Forces, Newton's Second Law

[physics_10]

Forces of 10.6 N north, 20.8 N east, and 14.3 N south are simultaneously applied to a 3.76 kg mass as it rests on an air table. What is the magnitude of its acceleration?


I tried
F=√(10.6²+20.8²)
a=F/m

..however this is incorrect. Can someone please tell me where I am going wrong?

 

[Delphi51]

Welcome to PF, physics_10. Glad to see another high school student here. I'm a retired high school teacher trying to keep that feeling of helping out.

It looks like you just forgot to include the 14.3 N south. Combine the north and south forces before you work out the combination of the north south and east west forces.

 

[Dadface]

You have 10.6N acting north and 14.3N acting South.The resultant of these two is not 10.6.

Hello Delphi 51.You beat me to it.

 

[physics_10]

okay thank you !
& how would i find out the direction of the acceleration in degrees? tan theta = 20.8 / 3.70?

 

[Delphi51]

Yes. Careful reporting that direction - it is the number of degrees east of south. The more standard way of reporting is to find the number of degrees east of north. That would be 180 minus the number you get from arctan(20.8/3.7).

 

[physics_10]

hmm for some reason i am getting the incorrect answer. The question states: What is the direction of the acceleration in degrees? (Take east to be 0 degrees and counterclockwise to be positive.)

 

[Delphi51]

Did you get 80 degrees for invTan(20.8/3.7)?
That's 80 degrees counterclockwise from south, right?
Same as 10 degrees clockwise from east.
Or -10 degrees counterclockwise from east.
Or 350 degrees counterclockwise from east.