- #1

logan3

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## Homework Statement

In a plane, find the resultant of a 300-N force at [itex]30^o[/itex] and a -250-N force at [itex]90^o[/itex]. Also, find the angle [itex]\alpha[/itex] between the resultant and the y axis.

## Homework Equations

[itex]x = F cos \theta[/itex]

[itex]y = F sin \theta[/itex]

[itex]r^2 = x^2 + y^2[/itex]

[itex]\theta = tan^{-1} (\frac {y} {x})[/itex]

## The Attempt at a Solution

[itex]x_1 = 300 cos 30^o = 259.80, x_2 = -250 cos 90^o = 0, x_{TOT} = 259.80[/itex]

[itex]y_1 = 300 sin 30^o = 150, y_2 = -250 sin 90^o = -250, y_{TOT} = -100[/itex]

[itex]r = \sqrt {259.80^2 + (-100)^2} = 278.38 \sim 280[/itex]

[itex]\theta = tan^{-1} (\frac {259.80} {-100}) = -68.947^o \sim -69^o[/itex]

Since this is the angle between the x-axis and resultant, then the angle [itex]\alpha[/itex] between the resultant and the y-axis must be [itex]-90 - (-68.947^o) = -21.053^o \sim -21^o[/itex]. However, the book solution for the angle [itex]\alpha[/itex] is given as [itex]69^o[/itex]. Why?

Thank-you