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Adding vectors algebraically to find displacement

  1. Aug 23, 2012 #1
    Hi everyone,
    I am taking a physics course this year and I find myself already looking for help unfortunately. I think I understand the basics of adding vectors using Ax = A cosθ , Ay = A sinθ, etc. I am getting confused with the 3rd leg of the trip though. Since it is due west wouldn't that mean there is no vertical displacement? How can work that into my equation?
    1. The problem statement, all variables and given/known data
    attachment.php?attachmentid=50133&stc=1&d=1345744862.jpg


    2. Relevant equations
    Ax = A cosθ
    Ay = A sinθ
    ...
    R = √Rx2 + Ry2


    3. The attempt at a solution
    I have tried using the previous formulas to solve it but i get an answer of ≈320km. the back of the book says it should be 245km. could someone walk me though this? Thank you!
     

    Attached Files:

  2. jcsd
  3. Aug 23, 2012 #2
    Hey Kevin ,
    In the picture I have uploaded you can see
    that
    (*note the symbols in bold are vectors)

    r =a + b

    , subsequently
    R= c+r
     

    Attached Files:

  4. Aug 23, 2012 #3

    CAF123

    User Avatar
    Gold Member

    What you want is [itex]\vec{R} = \vec{a} + \vec{b} + \vec{c}. [/itex]
    Write out the vector equations for each of [itex] \vec{a}, \vec{b}, \vec{c} [/itex] and then add them together. This will give an eqn for [itex] \vec{R} [/itex]

    Using [itex] |\vec{R}| = \sqrt{R_x^2 + R_y^2} [/itex] will give you it's length.
    Can you find the angle from this?
     
  5. Aug 23, 2012 #4
    thanks for the quick response but I am not quite following you guys. Still new to this.
    Here is my work, maybe you can show me what i am doing wrong that way. Once i get the right distance i'll have no problem finding the angle. I under stand how to do that part.

    Ax = A cos(30) = 175km (.866) = 151.55km
    Ay = A sin(30) = 175 (.5) = 87.5km
    Bx = B sin(20) =150km (.342) = -51.3km
    By = B cos(20) = 150 (.939) = 140.96km
    Cx = C cos(110) = -190km (-.342) = 64.98km
    Cy = C sin(110) = 190 (.939) = 178.54km

    √165.232 + 4072 = √192951.4 ≈ 439km
     
  6. Aug 23, 2012 #5
    Hey kelvin ,
    As the diagram and the question says , the vector C is parallel to X axis , which mean there is no change in the y component of total displacement .
    On the side note , What does the figure mean by 110 degrees ? what is it .
     
  7. Aug 23, 2012 #6
    Try solving
    Cx as 190
    and Cy as 0
     
  8. Aug 23, 2012 #7

    CAF123

    User Avatar
    Gold Member

    You have the components correct for vectors [itex] \vec{a} [/itex] and [itex] \vec{b},[/itex]however the angles are wrong for the components of [itex] \vec{c} [/itex]

    What angle does [itex] \vec{c} [/itex] make with the positive [itex] x [/itex] direction?
    There is no vertical component to [itex] \vec{c} [/itex]
     
  9. Aug 23, 2012 #8
    thank you so much. i understand that now. it will help me on a previos one i thought i had right too.

    vector c is at 0 or 180 degrees isn't it?
     
  10. Aug 23, 2012 #9
    180 you can call it .
     
  11. Aug 23, 2012 #10

    CAF123

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    Gold Member

    Vector [itex]\vec{c} [/itex] is at angle of 180o with respect to the positive [itex] x [/itex] axis. Therefore the vector eqn for [itex]\vec{c} [/itex] is,
    [itex] \vec{c} = (190cos180)\hat x = -190\hat x [/itex]
     
  12. Aug 23, 2012 #11
    oh okay! thats where that 110 came from. i was trying to figure out the angle so i decided to try the interior angle of ΔABC

    i had a very similar problem there i found the interior angle like that and coincidently the resultant was extremely close either way. I have now corrected both and i cant thank you two enough.
     
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