Two Horizontal Forces Working On an Object

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Two horizontal forces act on a 3.0 kg body on a frictionless floor: 9.0 N east and 9.4 N at 68° north of west. The net force calculation initially led to an acceleration of 1.83 m/s², but the correct answer is 3.34 m/s². The confusion arises from the interpretation of the forces, particularly the angle of the second force, which affects both x and y components. Despite the problem stating only horizontal forces, the angled force introduces a vertical component that complicates the analysis. The discussion highlights the need for clarity in problem wording regarding the forces' directions.
richard karn
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Homework Statement



Only two horizontal forces act on a 3.0 kg body that can move over a frictionless floor. One force is 9.0 N, acting due east, and the other is 9.4 N, acting 68° north of west. What is the magnitude of the body's acceleration?

Homework Equations



F=ma

The Attempt at a Solution



Fnet= 9 - 9.4cos(68) = 5.47869

F=ma
5.47869= 3.0 * a
a=1.826232

The right answer is 3.34 though.
 
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Your answer agrees with mine.
 
I disagree. Fnetx is that, yes, but you forgot about Fnety.

(I'm calling East-West the x direction, and North-South the y-direction).
 
thanks. i was thinking as the other force as pushing into the ground so it wouldn't be accelerating in that direction but then

but 9.4sin(68)/3=2.905

and sqrt(2.905^2+1.8^2) = 3.4
 
Hmm, the object with a 3 kg mass has a weight force of about 30 N acting down. How can it have acceleration in the y direction?? Then again the problem notes there are only 2 horizontal forces acting, but one of them is neither horizontal nor vertical, it is at an angle. Strangely worded problem. Maybe the object and floor are in a zero g or apparent zero g field? That still doesn't explain the 'horiziotal' force acting 68 degrees north of west.
 
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