Rod Hanging in Water: Find Line of Buoyancy Force

[Horvath Bela]

Homework Statement

Lets consider a rod of length l that is attached at a point to the cealing and relvolves freely. Half of the rod is immersed in water and half of it is out of the water. The rod is not vertical it has around a 60 degree angle to the water. Where is the line of action of the boyuancy force? Is it at the CM of the rod or is it at the center of ther part immersed in water? Why? I drew a picture hope it is understandible: https://imgur.com/a/6uxtO

Homework Equations

The Attempt at a Solution

 

[haruspex]

Homework Statement

Lets consider a rod of length l that is attached at a point to the cealing and relvolves freely. Half of the rod is immersed in water and half of it is out of the water. The rod is not vertical it has around a 60 degree angle to the water. Where is the line of action of the boyuancy force? Is it at the CM of the rod or is it at the center of ther part immersed in water? Why? I drew a picture hope it is understandible: https://imgur.com/a/6uxtO

Homework Equations

The Attempt at a Solution

You are required to complete the template. List any standard equations or laws that are likely relevant and show some attempt.

 

[Horvath Bela]

I am really qlueless as for the line of action of the force. I know the magnitude of the bouyancy force it is equal to the weight of the displaced water which is the half of the volume of the rod and times 1000 kg/m^3. The line of action is important because this a torque problem and since the rod is in equalibrium the net torque is zero, but to calculate the torque of the bouyancy force I would need to know its lever. I hope this shows you that I have considered the problem. Would you please tell me where the line of action of the bouyancy force is?

 

[haruspex]

I am really qlueless as for the line of action of the force. I know the magnitude of the bouyancy force it is equal to the weight of the displaced water which is the half of the volume of the rod and times 1000 kg/m^3. The line of action is important because this a torque problem and since the rod is in equalibrium the net torque is zero, but to calculate the torque of the bouyancy force I would need to know its lever. I hope this shows you that I have considered the problem. Would you please tell me where the line of action of the bouyancy force is?

No, but I will help you figure it out.
Consider removing the rod and filling in the space that leaves in the water with... more water. Clearly this would be in equilibrium. What is the weight of the added water, and where is the line of action of that weight?

 

[Horvath Bela]

The weight of the added water is equal to what the bouyance force was, and the line of action is where the center of the immersed part of the rod used to be.

 

[haruspex]

The weight of the added water is equal to what the bouyance force was, and the line of action is where the center of the immersed part of the rod used to be.

Yes.