What is the required force to push a box upwards on an angled surface?

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To determine the force required to push a box upwards on an angled surface, the box's mass is 15 kg with a friction coefficient of 0.3. The angle of the surface is 55°, and the gravitational force is calculated as mgcos(35). The frictional force opposing the motion is found to be 30.72 N, while the downward force is adjusted to 102.54 N. The necessary force X to move the box upwards is calculated to be 71.82 N after correcting the downward force value. The discussion emphasizes the importance of accurately calculating forces to assess the box's movement on the incline.
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Homework Statement


A box is placed on an angeled surface.

m = 15 kg
f = 0,3

[PLAIN]http://img5.imageshack.us/img5/8717/friction.png

Will the box slip downwards if X = 0?

How big does the force X have to be to push th box upwards?

Homework Equations


The Attempt at a Solution


My idea was that the only force working upwards is Ff.

A = 55°
g = 9,81
N = mg*cos(90-35)+18 = 102,40 N

Ff = 0,3*N = 30,72

Force downwards: mgcos35-18 = 102,54 N

The force down is greater than Ff therefore it would slip if X = 0.To get how big X needs to be to make the box move up i simply subtracted the force down with Ff and got = 71,82 N.Could this possibly be correct?
 
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Why is your downward force mgcos(35)-18? It should just be mgcos(35)
 
Pi-Bond said:
Why is your downward force mgcos(35)-18? It should just be mgcos(35)


Yea, okey. So the rest would be correct then?
 
After changing the value of the downward force, yes.
 

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