Physics on an Inclined Plane

1. Dec 1, 2011

Coca Cola

1. The problem statement, all variables and given/known data
A 4,535,924-kg ship rests on launching ways that slope down to the water at an angle of 10 degrees. If the coefficient of sliding friction is 0.18, how much force is needed to winch the ship down the ways into the water?

2. Relevant equations
Force of friction=μFn
Force parallel = mgsinθ
Force of friction in the y direction on an inclined plane = mgcosθ = Fn

3. The attempt at a solution
My first step is to solve for the force of friction using the given angle, mass and gravity.
Force of friction= (0.18)(4,535,924kg)(9.8 m/s^2)cos10°=7,879,811N

Now this is where I become confused. The problem states that the ship "rests," so I solve for the net force, right? The force needed to move the ship must be greater than this net force, correct? Or should it just be greater than the frictional force that I solved for?

ƩF=Fparallel - Force of friction
=mgsin10° - 7,879,811N
=7,719,018N - 7,879,811N=-160,973N

Conclusion:
My biggest problem is understanding what is required to make the ship move from rest. Comments and questions concerning my question would be appreciated.

Last edited: Dec 1, 2011
2. Dec 1, 2011

Staff: Mentor

That number doesn't look right. Check the math.
Presumably they're asking for the force that needs to be applies in order to keep the ship moving at a constant velocity down the slipway (I say this because only the coefficient of dynamic friction was given). The force applied would have to make the net force downslope zero.
Again, you'll have to fix your friction force value (above).
I think you'll have to assume that the ship is already moving, and that some winching force is required to keep it moving.

3. Dec 1, 2011

Coca Cola

Thank you for your help! I noticed the math errors and must have corrected them while you were typing your response.