Static Equilibrium with buoyant force. High difficulty.

In summary, the conversation discusses a problem involving a rod hanging from a wire at an angle with tension and the other end submerged in water. The conversation goes on to discuss various equations and solutions for unknown variables, ultimately leading to a surprising and legitimate solution of x/L = 1/2. The conversation also touches on the importance of considering the height and length of the rod, and the need for a balanced solution for the wire to be perfectly vertical.
  • #1
physninj
37
0

Homework Statement


I attached the original problem and a diagram I made with the variables on it. Basically we have a rod hanging from a wire at an angle theta with tension and the other end floating in water. So there is T=tension, weight=w=mg and buoyant force=F_b. I introduced symbols for total length L and submerged length x along with the height the rod hangs from h. The rod has a density half that of water or 500 kg/m^3.

Homework Equations


Ʃτ=0, about top of rod Ʃτ= (L/2)mgcosθ-(L-x/2)Fbcosθ=0
ƩFx=0,
ƩFy=0=T+Fb-mg

The Attempt at a Solution



The real difficulty I am having is that it's unclear what my answer should look like. I think it should depend upon the height and length of the rod. I really don't know what I'm expected to treat as known and unknown, and I only have two equations so I can only solve for two things correct? It asks for the fraction submerged which is x/L, but if those are the unknowns then the forces would have to be known. I keep going in circles on this one and it's driving me mad. Any guidance would be greatly appreciated.
 

Attachments

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  • #2
You should be able to write down the value of Fb in terms of m, g, L and x.
Just about everything except x and L should then cancel out in your torque equation.
 
  • #3
Thank you! I was getting worried that nobody would have any advice for me. I did come up for an expression for Fb in those terms, it was (2mgx)/L. (because twice the weight of the fraction of the rod submerged). but then the problem becomes that I have solved for x/L...in terms of x and L? Plugging and chugging that expression got me x/L= L/(2x)-1


I guess my questions still remain about the legitimacy of this. I never used the Fy equation and if L and x were known quantities then why couldn't I have just put fraction=x/L and QED!?
 
  • #4
physninj said:
Thank you! I was getting worried that nobody would have any advice for me. I did come up for an expression for Fb in those terms, it was (2mgx)/L. (because twice the weight of the fraction of the rod submerged). but then the problem becomes that I have solved for x/L...in terms of x and L? Plugging and chugging that expression got me x/L= L/(2x)-1
That's just a quadratic in x/L. Solve it!
I guess my questions still remain about the legitimacy of this. I never used the Fy equation and if L and x were known quantities then why couldn't I have just put fraction=x/L and QED!?
It's all legitimate and correct. You didn't need the other equations because you don't care about the tension in the string. You could have taken torque about some other point, which would have given you an equation involving that tension. Then you would have needed the Fy equation to eliminate it again. But why do that when you can finesse it by a smart choice of centre of moments?
 
  • #5
Well I'll be. I got everything on one side and multiplied it by 1/L to get it looking right and the solution came to be 1/2. I really expected that the answer would depend on the height it was hung from, I'm really suprised that there is a real solid solution for this problem, can you understand why that confuses me? I figure if it was hanging high enough the fraction would be necessarily smaller, not the same value no matter the length or height.

Well thank you for sticking with me anyways. Cheers:smile:
 
  • #6
physninj said:
Well I'll be. I got everything on one side and multiplied it by 1/L to get it looking right and the solution came to be 1/2. I really expected that the answer would depend on the height it was hung from, I'm really suprised that there is a real solid solution for this problem, can you understand why that confuses me? I figure if it was hanging high enough the fraction would be necessarily smaller, not the same value no matter the length or height.

Well thank you for sticking with me anyways. Cheers:smile:

Actually I DO get it. because I think it needs to be balanced that way for the wire to be perfectly vertical as the problem requires! that's awesome.
 
  • #7
physninj said:
I got everything on one side and multiplied it by 1/L to get it looking right and the solution came to be 1/2.
Hmmm... that's not a solution of the equation you posted before: x/L= L/(2x)-1.
That equation gives x/L = (√3 - 1)/2. Did your equation change?
 

Related to Static Equilibrium with buoyant force. High difficulty.

1. What is static equilibrium and how does it relate to buoyant force?

Static equilibrium refers to the state in which an object is at rest or moving at a constant velocity with no net force acting upon it. In the case of buoyant force, this refers to the balance between the upward force of buoyancy and the downward force of gravity on an object submerged in a fluid.

2. How is buoyant force calculated in high difficulty situations?

In high difficulty situations, buoyant force is calculated using Archimedes' principle, which states that the buoyant force on an object is equal to the weight of the fluid it displaces. This can be expressed mathematically as FB = ρVg, where FB is the buoyant force, ρ is the density of the fluid, V is the volume of the object submerged, and g is the acceleration due to gravity.

3. What factors affect the magnitude of buoyant force?

The magnitude of buoyant force is affected by the density of the fluid, the volume of the object submerged, and the acceleration due to gravity. It is also affected by the shape and surface area of the object, as well as the depth at which it is submerged.

4. Can an object be in static equilibrium with both buoyant force and external forces acting on it?

Yes, an object can be in static equilibrium with both buoyant force and external forces acting on it. In this situation, the net force acting on the object is equal to zero, which means that there is no acceleration and the object remains at rest or moves at a constant velocity.

5. How does the center of buoyancy affect an object in static equilibrium?

The center of buoyancy is the point at which the buoyant force acts on an object. In static equilibrium, the center of buoyancy must be in line with the center of gravity for the object to remain balanced. If the center of buoyancy and center of gravity are not aligned, the object will experience a torque and will not be in static equilibrium.

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