Friction will always be in the opposite side of the motion, so if you are sliding down the ramp, friction will be directed to the top of the ramp, if you are sliding up the ramp, friction will be down. Always.
In the real life you will have friction everytime, because you can't have a ramp 100% perfect without friction (but this in the real life).

And to put positive or negative will depend on the way of your motion. If you are down, the positive will be down and the negative will be up. Otherwise, if your box is upping in the ramp, then this is the positive and the negative is down.

To solve those problems without problems, try to think on forces separately. For example, you are trying to relate the component "x" with the positive or negative side and with friction ... Believe me, this is much more easier than that
First you think about the P, Px, Py. It is one work that you have to do.
Second, think about the friction. To solve that easily, just see the way of the motion. If it is upping the ramp, friction is down. If it is down the ramp, friction is up. Easy

The last question you got was about the 2nd law of Newton. Yes you are right, F = m.a, however the 2nd law of Newton is applied on motions that necessary vary the velocity, otherwise the motion will have a velocity constant, which has acceleration equal to zero, which makes the force equal to zero, because you will have F = m . 0 = 0
So, you will have acceleration when the sum of forces are different to zero :). If the sum are equal to zero, you won't have acceleration and the velocity will keep constant.

Now I'm gonna put the box sliding up the ramp with velocity constant:

First of all, if you just drop the box on a random ramp, in what way do you think the box would go? down or up? ofc down because of gravity. So, in order to have the box sliding up the ramp, you will have to have an external force acting on the box and pulling it to the top of the ramp.

So, in this case we will have these forces:



Look the force "T".
This force is responsible to pulling the box to the top of the ramp.

The velocity constant here is obtained the same way of the other case I've posted. Now you will just have to add the force "T" while summing the forces
And yes we have to hide force "P" and put two other forces that represent "P", as you know Px and Py.

Look the picture:



Well, now we have to obey again those two conditions together:

T = Px + Fr
N = Py

Py = P.cos(a)
Px = P.sin(a)
Fr = μ . N
P = m.g

T = P.sin(a) + μ . N
T = P.sin(a) + μ . P.cos(a)
T = P. [sin(a) + μ .cos(a)]

Later I'm gonna put two other cases with acceleration, thinking the box upping the ramp and downing the ramp.

Feel free to pm me for help too.

Now let's think about two more situations, but now with acceleration.

First option: you have the situation on the picture below and the box is sliding down with a random acceleration.



Well, If you the box has an acceleration, so its velocity is growning or lowering, depending on forces Px and Fr (friction).
Why ? Because if you have Px > Fr, the box will increase the velocity and the acceleration will be positive (assuming the direction of the motino to down)
Otherwise, if you have Px < fr, the box will drop the velocity and the acceleration will be negative. (assuming the direction of the motino to down)

I guess you are thinking this: "how can I know if Px is > or < Fr before start the problem?"
Well, really don't worry about that , I'm gonna show why**

I will call acceleration = a

Assuming a > 0 (positive) and box sliding down the ramp, we will have:

F = m.a (F is the sum of forces in this case that we are studying)
The box is down the ramp, so Px has the same direction of the motion and Fr has the opposite direction of the motion. "F" is the sum of forces, so F = Px + Fr, but Fr is negative because its direction is the opposite of the motion, so the correct formula is F = Px - Fr

So,
F = m.a
Px - Fr = m.a**
P.sin(a) - Fr = m.a
-Fr = m.a - P.sin(a)
Fr = -m.a + P.sin(a)
Fr = P.sin(a) - m.a
Fr = m.g.sin(a) - m.a
Fr = m. [g.sin(a) - a]
But Fr = μ . N and N = P.cos(a)

μ . N = m.g.sin(a) - m.a
μ . P.cos(a) = m.g.sin(a) -m.a
μ . m.g.cos(a) = m. [g.sin(a) - a]
now we cut m on both sides of the equation
μ . g.cos(a) = g.sin(a) - a
μ = g.sin(a) / g.cos(a) - a / cos(a)
μ = tan(a) - a.sec(a) or μ = tan(a) - a/cos(a)

**Look the formula above (Px - Fr = m.a). Let's say that you've put the direction of the motion to the top of the ramp, instead of the down. And after that you've decided to use Px - Fr instead of Fr - Px. Well, you can solve the whole problem without any problem because the only think you will get at the end is a negative acceleration, which indicates that you've put the direction of the motion in the opposite way, just that. The main value won't change at all.

I believe this is what you're gonna need to know in order to solve problems like that.

Now the last case, in other words, box up the ramp with acceleration

In order to up the ramp, you have to understand that is not normal, since is against the gravity. So, someone must apply a force in the opposite direction to have to box up the ramp without problems.

We will use this picture:



The direction of the motion is to the top of the ramp, so the force "T" must be bigger than the sum of "Px" and "Fr", otherwise we would not have the box sliding up the ramp.

Well, now we just have to apply the 2nd law of Newton:

F = m.a
T - (Px + Fr) = m.a
T - [m.sin(a) + μ . N] = m.a
T = m.a - [m.sin(a) + μ . N]
we know that N = P.cos(a), so...
T = m.a - m.sin(a) - μ . P.cos(a)
T = m.a - m.sin(a) - μ . m.g.cos(a)
T = m. [a - sin(a) - μ.g.cos(a)] or T = m. {a - [sin(a) + μ.g.cos(a)]}

If you find the acceleration positive, so the direction what we have chose in the begin of the exercise was right (direction to the top of the ramp). If we find a negative acceleration, so the direction of the motion is in the opposite side.

Again, feel free to count on me for help.

edit: changed some typo

This post was edited by betoggc on Mar 7 2015 05:00am