What is the Critical Density for a Wedge in Water?

[Brilli]

Homework Statement

If a beam with square cross-section and very low density is placed in water, it will turn one pair of its long opposite
faces horizontal. This orientation, however, becomes unstable
as we increase its density. Find the critical density when this
transition occurs. The density of water is ρw = 1000 kg/m3

Homework Equations

3. Attempt at solution
Taking the centre of mass of the square cross section of the wedge as the axis and buoyant force to rotate the wedge about this axis we can equate torques. But i don't know how to do this

 

[BvU]

Dunno isn't good enough in PF (see guidelines)
Perhaps it's good to turn this around: start with a high density, let it go lower and see if you can conclude something from that ...

 

[haruspex]

Taking the centre of mass of the square cross section of the wedge as the axis and buoyant force to rotate the wedge about this axis we can equate torques.

Not quite. A beam in the orientation described will always be in balance, and the torques will always be equal, by symmetry. It is a question of stability, which means that you need to consider a small perturbation from that position and find which way the net torque goes.

 

[Brilli]

Not quite. A beam in the orientation described will always be in balance, and the torques will always be equal, by symmetry. It is a question of stability, which means that you need to consider a small perturbation from that position and find which way the net torque goes.

Thank you for replying . I was confused in this term. Let me try ahead and see.

 

[Brilli]

Not quite. A beam in the orientation described will always be in balance, and the torques will always be equal, by symmetry. It is a question of stability, which means that you need to consider a small perturbation from that position and find which way the net torque goes.

But about what point should the perturbation be taken? ?

 

[haruspex]

But about what point should the perturbation be taken? ?

From the horizontal/vertical position. If it is tilted at a small angle to that, will the torque tend to oppose the tilt or increase it?

 

[Brilli]

From the horizontal/vertical position. If it is tilted at a small angle to that, will the torque tend to oppose the tilt or increase it?

I was asking about what point should be the axis passing through On the square?

 

[haruspex]

I was asking about what point should be the axis passing through On the square?

As you wrote in post #1, the centre of the square will do. But in principle, you can take an axis anywhere.

 

[Brilli]

As you wrote in post #1, the centre of the square will do. But in principle, you can take an axis anywhere.

But if taken at the centre the wedge is always rotataed back to original position. Could you explain your recommendation with a diagram? I am very confused with this question

 

[Brilli]

Also about the centre of the squares face what is balancing the bouyancy force ?

 

[BvU]

Gravity

 

[Brilli]

Gravity

If we take centre of square face of the wedge to be a point and axis passes through that weight of the wegde doesn't produce any torque about this axis. So what is balancing the torque due to bouyancy?

 

[haruspex]

But if taken at the centre the wedge is always rotataed back to original position.

Is it?
First question is, will the centre of mass be at the same height relative to the surface of the water? Probably, but needs to be checked.
Second, where is the centre of buoyancy?
Draw a couple of diagrams, one where most of the beam (it is not a wedge) is immersed and one where little is immersed. But in each case, since we are concerned with small perturbations, do not tilt it so far that a corner enters or leaves the water.

 

[BvU]

f we take centre of square face of the wedge to be a point and axis passes through that weight of the wegde doesn't produce any torque about this axis.

Correct. But the buoyancy force does -- when there is a small disturbance. If that is not counteracted the thing will start to tilt and stop at a different orientation.

Where does the word wedge come from ? You talk about a beam with a square cross section.

Make two drawings: one where the beam is horizontal and one where it's at an angle, partially above the water. mark center of mass and center of buoyancy

hey, Haru was a little quicker on the draw ! Good to see we agree

 

[Brilli]

So should i go on with the following idea- that if the beam turns by an angle then the polygon made by the part under water should have its centre of mass aligned with centre of mass of the square face of the wedge?

 

[haruspex]

So should i go on with the following idea- that if the beam turns by an angle then the polygon made by the part under water should have its centre of mass aligned with centre of mass of the square face of the wedge?

Not sure what you mean by "should" there. The point is that it will not be, but the question is, which side will it be? I.e. will the net torque accelerate the turn or restore it?

 

[Brilli]

Not sure what you mean by "should" there. The point is that it will not be, but the question is, which side will it be? I.e. will the net torque accelerate the turn or restore it?

The problem asks for the point at which the bouyancy torque does not turn it back to previous equilibrium position. So just at the time in which this happens there should be an equilibrium point. Thus the centre of bouyancy should pass through the centre of square face so that it doesn't produce any torque

 

[Brilli]

Could you say if i miss some torques? I take the torque due to bouyancy about the centre of the sqare face

 

[haruspex]

The problem asks for the point at which the bouyancy torque does not turn it back to previous equilibrium position. So just at the time in which this happens there should be an equilibrium point. Thus the centre of bouyancy should pass through the centre of square face so that it doesn't produce any torque

Ok, but that need not be exactly true for a finite rotation. You just need the first order term to vanish.
Don't forget to make sure the same volume is immersed in the rotated position.