1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Force in different directions, magnitude of acceleration?

  1. Sep 20, 2011 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant equations

    [tex]a = \frac {F_{net}}{m}[/tex]

    3. The attempt at a solution

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

    Not sure if this is correct, but this is my guess.
     
  2. jcsd
  3. Sep 20, 2011 #2
    It looks like you are correct.
     
  4. Sep 20, 2011 #3
    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!
     
    Last edited: Sep 20, 2011
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook