Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Vector Question of two workers

  1. Jan 8, 2007 #1
    1. The problem statement, all variables and given/known data

    Two workers pull horizontally on a heavy box, but one pulls twice as hard as the other. The larger pull is directed at 25.0 degrees west of north, and the resultant of these two pulls is 350.0 N directly northward.

    Use vector components to find the magnitude of each of these pulls and the direction of the smaller pull.

    2. Relevant equations

    Vector A + VectorB = VectorR
    Magnitude of R = 350

    3. The attempt at a solution
    Well, Since Rx = 0 = Ax + Bx then Ax=-Bx

    and since Ay+By=350 then Ay=350-By and By=350-Ay

    I know that Ay=A*cos(25)
    and that Ax=A*sin(25)
    Also that By=B*sin (x)
    and Bx=B*cos(x) Where x is the angle from the x axis to the vector B.

    A=2B since A is twice that length of B

    So, the problem here is trying to get these things in terms of each other so that i can solve for only one variable.

    I've figured out that since By=350-Ay then By=350-A*cos(25) which is equal to By=350-2B*cos(25).

    So that's pretty much where I'm stuck. I know that there has to be some way to relate the fact that Ax=-Bx and A=2B and that the angle of 25 degrees for the C vector, that I have sufficient information to figure this problem out. It seems that I can never get it down to one variable. Maybe there's an equation that I'm forgetting. I've spent more than a few hours on this problem and I'm getting frustrated, it's the last problem in the assignment. I appreciate any help.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jan 8, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    Well, you know that Ax=Bx (note no - sign, since the small pull must be to the easterly side of the resultant). The above gives, in your notation, Asin(25)=Bcos(x), but then, since A=2B, this becomes, 2Bsin(25)=Bcos(x), which you can solve for the angle.

    Does this help?
  4. Jan 8, 2007 #3
    Well I thought since the pulls were in opposite directions then one would be going West (-) and the other East (+) and because the resultant is on the y-axis my understanding is that since Rx=(Ax+Bx) and Rx equals zero then Ax=-Bx. This is new to me... and for 2Bsin(25)=Bcos(x) I don't understand how to solve if there is two variables B and X.???

    thanks for the reply
  5. Jan 8, 2007 #4


    User Avatar
    Staff Emeritus
    Science Advisor

    Rx=Ax-Bx=0 (since, as you said, the two components are in opposite directions.) The easier way to see this is if you simply resolve the vectors into horizontal components then you see that they must equal one another, otherwise there would be some resultant force in the x direction.

    Divide both sides by B (we can do this, since clearly B is non-zero). The you will obtain 2*sin(25)=cos(x). Can you solve this?
  6. Jan 8, 2007 #5
    Sweet ass... Thanks alot man, that really helped alot. I didn't even think to divide out a variable like B that was on both sides. I kept getting two variables and thinking that I was stuck. I should go back and take algebra again. Thanks again cristo. What I got was that x=32.30 and the mag of A=298.28 and B=149.12.
  7. Jan 28, 2008 #6
    arccos (2.sin 25) = 32.30, but how did you get A and B?
  8. Aug 31, 2010 #7
    What if it asks to find south east instead of north east? How would you solve it then?
    Last edited: Aug 31, 2010
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook