Exercesie about force with friction

[Rukawa0320]

here is the task:
A ship whose mass is 2.0 X 10(to the seventh) kg rests on launching ways that slope down to the water at an angle of 6 degrees. The ways are greased to reduce the coefficient of kinetic friction to 0.11 . Will the ship slid e down the ways into the water by itself? If not, find the force needed to winch the ship into the water.

Ive been dealing with this question for 2 days. The first thing that I want to ask is, how can I know that will it slide down by itself or not. I've calcluated its acc. but I got a negative number. My teacher said it is not going to work in that way, so I have to find another way to answer this question.
I also don't understand the second part of the question, how could it possible to lift an object and make it to slide down? I am not a native speaker, but I've checked the meaning of "winch" like 4-5times and it means to lift something. I just can't see the point. Could anybody help me?

 

[mrjeffy321]

The ship exerts a force straight downward due to gravity, but since the ship is on a ramp, the force due to gravity is divided up between a force pointing down into the ramp (Normal force) and a force pointing parallel to the ramp (parallel force).
Initially, the parallel force is the only force acting on the object that will cause it to move, so the question comes down to...Will the parallel force be enough to over come friction and move the box down the ramp?

Since the box is initially at rest, the force of friction it must first overcome is the force of static friction, which we are not given, so we will assume that some nice person has already brought the ship to the brink of motion for us.

To calculate the force of kinetic friction, you need to know the normal force the ship is exerting onto the ramp. This is calculated by,
F_normal = mg*cos(angle)
Afterward, to calculate the force of kinetic friction,
F_friction = F_normal * coefficient of friction

To calculate the parallel force pushing the ship down the ramp,
F_parallel = mg*sin(angle)

Is the parallel force equal to or higher than the frictional force? If so, the box will continue to move down the ramp, If not, the box will slow to a stop unless an extra force is applied to keep it going.

[If an extra force is needed, then to calculate it, simply find the minimum amount of extra force needed to be applied parallel to the ramp in order for the parallel force to equal the frictional force, thus the net forces being equal, there will no acceleration and the box will continue to move down the ramp (because remember, some nice person had already started the ship in motion for us)].

 

[Rukawa0320]

appreciate for your help, I've thought about this idea before but I got a result of 0,096N and I believed that it can't be, but after your explanation I see the whole thing
thx for your help