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

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SUMMARY

The discussion centers on calculating the force required to winch a 4,535,924-kg ship down an inclined plane at a 10-degree angle, considering a coefficient of sliding friction of 0.18. The force of friction was initially calculated incorrectly, with a corrected value of 804,062 N. The net force required to maintain constant velocity down the slope was determined to be 6,914,956 N, emphasizing the need for clarity on whether the ship is at rest or already in motion.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the concepts of friction and inclined planes
  • Basic knowledge of trigonometry as applied to physics problems
  • Ability to perform calculations involving mass, gravity, and forces
NEXT STEPS
  • Review the principles of static and dynamic friction
  • Learn about the calculations involved in inclined plane problems
  • Study the effects of different angles on force calculations
  • Explore real-world applications of winching systems in marine engineering
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Students studying physics, engineers involved in marine design, and anyone interested in the mechanics of launching large vessels.

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.
 
Last edited:
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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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