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

[Waelroe]

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?

 

[Pi-Bond]

Why is your downward force mgcos(35)-18? It should just be mgcos(35)

 

[Waelroe]

Why is your downward force mgcos(35)-18? It should just be mgcos(35)

Yea, okey. So the rest would be correct then?

 

[Pi-Bond]

After changing the value of the downward force, yes.

 

[rwisz]

Your calculations are correct. To push the box upwards on the angled surface, a force of at least 71.82 N is required. This is because the force of friction (30.72 N) is not enough to counteract the downward force of gravity (102.54 N). Therefore, an additional force (X) is needed to overcome the difference between the two forces and push the box upwards. If X is less than 71.82 N, the box will not move upwards and will slip downwards.