Equilibrium Question - at what rate does the block accelerate?

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SUMMARY

The discussion centers on calculating the acceleration of a box subjected to a force of 120N at an angle of 60 degrees from the horizontal on a frictionless surface. The box has a mass of 5kg, leading to a net force calculation using Fnet = ma. The correct approach involves resolving the applied force into its horizontal and vertical components, resulting in an acceleration of 12 m/s². The confusion arises from understanding the balance of vertical forces, specifically the normal force and gravitational force.

PREREQUISITES
  • Understanding of Newton's Second Law (Fnet = ma)
  • Knowledge of force decomposition into components
  • Familiarity with gravitational force calculations (Fg = mg)
  • Concept of normal force in physics
NEXT STEPS
  • Study vector decomposition of forces in physics problems
  • Learn about normal force calculations in inclined scenarios
  • Explore frictionless surface dynamics in classical mechanics
  • Review examples of force balance in two-dimensional motion
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Students studying physics, particularly those focusing on mechanics and force analysis, as well as educators seeking to clarify concepts related to force components and acceleration calculations.

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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