Force in different directions, magnitude of acceleration?

  • Thread starter deezy
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  • #1
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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



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

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.
 

Answers and Replies

  • #2

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



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

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.
It looks like you are correct.
 
  • #3
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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!
 
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