Derive a 2 order ODE from a physical problem

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Homework Help Overview

The discussion revolves around deriving a second-order ordinary differential equation (ODE) related to the motion of objects submerged in water, based on Newton's second law (F = ma). The forces considered include gravitational force, frictional force due to water, and buoyant force as described by Archimedes' principle.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the correct sign conventions for forces acting on the object, particularly regarding friction and buoyancy. Questions arise about the dependence of volume on depth and how to express it in the context of the ODE.

Discussion Status

There is an ongoing examination of the sign conventions and the relationships between the forces involved. Some participants provide insights into the treatment of friction and buoyancy, while others seek clarification on the definitions and implications of volume in the equation.

Contextual Notes

Participants are navigating potential confusion regarding positive and negative signs in their equations, as well as the implications of defining volume in relation to depth. The discussion reflects a need for clarity on these assumptions without reaching a definitive conclusion.

lo1206
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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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