# Basic Vector Problem

1. Jun 16, 2015

### logan3

1. The problem statement, all variables and given/known data
In a plane, find the resultant of a 300-N force at $30^o$ and a -250-N force at $90^o$. Also, find the angle $\alpha$ between the resultant and the y axis.

2. Relevant equations
$x = F cos \theta$
$y = F sin \theta$
$r^2 = x^2 + y^2$
$\theta = tan^{-1} (\frac {y} {x})$

3. The attempt at a solution
$x_1 = 300 cos 30^o = 259.80, x_2 = -250 cos 90^o = 0, x_{TOT} = 259.80$
$y_1 = 300 sin 30^o = 150, y_2 = -250 sin 90^o = -250, y_{TOT} = -100$
$r = \sqrt {259.80^2 + (-100)^2} = 278.38 \sim 280$

$\theta = tan^{-1} (\frac {259.80} {-100}) = -68.947^o \sim -69^o$
Since this is the angle between the x-axis and resultant, then the angle $\alpha$ between the resultant and the y-axis must be $-90 - (-68.947^o) = -21.053^o \sim -21^o$. However, the book solution for the angle $\alpha$ is given as $69^o$. Why?

Thank-you

2. Jun 16, 2015

### Staff: Mentor

Nice work. I've bolded the part where you transposed the y & x components when calculating theta...

3. Jun 21, 2015

### FrankJ777

Is the problem that you mixed up your terms in the formula for θ.
I think the formula for θ, the angle with respect to the x axis is: tan -1( y/x).
It looks to me like you have x and y transposed, as y=-100 and x = 259.8, so should the expression be:
θ= tan -1(-100/259.8)?
If i use that expression I get θ ≈ -21° so with respect to the y axis ∠ = -69.