How Much Force is Required to Launch a Ship on an Inclined Plane?

AI Thread Summary
To determine the force required to launch a 4,535,924-kg ship on a 10-degree incline with a friction coefficient of 0.18, the force of friction must first be calculated. The frictional force is derived from the equation F_friction = μFn, resulting in a value of 804,062N. The net force needed to keep the ship moving at a constant velocity down the incline is calculated by subtracting the frictional force from the gravitational force acting parallel to the incline. The final calculation shows that a force of 6,914,956N is necessary to maintain motion. Understanding the distinction between static and dynamic friction is crucial for solving this problem effectively.
Coca Cola
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Homework Statement


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?

Homework Equations


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

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.
 
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Coca Cola said:

Homework Statement


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?


Homework Equations


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

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°=804,062N
That number doesn't look right. Check the math.
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?
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.
ƩF=Fparallel - Force of friction
=mgsin10° - 804,062N
=7,719,018N - 804,062N=6,914,956N
Again, you'll have to fix your friction force value (above).
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.
I think you'll have to assume that the ship is already moving, and that some winching force is required to keep it moving.
 
Thank you for your help! I noticed the math errors and must have corrected them while you were typing your response.
 
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