A cubical object of side 100cm floats over a fluid of specific gravity 1.6. The depth of immersion is 75 cm. The centre of buoyancy lies 25 cm below the surface of the fluid. When measured from the top surface of the object, identify the location of the centre of gravity from the top surface, which will cause the object to overturn.

a. 100 cm

b. 95 cm

c. 75 cm

d. 50cm

e. 25 cm

f. None of the above

This is a question from a fluid mechanics course I'm taking. The answer is 50cm and I'm having a hard time visualizing it.

What I've done:

enter image description here

The blue line being water level, yellow is centre of buoyancy, and orange is centre of gravity. From my sketch, it looks that the answer should be 25cm instead.

Could someone please explain what I've missed? This is also all the information I was given and not sure if there any assumptions I need to make?

Thanks in advance.

  • $\begingroup$ Is the mass of the cube uniformly distributed? $\endgroup$ – Harshit Joshi Jan 5 '19 at 5:51
  • $\begingroup$ I was wondering the same. But like I said this is all the info I have and I couldn't make sense of the solution even considering the case when it's not uniform. $\endgroup$ – Lin Jan 5 '19 at 14:57
  • $\begingroup$ If the weight is uniform, then the weight and buoyant force would be along the center of mass. So why should the block overturn? $\endgroup$ – Harshit Joshi Jan 5 '19 at 14:59
  • $\begingroup$ Also, why would the center of gravity change whether or not you put this in water. It's a fixed point which is located at the geometrical center of the body. $\endgroup$ – Harshit Joshi Jan 5 '19 at 15:05
  • $\begingroup$ When the block is tilted in the water, the centre of buoyancy would be to the side? And I was assuming the density is somewhat uniform (only shifting up and down) else it could be anywhere in the body and the question wouldn't make sense. But that got me to what I described above which doesn't make sense either. $\endgroup$ – Lin Jan 5 '19 at 15:05

The centre of Gravity of a cube is the point where its diagonals meet. Instead of inverting the cube, you should have sketched out the problem in such a way that the sides of the cube are perpendicular to the water surface (because that's the general method of solving these types questions in Fluid Mechanics). Doing this, you can now will be easily able visualise that, without any calculations relating any of the principles of Fluid Mechanics, the centre is half the side, that is, 50 cm.

Now, even if you keep the cube in a overturned position, the answer coming out is 50√2 ≈ 50 cm

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  • $\begingroup$ But it is given that the centre of buoyancy is 25cm below the water surface and the tube has 75cm immersed in water, which would not make sense if the sides of the cube are perpendicular to the water surface. $\endgroup$ – Lin Jan 4 '19 at 20:28
  • $\begingroup$ $50\sqrt 2 \approx 50$. What is this? $\endgroup$ – Harshit Joshi Jan 5 '19 at 5:53
  • $\begingroup$ I've mentioned the case in the end. $\endgroup$ – Aniruddha Mukherjee Jan 5 '19 at 11:29
  • $\begingroup$ If you want me to read your comment use @harshit54 before it. $\endgroup$ – Harshit Joshi Jan 5 '19 at 15:17
  • $\begingroup$ Also what you mention is called the center of mass and not the center of gravity. $\endgroup$ – Harshit Joshi Jan 5 '19 at 15:17

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