Statics: two unknown angles and resultant force

AI Thread Summary
The problem involves determining the angles that two forces, P (100N) and Q (200N), make with their resultant force R (250N) directed along the +x axis. The equations derived from the force components are ΣRx = 100cos(θ) + 200cos(φ) = 250N and ΣRy = 100sin(θ) - 200sin(φ) = 0. The discussion highlights confusion with algebraic manipulation and the potential use of trigonometric identities to solve for the angles. Suggestions include using a force triangle to visualize the problem and considering the relationships between the angles through tangent functions. Ultimately, the solution requires careful algebraic handling of the equations to isolate the angles.
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Homework Statement



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.


Homework 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

The Attempt at a Solution



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

\SigmaRx = 100cos(\theta) + 200cos(\phi) = 250N

\SigmaRy = 100sin(\theta) - 200sin(\phi) = 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?
 
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How about tan(\theta) does that equal anything?
 
I tried that, but still had phi in the expression.

Rx: sin(\theta) = 2sin(\phi)

Ry: 2.5 - 2cos(\phi)


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\phi) - tan(\phi)]
 
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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