Force in different directions, magnitude of acceleration?

[deezy]

Homework Statement

Forces of 10 N north and 20 N east are simultaneously applied to a 10-kg object as it rests on a frictionless horizontal table. What is the magnitude of the object's acceleration?

Homework Equations

a = \frac {F_{net}}{m}

The Attempt at a Solution

a_{north} = \frac {10}{10} = 1 \; m/s^2
a_{east} = \frac {20}{10} = 2 \; m/s^2
a_{northeast} = \sqrt{1^2+2^2} = 2.24 \; m/s^2
magnitude = tan^{-1}(1/2) = 26.57 ^\circ \; northeast ?

Not sure if this is correct, but this is my guess.

 

[Aggression200]

Homework Statement

Forces of 10 N north and 20 N east are simultaneously applied to a 10-kg object as it rests on a frictionless horizontal table. What is the magnitude of the object's acceleration?

Homework Equations

a = \frac {F_{net}}{m}

The Attempt at a Solution

a_{north} = \frac {10}{10} = 1 \; m/s^2
a_{east} = \frac {20}{10} = 2 \; m/s^2
a_{northeast} = \sqrt{1^2+2^2} = 2.24 \; m/s^2
magnitude = tan^{-1}(1/2) = 26.57 ^\circ \; northeast ?

Not sure if this is correct, but this is my guess.

It looks like you are correct.

 

[vweltin]

You've calculated the magnitude correctly, but you may have made a mistake in calculating the direction. Cardinal directions typically hold due north as 0º and increase going clockwise. The "triangle" formed by these acceleration vectors-- if represented on the x-y plane where +y is north and +x is east-- would have an x-component of 2 and a y-component of 1. The point of this triangle rests at the origin and the base (x-component) runs parallel to the x-axis but one unit in the +y direction. Using this you should be able to calculate the angle between the y-axis and the acceleration vector --SOHCAHTOA ;). You're very close!