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Summersnow8
 one year ago
Find the resultant of each set of displacements: 60 km due south, then 90 km at 15 degrees north of west, and then 75 km at 45 degrees north of east. I know the answer is 38 km at 25 degrees north of west.
Summersnow8
 one year ago
Find the resultant of each set of displacements: 60 km due south, then 90 km at 15 degrees north of west, and then 75 km at 45 degrees north of east. I know the answer is 38 km at 25 degrees north of west.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So what do you need? How to get there?

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.1yeah, I need help with how to solve it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This seems like physics ._. @hba You good in Physics?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im new to physics lol

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.1ok, @undeadknight26 thanks for trying

princeharryyy
 one year ago
Best ResponseYou've already chosen the best response.0can you draw the figure

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.1a figure isn't given, but this is the figure i have drawn

princeharryyy
 one year ago
Best ResponseYou've already chosen the best response.0until you can't draw it you can't solve it.

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436290559148:dw

princeharryyy
 one year ago
Best ResponseYou've already chosen the best response.0try solving it now!

princeharryyy
 one year ago
Best ResponseYou've already chosen the best response.0ok! leave it! HOW DID YOU SOLVED IT!

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.1I don't know how to solve it, that's why I am asking for help

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1woops let fix your graph a sec :) 15 `north` `of west` means we start by facing west, and rotating up 15 degrees.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1So what are we doing? Adding up these vectors? :O

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1That resultant shinanigans and all that? +_+

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Let's figure out the components to each vector or force as I've written them :) For F_1, there is no x component, we've gone zero to the left or right. So our F_1 is simply \(\Large\rm \vec F_1=\left<0,60\right>\) Is this notation ok? Or do you prefer the i's and j's? :3

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1SummerPotatooooo, just chat in here XD don't need private message box lol

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.1what are i's and j's? I guess whatever is easiest

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436292562770:dwLet's say we wanted to get to this point. We would go some distance to the right, and some distance upward. This is i that I've labeled, it's 1 unit long, in the x direction.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436292657672:dwSo to get to that point, we'll need 3 i's. We moved 3 units horizontally.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1And j is 1 unit vertically. So we need 4 j's.

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.1I've never learned that way....I know to use trigonometry

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436292736659:dwSo what I've done is, I've written this vector in component form.\[\Large\rm v=3\hat i+4\hat j\] Our problem is kinda tricky. We're only given the length of the vector, so yes, you have to use trig to help you find each component of the vector.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Haven't learned that? :[ hmm

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.1I use x and y no i and j

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Let's look at the F_3 that I labeled on the graph. That one will be easiest so maybe we should start there.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436292975469:dwIf we think of this as a triangle

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1We can use our trigonometry to get our x and y values.

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.1so x = 53 and y = 53

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Mmm good good that sounds right! :) So here is a way we can write our F_3,\[\Large\rm \vec F_3=53x+53y\]

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.1This was my attempt at solving

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Ok your process looks great! Let's look at displacement B a second though.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Notice that arrow is pointing to the left. That tells us that our x coordinate should come out to be negative, ya?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1mmm yah that looks correct!

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.1now A really confuses me

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436293376046:dwAh yes :) kind of hard to make a triangle with this one, huh?

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.1yeah... that's where i get stuck. one I solve that then I know to add up all the x's and y's and take the square root of x squared + y squared

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1If you can get used to this shortcut, then it might help when we get stuck. In the first triangle, you were able to find x by this formula: \(\large\rm x=A\cos(\theta)\) Where A is the length of A, and theta is the angle formed. The reason your sign came out wrong was because you measured your angle from the `negative branch of the xaxis`.dw:1436293621678:dwIf we had measured from the correct axis, using this large angle, then the negative sign would have come out correctly.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1So for triangle A, we could have put \(\large\rm x=A\cos(\theta)\) \(\large\rm x=90\cos(165^o)\) to get that 86.9

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1AHHH that was all for letter B sorry sorry :( forgot to look back at the letters

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1So for A, if we measure from the positive side of the xaxis,dw:1436293784227:dw

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Do you see how I got that angle? :O

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large\rm x=A\cos(90^o)\]\[\large\rm x=60\cos(90^o)\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Similarly,\[\large\rm y=A\sin(\theta)\]\[\large\rm y=60\sin(90^o)\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1This shortcut comes in handy when you're not able to draw a triangle :P that made things kinda tricky with it being on the axis hehe

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.1so x = 0 and y = 60

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1And that makes sense if you look at the picture, right? we went 0 to the left or right, and we went 60 down.

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.1yeah that makes sense

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.1\[\sqrt{33.9^{2}16^{2}} = 37.48\]

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.1then tan (16/33.9) = 25

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1The length of your resultant looks correct!\[\sqrt{33.9^{2}\color{red}{+}16^{2}} = 37.48\] Angle um um um thinking :3

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.1I know the answer is 38 km at 25 degrees north of west. but how do i know the direction and i got 37 and not 38

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1You probably just didn't carry around enough decimals to get the correct rounded value :) no big deal.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1For the angle,\[\Large\rm \tan^{1}\left(\frac{y}{x}\right)=\tan^{1}\left(\frac{16}{33.9}\right)=25.69\]Think about WHERE this angle is located, and then think about WHERE your x and y are located for your resultant.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436294373840:dwnegative 25 degrees is somewhere around here

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436294434921:dwBut notice that our resultant vector should be somewhere in quadrant 2.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1So tangent gave us the reference angle, we need the NEXT one though. So we have to ADD 180 degrees to the angle that we ended up with.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436294610322:dwmm yessss

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1mmm what do you think? too confusing? :O

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1when dealing with inverse tangent things can get a little bit weird, you really have to pay attention to which quadrant your resultant is in

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.1yeah i get how we get the resultant but the answer has +25 degrees. So i really dont understand how we got NW

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436294829972:dwSo from facing West, to get to our resultant vector, we have to rotate north 25 degrees.

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.1but how do you go from tan = 25 to the answer being 25? and knowing it will be NW, I don't understand. is there a different way of explaining

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436295041858:dwI guess you can just remember that `vertical angles are always congruent`, so these two angles I labeled will always be the same. That goes back to geometry though :P

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Keep in mind though, we didn't HAVE TO use NW for directional. Another acceptable answer would have been to use ... 65 degrees west of north.dw:1436295194403:dw

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.1so because the tan is 25 and we cant have a  degree you go congruently to the + degree?

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.1so instead of 25 in the SE it is 25 in the NW

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1ya lemme try to clarify that a little bit :) the problem was NOT that the angle was negative. it was the fact that the angle was in quadrant 4, and our resultant vector was in quadrant 2. the problem was that the angle wasn't where it was supposed to be. But yes, that's a good little shortcut to remember maybe! :) This might come up in other problems as well... like... maybe inverse tangent spits out an angle in quadrant 1,dw:1436295356582:dwBut your resultant x and y are clearly in the third quadrant.

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.1how did we know our resultant was in that quadrant though?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436295443084:dwIt looks like your trick will work nicely for this type of problem. So instead of 19 degrees North of East, we would say our resultant is located 19 degrees South of West.

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.1because x= a  number? and y was +?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Because of the `x` and `y` that you end up with. The `signs` of those values tells you which quadrant you are in. Yes, good :)

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.1okay, I suppose I understand now, thank you :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1cool cool. i know i know, the math is kinda crazy \c:/ just keep up the hard work lady!
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