Derive a 2 order ODE from a physical problem

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SUMMARY

The discussion focuses on deriving a second-order ordinary differential equation (ODE) to model the motion of objects bobbing in water, based on Newton's second law (F = ma). Key forces include gravity, friction from water, and buoyancy as described by Archimedes' principle. The derived ODE is z'' + (w/m)z' - (pVg)/m + g = 0, where z represents the depth of the object, w is the coefficient of friction, p is the water density, V is the submerged volume, m is the mass, and g is the acceleration due to gravity. The discussion emphasizes the importance of sign conventions in force calculations, particularly for friction and buoyancy.

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  • Understanding of Newton's second law (F = ma)
  • Familiarity with Archimedes' principle and buoyancy
  • Knowledge of differential equations, specifically second-order ODEs
  • Basic concepts of fluid dynamics, including frictional forces in fluids
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Homework Statement



The motion of various shaped objects that bob in a pool of water can be modeled by a second order differential equation derived from F = ma.

The forces acting on the object are:
1)force due to gravity,
2)a frictional force by water,
3)and a buoyant force based on Archimedes' principle: An object that is completely or partially submerged in a fluid is acted on by an upward force equals to the weight of the water it displaces.

Derive the ODE by:
1)the dependent variable is the depth z of the object's lowest point in water,take z to be negative downward so that z= -1 means 1 ft of the object has submerged.
2)Let V(z) be the submerged volume of the object
3)m be the mass of the object
4)p be the density of the water
5)g be the acceleration due to gravity
6)w be the coef of friction for water(friction is proportional to the vertical velocity of the object)


Homework Equations





The Attempt at a Solution


I am not sure whether my answer is correct since I had confusion in treating positive and negative sign.

z''+ (w/m)z'-(pVg)/m +g =0

Thx very much for your attention
 
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With a force pointing up as positive the net force, F, acting on the object is

F = -Ff + Fb - Fg

So, is your sign convention correct?
 
chrisk said:
With a force pointing up as positive the net force, F, acting on the object is

F = -Ff + Fb - Fg

So, is your sign convention correct?

but I don't think that friction force must be negative since it opposes the direction of velocity,so I propose that

F= -wv + pVg - mg (negative sign for friction force since it opposes the velocity, pVg is positive since it points upward, mg is downward hence it is negative)

Does my argument make sense?

Moreover, V depends on the depth z, if we consider the volume of a cube of length L,
would it be V=L2z or V= -L2z ?
in case we define volume as positive
 
Last edited:
Your statement about friction opposing the direction of velocity is correct. Since Ff = wv the minus sign for Ff provides the correct direction of the frictional force. Regarding the volume, it's part of the buoyant force expression. You can state that V(z) = V(-z) in the ODE and leave Fb as positive (upward force) in the force equation, or your can place a negative sign in front of the buoyant force expression if V(z) = -V(z) as you suggested.
 

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