Summersnow8
  • Summersnow8
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.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
undeadknight26
  • undeadknight26
So what do you need? How to get there?
Summersnow8
  • Summersnow8
yeah, I need help with how to solve it.
undeadknight26
  • undeadknight26
This seems like physics ._. @hba You good in Physics?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

undeadknight26
  • undeadknight26
Im new to physics lol
Summersnow8
  • Summersnow8
ok, @undeadknight26 thanks for trying
princeharryyy
  • princeharryyy
can you draw the figure
Summersnow8
  • Summersnow8
a figure isn't given, but this is the figure i have drawn
princeharryyy
  • princeharryyy
until you can't draw it you can't solve it.
Summersnow8
  • Summersnow8
|dw:1436290559148:dw|
princeharryyy
  • princeharryyy
try solving it now!
princeharryyy
  • princeharryyy
ok! leave it! HOW DID YOU SOLVED IT!
Summersnow8
  • Summersnow8
I don't know how to solve it, that's why I am asking for help
princeharryyy
  • princeharryyy
oh!!!!
princeharryyy
  • princeharryyy
ok
zepdrix
  • zepdrix
woops let fix your graph a sec :) 15 `north` `of west` means we start by facing west, and rotating up 15 degrees.
zepdrix
  • zepdrix
|dw:1436292130604:dw|
zepdrix
  • zepdrix
So what are we doing? Adding up these vectors? :O
zepdrix
  • zepdrix
That resultant shinanigans and all that? +_+
zepdrix
  • zepdrix
|dw:1436292221555:dw|
zepdrix
  • zepdrix
Let'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
  • zepdrix
SummerPotatooooo, just chat in here XD don't need private message box lol
Summersnow8
  • Summersnow8
what are i's and j's? I guess whatever is easiest
zepdrix
  • zepdrix
|dw:1436292562770:dw|Let'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
  • zepdrix
|dw:1436292657672:dw|So to get to that point, we'll need 3 i's. We moved 3 units horizontally.
zepdrix
  • zepdrix
And j is 1 unit vertically. So we need 4 j's.
Summersnow8
  • Summersnow8
I've never learned that way....I know to use trigonometry
zepdrix
  • zepdrix
|dw:1436292736659:dw|So 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
  • zepdrix
Haven't learned that? :[ hmm
Summersnow8
  • Summersnow8
I use x and y no i and j
Summersnow8
  • Summersnow8
not*
zepdrix
  • zepdrix
ah ok :3
zepdrix
  • zepdrix
Let's look at the F_3 that I labeled on the graph. That one will be easiest so maybe we should start there.
Summersnow8
  • Summersnow8
okay
zepdrix
  • zepdrix
|dw:1436292975469:dw|If we think of this as a triangle
zepdrix
  • zepdrix
We can use our trigonometry to get our x and y values.
Summersnow8
  • Summersnow8
so x = 53 and y = 53
zepdrix
  • zepdrix
Mmm good good that sounds right! :) So here is a way we can write our F_3,\[\Large\rm \vec F_3=53x+53y\]
Summersnow8
  • Summersnow8
This was my attempt at solving
zepdrix
  • zepdrix
Ok your process looks great! Let's look at displacement B a second though.
zepdrix
  • zepdrix
Notice that arrow is pointing to the left. That tells us that our x coordinate should come out to be negative, ya?
Summersnow8
  • Summersnow8
okay
Summersnow8
  • Summersnow8
so x= - 86.9 ?
zepdrix
  • zepdrix
mmm yah that looks correct!
Summersnow8
  • Summersnow8
now A really confuses me
zepdrix
  • zepdrix
|dw:1436293376046:dw|Ah yes :) kind of hard to make a triangle with this one, huh?
Summersnow8
  • Summersnow8
yeah... 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
  • zepdrix
If 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 x-axis`.|dw:1436293621678:dw|If we had measured from the correct axis, using this large angle, then the negative sign would have come out correctly.
zepdrix
  • zepdrix
So 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
  • zepdrix
AHHH that was all for letter B sorry sorry :( forgot to look back at the letters
zepdrix
  • zepdrix
So for A, if we measure from the positive side of the x-axis,|dw:1436293784227:dw|
zepdrix
  • zepdrix
Do you see how I got that angle? :O
Summersnow8
  • Summersnow8
yes, I understand
zepdrix
  • zepdrix
\[\large\rm x=|A|\cos(-90^o)\]\[\large\rm x=60\cos(-90^o)\]
zepdrix
  • zepdrix
Similarly,\[\large\rm y=|A|\sin(\theta)\]\[\large\rm y=60\sin(-90^o)\]
zepdrix
  • zepdrix
This 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
  • Summersnow8
so x = 0 and y = -60
zepdrix
  • zepdrix
good good
zepdrix
  • zepdrix
And that makes sense if you look at the picture, right? we went 0 to the left or right, and we went 60 down.
Summersnow8
  • Summersnow8
yeah that makes sense
Summersnow8
  • Summersnow8
\[\sqrt{33.9^{2}16^{2}} = 37.48\]
Summersnow8
  • Summersnow8
then -tan (16/33.9) = 25
Summersnow8
  • Summersnow8
right?
zepdrix
  • zepdrix
The length of your resultant looks correct!\[\sqrt{33.9^{2}\color{red}{+}16^{2}} = 37.48\] Angle um um um thinking :3
Summersnow8
  • Summersnow8
I 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
  • zepdrix
You probably just didn't carry around enough decimals to get the correct rounded value :) no big deal.
zepdrix
  • zepdrix
For 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
  • zepdrix
|dw:1436294373840:dw|negative 25 degrees is somewhere around here
zepdrix
  • zepdrix
|dw:1436294434921:dw|But notice that our resultant vector should be somewhere in quadrant 2.
zepdrix
  • zepdrix
So 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.
Summersnow8
  • Summersnow8
so 155 degrees
zepdrix
  • zepdrix
|dw:1436294544290:dw|
zepdrix
  • zepdrix
|dw:1436294610322:dw|mm yessss
zepdrix
  • zepdrix
mmm what do you think? too confusing? :O
zepdrix
  • zepdrix
when 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
  • Summersnow8
yeah i get how we get the resultant but the answer has +25 degrees. So i really dont understand how we got NW
zepdrix
  • zepdrix
|dw:1436294829972:dw|So from facing West, to get to our resultant vector, we have to rotate north 25 degrees.
Summersnow8
  • Summersnow8
but 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
  • zepdrix
mmm thinking :)
Summersnow8
  • Summersnow8
okay thanks
zepdrix
  • zepdrix
|dw:1436295041858:dw|I 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
  • zepdrix
Keep 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
  • Summersnow8
so because the -tan is -25 and we cant have a - degree you go congruently to the + degree?
Summersnow8
  • Summersnow8
so instead of -25 in the SE it is 25 in the NW
zepdrix
  • zepdrix
ya 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:dw|But your resultant x and y are clearly in the third quadrant.
Summersnow8
  • Summersnow8
how did we know our resultant was in that quadrant though?
zepdrix
  • zepdrix
|dw:1436295443084:dw|It 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
  • Summersnow8
because x= a - number? and y was +?
zepdrix
  • zepdrix
Because 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
  • Summersnow8
okay, I suppose I understand now, thank you :)
zepdrix
  • zepdrix
cool cool. i know i know, the math is kinda crazy \c:/ just keep up the hard work lady!

Looking for something else?

Not the answer you are looking for? Search for more explanations.