Spring constant for a solid not in equilibrium in a fluid

In summary, the conversation discusses a floating cylinder in a liquid and finding the expression for the net force in the y direction. The problem also asks for the "spring constant" k and the conversation goes on to prove that h=(RhoO/RhoF)L. The simplified expression for Fnet is -ky, where k = A*g*ρf.
  • #1
Twoism
17
0

Homework Statement


A cylinder of density RhoO, length L, and cross-section area A floats in a liquid of density RhoF with its axis perpendicular to the surface. Length h of the cylinder is submerged when the cylinder floats at rest. Suppose the cylinder is distance y above its equilibrium position. Find an expression for Fnet (in the y direction). Use what you know to cancel terms and write this expresion as simple as possible. What is the "spring constant" k? I go on to prove h=(RhoO/RhoF)L


Homework Equations


Fnet=F(buoyant) + -(F(cylinder))
F(buoyant)=RhoF*VolumeInWater*g
F(cylinder)=RhoO*TotalVolume*g
Fnet=-ky
VolumeInWater=A(h-y)
TotalVolume=AL
h=(RhoO/RhoF)L
k= Spring constant

The Attempt at a Solution


I get to a point where the equation goes as follows
-ky=Ag(RhoF(h-y) - (RhoO*L))
I tried substituting in h=(RhoO/RhoF)L and L=(RhoF/RhoO)h, but neither seem to make the expression "simple" like the problem states, and when I solve for k, it's just messier. Thanks for any help. Oh, and the LaTeX kept messing up, so I had skip it, sorry.
 
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  • #2
From the equilibrium position if you press the cylinder fro a small distance dy,
weight of the displaced liquid is A*g*ρf*dy = buoyant force = -k*dy.

So k = A*g*ρf.

Put it in the final expression and simplify.
 
  • #3
rl.bhat said:
From the equilibrium position if you press the cylinder fro a small distance dy,
weight of the displaced liquid is A*g*ρf*dy = buoyant force = -k*dy.

So k = A*g*ρf.

Put it in the final expression and simplify.

You're pulling it up though, not pressing it down, if that makes a difference. Also, I'm slightly confused because how can the buoyant force equal the restoring force if the there's already part of the cylinder in the water. The length of the cylinder in the fluid isn't just dy. Wouldn't it be h+dy (if you're pressing it down), which would make
A*g*ρf*(h+dy)=-k*dy

Right?

EDIT: Nevermind, I understand now. If you did h+dy, that would be implying the equilibrium point is at the surface of the fluid, which it is not. I don't know why I was thinking that the rest of the buoyant force would contribute to the restoring force if it was in equilibrium at length dy away. Thank you!
 
Last edited:

Related to Spring constant for a solid not in equilibrium in a fluid

1. What is the definition of spring constant for a solid not in equilibrium in a fluid?

The spring constant, also known as the stiffness constant, is a measure of how easily a solid material can be deformed by a force. It is a property that describes the relationship between the applied force and the resulting displacement of the material.

2. How is the spring constant calculated for a solid not in equilibrium in a fluid?

The spring constant can be calculated by dividing the applied force by the resulting displacement of the material. This can also be represented by the slope of the force-displacement graph, where the steeper the slope, the higher the spring constant.

3. What factors can affect the spring constant for a solid not in equilibrium in a fluid?

The spring constant is affected by the material's properties, such as its elasticity, density, and shape. It can also be influenced by external factors such as temperature, pressure, and the presence of a fluid surrounding the material.

4. How does the spring constant for a solid not in equilibrium in a fluid differ from that of a solid in equilibrium?

The spring constant for a solid not in equilibrium in a fluid may be different from that of a solid in equilibrium due to the presence of an external force, such as gravity or fluid pressure, acting on the material. This can cause a change in the material's stiffness and affect the spring constant.

5. Why is the spring constant important for understanding the behavior of solids in fluids?

The spring constant is important because it helps scientists and engineers understand the relationship between applied forces and resulting displacements in solid materials. This knowledge is crucial for designing structures and predicting the behavior of materials in various environments, including fluids.

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