Statics: two unknown angles and resultant force

  1. 1. The problem statement, all variables and given/known data

    Two forces P and Q with respective magnitudes 100N and 200N are applied to the upper corner of a crate. The sum of the two forces is to the right (+x direction) with a magnitude of 250N. Find the angles that P and Q make with their sum - that is, with the horizontal line through +x axis.

    2. Relevant equations

    R = P + Q where R is the resultant vector and P and Q are vectors given in the problem

    Rx = Px + Qx = 250

    Ry = Py + Qy = 0

    3. The attempt at a solution

    All the information is given except the two angles. Plugging in the given values gives me the equation:

    [tex]\Sigma[/tex]Rx = 100cos([tex]\theta[/tex]) + 200cos([tex]\phi[/tex]) = 250N

    [tex]\Sigma[/tex]Ry = 100sin([tex]\theta[/tex]) - 200sin([tex]\phi[/tex]) = 0

    where theta is the angle between P and R and phi is the angle between Q and R.

    Basically this comes down to confusion on algebra for me. I attempted substitution and that got me nowhere. I know there is some kind of trick to solving this, but I cannot figure it out. There are two equations and two unknowns so there must be a way to do it. I have worked a similar problem where one of the angles is known and the other was supposed to be at a maximum and calculus could be used there. Is that a possibility in this problem or is there just a little trick that I am missing?
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. djeitnstine

    djeitnstine 619
    Gold Member

    How about [tex]tan(\theta)[/tex] does that equal anything?
  4. I tried that, but still had phi in the expression.

    Rx: sin([tex]\theta[/tex]) = 2sin([tex]\phi[/tex])

    Ry: 2.5 - 2cos([tex]\phi[/tex])

    I thought that since magnitude of R is sqrt[Rx^2 + Ry^2] = 250 things might cancel out. Well all the angles canceled out so i just got an incorrect expression.

    I tried Ry/Rx just for fun, to get a tan expression and that was not beneficial because nothing canceled out there either. I just thought about this though: if you take tan inverse of Ry/Rx the resultant angle will be 0 since the resultant is about the x-axis. However, that means little as I don't know if you can evaluate arctan[0.8sin[tex]\phi[/tex]) - tan([tex]\phi[/tex])]
  5. Since vectors add head-to-tail, could you maybe draw a force triangle with P, Q, and R and find the angles that way?
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