Solve Basic Vector Problem: Resultant & Angle \alpha

[logan3]

Homework Statement

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.

Homework Equations

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

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

 

[berkeman]

Homework Statement

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.

Homework Equations

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

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

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

 

[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.