Box being pulled up a slope with friction

[alikim]

If I solve a problem of a box sliding down a slope or standing still on the slope, the force of friction is directed up along the slope.

What happens if there is a force F pulling the box up along the slope, but it's unknown if it overcomes gravity and box is moving up or it only slows the box down and it is still sliding down.

Since the force of friction is directed opposite the movement, is it directed up or down?

If the box is moving down:
ma = mg sinθ - F - μmg cosθ, a >= 0

If the box is moving up:
ma = mg sinθ - F + μmg cosθ, a < 0

Is there a way to write one "universal" equation to solve? Or do I check a conditions first, and then choose which equation to use?

 

[PeroK]

If I solve a problem of a box sliding down a slope or standing still on the slope, the force of friction is directed up along the slope.

What happens if there is a force F pulling the box up along the slope, but it's unknown if it overcomes gravity and box is moving up or it only slows the box down and it is still sliding down.

You have to calculate the net force to determine the direction in which motion is possible, before including the forces that resist motion.

Since the force of friction is directed opposite the movement, is it directed up or down?

If the box is moving down:
ma = mg sinθ - F - μmg cosθ, a >= 0

If the box is moving up:
ma = mg sinθ - F + μmg cosθ, a < 0

Is there a way to write one "universal" equation to solve? Or do I check a conditions first, and then choose which equation to use?

You have to solve the problem in two stages. Note that there is the intermediate case where the net force is insufficient to overcome friction, in which case the box remains at rest; and, the friction force is not the maximum.

Moreover, it's often the case that the coefficient of static friction is greater than the coefficient of kinetic friction, which adds another layer of complexity - and another step in the solution.