An angle at which the body cannot be moved

[Mad_Eye]

Homework Statement

While moving in, a new homeowner is pushing a box across the floor at a constant velocity. The coefficient of kinetic friction between the box and the floor is 0.41. The pushing force is directed downward at an angle ALPHA below the horizontal. when ALOHA is greater than a certain value, it is not possible to move the box, no matter how large the pushing force is. Find that value of ALPHA.

Homework Equations

friction = coefficient * normal force

The Attempt at a Solution

I feel really stupid not to succeed doing this..
this is what I've tried to do:
a is ALPHA
u is drag coefficient
N is the normal force (between floor and box)
F is the pushing force)

F * cos (a) <= u * N
N = mg + F sin (a)

F (cos (a) - usin (a)) < umg

this should be true no matter what the value of F is..
as F is getting bigger, the right side value is staying the same,
so i thought, the only case this is true for every positive F, is when
cos(a) - usin(a) <= 0

math allow us to go here:

(cos ² (a)) * (1+u²) = u²

and here because u is known, (0.41)

a = 67.7 (approx.)

but it doesn't seems to be the right answer..

please to explain how'd you do that.thanks
p.s.
mathematically, another answer is (-67.7), which doesn't make any sense at all... can anyone explain it to me?

 

[willem2]

Your normal force is wrong. Gravity, the normal force and the vertical component of the pushing force must cancel.

 

[Mad_Eye]

Your normal force is wrong. Gravity, the normal force and the vertical component of the pushing force must cancel.

oh sorry, by this:
N = mg * F sin (a)

i ment:
N = mg + F * sin (a)