# Statics: two unknown angles and resultant force

1. ### drnoisewater1

2
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:

$$\Sigma$$Rx = 100cos($$\theta$$) + 200cos($$\phi$$) = 250N

$$\Sigma$$Ry = 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?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. ### djeitnstine

619
How about $$tan(\theta)$$ does that equal anything?

3. ### drnoisewater1

2
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($$\phi$$)]

4. ### mplayer

154
Since vectors add head-to-tail, could you maybe draw a force triangle with P, Q, and R and find the angles that way?