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North of west?

  1. Oct 14, 2014 #1
    Hi, really I just want to know if I have drawn this out correctly. I have an object with a velocity going North of West.
    It is going against a current South. As you can see from the chart I have mapped out the degree the object is heading North of West. How would I find net velocity from this information?
    This is what I have drawn so far - would I be solving for the purple line or is this incorrect?
    If so, I would need to find the components?
    Vy = Vsin 0
    Vx = Vcos 0

    Vy for current is on the negative y axis. So this would end up being -3.0....?
    We don't know the distance, but we need to find net velocity, so we can't use displacement.

    This is where I get stumped.
    Thank you for setting me on the right path!
  2. jcsd
  3. Oct 14, 2014 #2


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    Does the problem say specifically 55 degrees "north of west"? I don't see that in your statement of the problem but, if so, then your picture is correct. "North of west" without a specific angle measure would be peculiar. "Northwest", alone, specifically means "45 degrees north of west" (which is also "45 degrees west of north", of course).

    Also, if you are told the object is going "55 degrees north of west at 5 m/s" then your "5 m/s" should be on the black arrow, not the blue one. Please give the full statement of the problem.
  4. Oct 14, 2014 #3


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    Staff: Mentor

    Hi LeighV, Welcome to Physics Forums.

    When you want to find the resultant of two vectors that are being added you place the vectors tip to tail or complete the parallelogram, or algebraically sum their X-Y components:
  5. Oct 14, 2014 #4
    Hi, sorry if I wasn't clear enough. Yes it says 55 degrees north of west. Thank you. The blue/purple arrow was just one I added in to try and figure out if that was what I needed to answer (find the value for), the black arrow represents the direction not necessarily the speed. Thank you for your response!
  6. Oct 14, 2014 #5
    Thank you! So what does the purple arrow in your diagram stand for? Is that what I need to find or is that represented by those dashed lines?

    Also, TARDIS. You are awesome.
  7. Oct 14, 2014 #6


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    Remember to sum the velocities separately by components. And please note that, in problems where there is something moving within a moving medium, it's useful to remember the rule:

    absolute motion = relative motion + dragging motion

    For example, you step in an escalator that moves with velocity w and start going up the steps at a velocity v. Your absolute motion (with respect to the building) is the sum of your relative motion (yours with respect to the escalator, that is, v) and the dragging velocity (that of the escalator wit respect to the building).

    In your problem, you're trying to find the absolute velocity; you know the relative velocity (with respect to the medium, water, probably, but that doesn't matter) and you know also the... [here my help has to stop]
  8. Oct 14, 2014 #7
    That really helps me NTW, thank you. The relative motion is 5.0 "you know also the...dragging motion?" Being 3.0?

    So when I find the components of the velocities, I just add them together?
  9. Oct 14, 2014 #8


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    The dragging motion is the motion of the medium with respect to the absolute reference. In the case of the escalator example, the absolute reference is the building. In the case of a sea or lake, the reference is the coast... The dragging motion, in the case of a river, would be that of the river water with respect to the river bank...

    Just split the vector of each velocity by components and then add them, minding the signs...
  10. Oct 14, 2014 #9


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    LeighV, the graphic addition of your vector was already explained above, by gneill.

    Now, to do it 'analytically', you should start by stating the vectors: you have, for example, v (the velocity of the object in the current), w (the velocity of the current with respect to the shore, or to distant, non-moving waters...). And -as you want the sum of those two vectors- you should start by splitting into components. You are in the plane, with two dimensions, and your vectors are defined by its two components, horizontal and vertical.

    Let's start with v. If the angle that it forms with the left x-axis angle is really 55º (is it, really?), then the horizontal component is

    v(x) = v*cos 55º = 5*0,57 = 2,87 => -2,87

    and the vertical one:

    v(y) = v*cos 35º = v*sin 55º = 5*0,82 = 4,10

    Now we go to vector w. This is easier, since it has no horizontal component. Only vertical, and 'down' (negative). Hence:

    w(x) = 0

    w(y) = -3

    Now, the only thing you are left with is to find the two components of vector z, that is, z(x) and z(y)...

    I shouldn't complete the problem... It's your job to do it... And please, confirm that the angle is 55º... As pointed out above in the thread, it could be 45º...
  11. Oct 14, 2014 #10


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    Staff: Mentor

    I like to write out each velocity as "something with respect to (w.r.t.) something else", and then write out the sum in a way that visually "chains them together":

    velocity of you w.r.t. building = velocity of you w.r.t escalator + velocity of escalator w.r.t. building

    See how the position of the word "escalator" sort of chains the two velocities together on the right side?
  12. Oct 14, 2014 #11


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    Staff: Mentor

    Yes, the purple arrow is the resultant of the the velocities being added (Black and Red vectors).
    What can I say? I'm a fan :smile:
  13. Oct 14, 2014 #12
    Thank you all for your help!! I think I get it now! :D
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