Statics: two unknown angles and resultant force

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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?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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