Equilibrium Question - at what rate does the block accelerate?

AI Thread Summary
A force of 120N is applied to a 5kg box at a 60-degree angle on a frictionless surface, prompting a discussion on how to calculate the block's acceleration. The net force equation, Fnet = ma, is used, with the horizontal component of the applied force being Fcos60. The confusion arises regarding the vertical forces, where the gravitational force (mg) and the vertical component of the applied force (Fsin60) must be considered. The normal force must balance the vertical forces, leading to the conclusion that the upward forces include both the normal force and the vertical component of the applied force. Clarification is provided that the downward component of the applied force must also be factored into the calculations for accurate results.
tobya93
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Homework Statement



A force is applied to a box 60 degrees from the horizontal. The box weights 5kg. If the force is 120N, and the surface upon which the box rests upon is frictionless, at what rate does the block accelerate?




Homework Equations



Fnet = ma
Fg= mg



The Attempt at a Solution



I understood the problem initially got the correct answer. The only forces are horizontal so Fcos60 = ma --> 120Ncos60/5 kg = a --> a=12

But when I looked at it again, I wasn't sure how the vertical forces canceled. Fg is 50N. So the normal force has to equal that in the opposite direction to cancel.

Down forces: mg
Up forces: mg + Fsin60

So it seems that the up forces are greater since it's mg + Fsin60 and the down is only mg..little confused.


IDK if this will allow me to show it but this picture shows all the forces:
https://dl.dropboxusercontent.com/u/48340065/20130615_180533.jpg

Thanks!
 
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tobya93 said:
But when I looked at it again, I wasn't sure how the vertical forces canceled. Fg is 50N. So the normal force has to equal that in the opposite direction to cancel.
No, the normal force must equal whatever it needs to be to cancel all the vertical forces.

Down forces: mg
That's one force. But the applied force also has a downward component (as shown in your diagram).

Up forces: mg + Fsin60
The only "up" force is the normal force of the surface on the box. (Which does equal mg + Fsin60.)

So it seems that the up forces are greater since it's mg + Fsin60 and the down is only mg..little confused.
You're just missing the downward component of the applied force.
 
I feel dumb. Thanks for pointing that out.
 
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