Normal force, friction, Newton's law

[jehan4141]

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 θ below the horizontal. When θ 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 θ.

Homework Equations

F_N = mg + FsinƟ

F_friction = 0.41(mg + FsinƟ)

F_friction = FcosƟ in our problem.
0.41(mg+FsinƟ) = FcosƟ
0.41mg + 0.41FsinƟ = FcosƟ

I think I am correct and that my logic is sound so far but I just do not see how I can find Ɵ? Hints please?

The Attempt at a Solution

 

[ehild]

F_friction = 0.41(mg + FsinƟ)

F_friction = FcosƟ in our problem.
0.41(mg+FsinƟ) = FcosƟ
0.41mg + 0.41FsinƟ = FcosƟ

The second equation gives the horizontal component of the pushing force, say, Fx, which has to be equal to the force of friction.
F_x = FcosƟ, F_x=F_friction. But the logic is correct otherwise.

You need to apply some trigonometry to find the angle. Express sin and cosine with the same function: cos with sine, or both cosine and sine with tangent...

ehild