to remember in this is that the clock is backwards from the unit circle or vice versa. Quadrants we grow this way starting from 0 here so this is first quadrant second quadrant third quadrant fourth quadrant you should know that this is the positive Y axis negative Y axis positive X axis negative X axis and one consequence of that is if I have a vector sort of going this way and I want to decompose let's call this a force F and I want to decompose this force this is my F X right here right so flat against the X axis and then this is my F Y flat against the Y axis F Y is positive because it's going up and F X is negative its going to the left and remember that going up is positive going to the right is positive this is the convention for vectors. Triangles and complimentary angles I already talked about this briefly earlier, if this is 60 then this over here is 30 you want to use the blue angle not the red angle we'll talk more about this later. If this is 60 then this over here would be a 60 as well. So notice that this guy is between this angle is between the vector and the X axis this angle is between the vector and the X axis as well and this guy is with the Y axis so this guy with the Y axis over here is 30 as well. If I make a triangle out of this I can make a triangle this way or this way and actually from the rectangle but if I do that then this is 60 and this is 30 over here this is what the Y axis this is what the Y axis they're both 30 so you need to know these kinds of things and how they work.

The last thing is I that I want to talk about angles using north south west and east you know put it here north is this way south is this way west is this way east is this way and how do we do this 30 degrees north of east you might see that well the strategy I use is to start with east the second word and go back to the first word let me show you what I mean east is to the right so I'm going to go east and then I'm going pull towards north so I'm here and I'm going to pull towards north and this is my angle so I'm going to pull towards north then east this is my eagle right there that's the 30 degrees that I'm talking about now let's do this one real quick south of west again I'm going to go this way I'm going to start going south and I then I'm going to pull towards west. So I started at I started south and I'm going to go west like that, so it's going to look like this theta equals 30 and that's it now there's 8 variations of these its kind of annoying so one two three four five six seven eight possibilities which are the combinations of you know north of east and east of south and all kind of difference once and you can hardly get that. So the next thing I want to talk about is when you take the arc tangent of a function you have to be very careful there's a few rules to remember and you're going to do this when you're putting together a vector remember theta equals the arc tangent of Y over X so you are going to use that a lot the arc tangent function you need to know always gives you an angle in either the first or fourth quadrants first or fourth quadrant this is first this is fourth. Even if U vector is somewhere over this way the calculator is always going to give you an angle that is relative to the X axis either above or below the X axis. If a vector happens to be in the second or third quadrant you are going to do the adjusting yourself you're going to have and to do that you can just add or subtract 180 the reason I say add subtract you pick is because it doesn't matter if you add or subtract 180 you end up in the same place for example if I'm here at 10 degrees and I add 180 I end up at 190 but if instead you subtract 180 from here you end up at -170 and guess what those two numbers are the same. Both end up in that thick blue line.

Alright lets do a problem for each set of vectors find the absolute angle of the vectors so A X equals 4, A Y equals -3 so A X equals 4 looks like this A Y equals 3 looks like this. You could've gone Y first X after it would've given you the same thing in fact it would look like this A Y, A X you still end up having the same arrow doesn't matter and this is your A vector right here those are the legs of that vector. Now remember the angle that you want from the starting position here is against the X axis so you want the angle against the X axis, you want this angle this is the angle with the Y you want that's bad you want this angle right here that is your theta. Actually we want the absolute angle so it's either going to be the total angle going this way or the angle going this way sorry about that so but definitely not this guy right so let's see if we can figure this out I have the two legs so I can just say that theta is the arc tangent of A Y over A X, A Y we plug in with the signs A Y is -3 A X is 4 and I'm going to get a -37. This angle is in the fourth quadrant which is ok that's a good quadrant for Artesian function so you don't have to make any adjustments -37 is the correct answer if you want to make this a positive angle you can just add 360 but you don't have to right you can add 360 if you want to but you don't have to absolute just means that it's relative to 0 right it's coming from 0 in either directions so that's the right answer. Let's do the next one so here A X is -4 so I'm going to go -4 to the left or 4 to the left and A Y is 3 so it goes up like that so your vector A vector looks like this and typically you would want this angle here but in this situation I actually want the absolute angle which is going to be this angle right here from 0 from the positive X axis 0 degrees. So I'm going to call this angle theta absolute here by the way the theta absolute was the same as the theta that we of that we got a diagram. So the first I want to do is find this theta right here and then make the adjustments so theta is the arctangent of Y over X, A Y over A X. Its the arctangent of A Y 3, A X -4 and if you plug this in you're going to get -37 degrees now you are on the second quadrant which is bad you need to adjust and to adjust you're going to have to add or subtract 180 degrees to find the correct angle so from -37 I'm going to add 180 degrees and I'm going to get 143. That is the correct angle right that's the absolute angle now that should make sense because if you even count this slowly here you will get that this is 90 this 37 is this angle here except that it's positive so other guy here needs to be 53 so 90 plus 53 that's the other way of doing it but it's a little bit more work and more thinking is 143 or you can just know that whenever you're in the second quadrant or the third quadrant you can just add or subtract 180 to get the actual absolute angle that you're supposed to have. Alright so that's it for the one.

Concept #3: Using Sine and Cosine

**Transcript**

Hey guys so 2 out of 3 through Trig functions are going to be using all the time are sine and cosine and there are specific things you need to know about those functions when you're using them, now before we start I want to talk about angles and 3 basic types of angles you can have and angles can be absolute or they can be relative, relative to the X axis or Y axis, absolute means that it's measured from 0 degrees or the positive X axis which is right here, relative which can also be before it was against or measured from so I can so for example that this angle right here for this vector is against the X axis, I'm going to call this theta X or I can say that this angle over here is measured from the Y axis. Relative, measured from, against same thing those are the 3 types of guessing the types of angles and you should be able to convert from one to the other so for example let me say let's say that this angle right here is 30 degrees, this angle right here is 30, now this angle is against the X axis because its closest to the X axis now it's negative X axis but that is the closest X you get so that is the angle that is relative to the X axis so I'm going to make it blue and let's say that this is 30 degrees, if that angle is 30 then this angle over here is 60 that angle is the one that goes with the Y axis so it's measured from the other Y axis so this has to be 60, I'm going to make this little bit smaller here and the absolute angle that comes all the way from a 0 I'm going to make that green and it's going to be this guy here, OK? Now this is angle is obviously 90+60 so it is 150 degrees so you should be able to get any one of these angles and convert them to the others and I know it's going to be more clear as to what you're doing with that so here let's find the angles relative to X and Y that are equivalent 210 so first of all I hope you remember this is 0, 90, 180, 270 so 210 is somewhere here, in fact 210 is 30 more than 180 this is 180, right here negative X axis so 210 is simply 30 more than that so that 30 degrees is my angle with the X axis that's the angle relative to the X axis closest to the X axis or the angle against the closest X axis, the angle against the closest Y axis which is axis right here is going to be this one, that is a 60 and the absolute angle starts from here and it goes all the way around so obviously that angle is going to be 180+30, 90+90+30 so this angle is 180+30=210, alright? It's just kind of working our way back but this is your angle with the X, this is your angle with the Y over here let's do another one, so a vector is directed 37 degrees off the negative Y axis counterclockwise, so negative Y axis is this guy and I'm going to go off the Y axis plus 37 counterclockwise, remember that the unit circle goes like this that's the positive direction and they come from here counterclockwise so I'm going to be counterclockwise so I'm going to be 37 degrees off the Y axes in this direction that's the angle right there, so this is the vector and the angle that I'm given is 37 right here that is the angle with the closest Y axis right there so this angle here which is the angle closest with the closest X axis is going to be 90 minus that so 90-37=53 and when I do the absolute angle there's two ways I can go about it, I can go all the way around but then it's a lot of adding or I can just go in the shortest way from 0 right here which would be just this way and the angle the absolute angle here is simply -53, OK?

Now notice how for the relative angles I don't have to put signs because I know that they're relative to those but the absolute angle you need to put a sign on it, OK? Cool so the reason we're talking about this is because you need to know what kind of angles you can use, what kind of angles are better if the angle is absolute or the angle is against the X axis those are the two good angles they're good angles then you AX is going to be A cosine of theta and your Ay is A sine of theta and I've mentioned this a few times, X goes with cosine if you remember that but hopefully you remember that the other variable Y goes with the other Trig function sine, right? So if X goes with cosine then the other variable Y goes with the other function sine so we actually only have to remember one of them if your angle happens to be against the Y axis and you can't change it for whatever reason then it's basically the opposite here, A will be going with sine.....I'm sorry AX is going to go with sine and Y is going to go with cosine, so if you have that bad angle then you have to remember to flip the Trig functions, my recommendations is that you stick with the good angle if you are given an angle with Y axis just replace it so again for example if I know that this is 30 I get rid of it and I make this a 60 and I can use that your professor is going to be fine with that unless he specifically tells you to use the angle that's given which is very rare, right? Usually he doesn't care in fact he might even be proud of you that you know how to flip the angles, Cool.

let's see decompose each vector they both have size 5 let me write that here using the given angle, don't change and so don't swap the angles lets decomposer them, so decomposing the vectors means that I'm going to have let's call this A and this is going to be B so I have AX down here and then I have AY up here and to decompose these things I just have to do AX is A cosine of theta we've done this before, A is 5 cosine of 37 so that is a 4 and Ay is A sine of 37 or 5 sine of 37 which is a 3, notice how the calculator gave us positive values that's good news and they are positive values, up and to the right, OK? There's nothing fancy here in fact we've done that very problem already have few times so here I'm giving you the wrong angle and the point here, is just that since you can't change the angle you can't flip the angle, right? This would be 37 right here which happens to be the same angle but you can't use it you're going to have to use the wrong bad angle so you have to flip the Trig functions, if you can't flip the angles you have to flip the Trig functions so now AX... sorry BX is going to be B sine of theta and BY is going to be being cosine of theta notice how they're flipped B is 5 sine of 53, that is a positive 4 and if you plug these numbers here you get a positive 3 and if you notice these are the same exact numbers guess what because the angle was exactly the same angle they're both point in the same direction, I just gave you in one problem this measurement and the other problem this measurement up here but it's obviously the same exact thing and that was that example just to show you how to use the wrong one if you have to and how it's basically the same thing, OK?

So, if your angle is absolute, right? So, this is the conclusion of this if your angle is absolute if you have the absolute angle then the sine cosine function when you plug it into the calculator is already going to give you the correct sign whether it's positive or negative, right? If it's positive it's going up negative to the left and all that kind of stuff, it will give you the correct sign which is awesome if you don't use that you're going to have to add the signs yourself so either use the absolute angle and it give you the correct sign or you should be using the angle that is relative to the X axis, OK? So for example if I gave you that this is 60, you're going to use either the 60 or you're going to use this angle right here which is 120, you're going to use either one of these 2 angles which you're not going to use with the Y axis don't use the Y angle as much as possible because then you're going to have to flip your functions but if you use the absolute angle or the X angle then you're good to go the only difference is for the absolute angle you don't have to worry about signs and with the relative angle you have to look at your vector and make sure that you plug in the signs we'll do an example just now so the question is why not just always use the absolute angle? Well they're both just as much work for you to use the absolute angle you have to find the absolute angle which my required adding 90 or adding 180 or whatever and so either you want to worry about finding the absolute angle but not the signs or you don't want to worry about finding the absolute angle but then you got to do the signs it's a matter of preference, different professors do it different ways I recommend you go with whatever your professor prefers and if it doesn't seem like he's very picky about that then you just go with whatever you prefer and here I'm going to do the two different ways to illustrate this point so this is 5 and I got 37, notice that this angle is with the Y axis this is a bad angle, right? The reason again I say bad is because I want you to just memorize that X goes with cosine and not have to worry about flipping those instead we're going to fix the angle and not the Trig function so I want to do this using the absolute angle, absolute angle looks like this, OK? If this is a 37, this guy over here is 90-37=53 so the absolute angle is going to be 180+53 so theta absolute is 180+53 and that is 233 degrees, the angle with the X axis it's negative 53 there or you could have just used the 53 because you're going to have to fix the signs anyway, yea let's just use 53 because relative angles should just be positive though again it doesn't you're going to have to go back and check the signs anyway, so the first one is to do this with the absolute angle and I told you that if you're decomposing this, Let's say that this is a force so you have Fy and Fx, again it does doesn't matter where I draw these things I could have drawn Fx over here as well let's leave it there makes sense more with the little triangle over here, cool. So, remember if you're using the absolute value then X goes with cosine and Y goes with sine if you plug this into the calculator you get 5 cosine of 233 and 5 sine of 233, your Fx is going to come out to be -3 and the Fy is going to come out to be -4 if you look this is confirmed the signs are correct, this is a negative because it's going to the left and this is a positive because it is going down, OK? So that makes sense as it should the calculator knows because you're telling it exactly what I need to go to, when you give the calculator instead of 53 the calculator doesn't know for sure where that angle is, in fact the calculator thinks that you're talking about this 53 over here, right? So that's why it's going to give you a positive and you're going to have to change them to negative so now with the absolute angle with the relative angle Fx=F cosine of theta, 5 cosine of 53, Fy= F sine of theta, 5 sine of 53 and if you do this you get a positive 3 and a positive 4 put them into the calculator but then you look at your diagram and realize actually they're negative and since I'm using the relative angle not the absolute angle I'm going to have to verify my signs and then that's when you're going to say actually these guys are just put a negative, -3, -4 either way whenever you get an answer you should always be double checking with the vector diagram anyways to be safe but again the point here is you can use either one and the absolute value will have the benefit of not having to find the correct sign if you do the actual work of finding the absolute value, alright? That's it for.

Concept #4: Vector Problems (Practice Intro)

**Transcript**

Hey guys, so here I have four more practice problems for you guys to work on, let's take a look. So before we start I'm going to remind you that we have four vector equations this is your A, this is your A X, this is your A Y and this is your theta and you can use these four questions to help you solve these problems, you want to use the tip to tail method to connect different segments and you want to remember the rules we talked about for sine cosine and arctangent. Alright so give this a shot and hopefully you get them right.

Practice: PRACTICE 1: You move 15 m due east, then 30 m north, then 25 m east. Draw displacement vectors and find the magnitude and direction (use the *absolute* angle) of your total displacement.

Practice: You move 4 m due east, 5 m due north, and 6 m due south. What is the magnitude and direction (use the absolute angle) displacement would you need to *return* to your point of origin?

Practice: You move 5 m along the positive X axis, then 10 m directed at 53 degrees above the negative X axis. Find the magnitude and direction (use the absolute angle) of your displacement.

Practice: You move 2 m due north, then 5 m directed at 53 degrees east of north. What magnitude and direction (use the absolute angle) displacement would you need to *return* to your point of origin?

0 of 8 completed

For the three vectors shown in (Figure 1), A + B + C = 1 ĵ. (Not necessarily to scale.)(a) Write B in component form. Express your answer in terms of the unit vectors î and ĵ.(b) What is the magnitude of B ? Express your answer using two significant figures.(c) What is the direction angle of B , measured clockwise from the negative x axis? Express your answer using two significant figures.

The diagram below shows a block of mass m = 2.00 kg on a frictionless horizontal surface, as seen from above. Three forces of magnitudes F4.00 N, F2 6.00 N, and F3 8.00 N are applied to the block, initially at rest on the surface, at angles shown on the diagram. (Figure 1) In this problem, you will determine the resultant (total) force vector from the combination of the three individual force vectors. All angles should be measured counterclockwise from the positive x axis (i.e., all angles are positive).

a) Write the vector A in the figure (Figure 1) in terms of the unit vectors î and ĵ.b) Write the vector B in the figure in terms of the unit vectors î and ĵ.c) Use unit vectors to express the vector C , where C = 3.00A −4.00B .d) Find the magnitude of C .e) Find the direction of C .

You are given two vectors: A = −3.00 î + 5.00 ĵ and B = 8.00 î + 2.00 ĵ. Let the counterclockwise angles be positive. Vector C is the sum of A and B , so C = A + B. What angle θC, where 0° ≤ θC < 360°, does C make with the +x-axis?

You are given two vectors: A = −3.00 î + 5.00 ĵ and B = 8.00 î + 2.00 ĵ. Let the counterclockwise angles be positive. What angle θB, where 0° ≤ θB < 360°, does B make with the +x-axis?

You are given two vectors: A = −3.00 î + 5.00 ĵ and B = 8.00 î + 2.00 ĵ. Let the counterclockwise angles be positive. What angle θA, where 0° ≤ θA < 360°, does A make with the +x-axis?

Find the components Nx and Ny of N in the tilted coordinate system. Express your answer in terms of the length of the vector N and the angle θ.

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