How far above the water is the top surface of the cube?

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The discussion centers on the buoyancy of a wooden cube measuring 0.55 m on each side and weighing 133.1 kg. The cube's interaction with water is analyzed through the principles of buoyant force and equilibrium. The gravitational force acting on the cube is equal to the buoyant force when the cube floats, leading to the conclusion that if the density of the cube (P2) is greater than the density of water (P1), the cube will sink. The discussion also poses questions regarding the percentage of a solid cube that will sink based on density comparisons and the floating capability of large steel ships versus small steel pieces.

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A cube of wood measures 0.55 m on each side and has a mass of 133.1 kg. How far above the water is the top surface of the cube?

Can someone help me out with this one
 
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Cheman said:
But why does the water exert a force UPWARDS on the cube? It makes sense to say that Fdownwards = h1gpA, but why should the water below the cube exert a force of h2gpA upwards?
Thanks. :smile:

I see that there have been replies to your question. I would just like to add another way of looking at it. So here goes...

Imagine a trough filled with liquid till the brim. Let the liquid be in equilibrium. Consider a small cubic section of the liquid, of dimensions A x A x A and mass M=A^3 x P1(density), on the surface of the liquid, roughly at the centre. This is in equilibrium and it does not move (it is at rest initially). Which means that the net force on the cube is equal to zero. There are four surfaces of the cube that are vertical planes and two surfaces which are horizontal planes.

Forces on the cube
The four vertical planes experience a resultant force which is zero and so there is no horizontal movement.

There are two vertical forces acting on the cube.
1. Gravitational force = Mg, acting vertically downwards.

2. As we observe no vertical motion by the liquid section downwards, we are to conclude that there is an upward vertical force on the liquid section whose magnitude is = Mg = A^3 x P1 x g . This is what we call as buoyant force.

Under equilibrium conditions this is the maximum buoyant force (Mg=A^3 x P1 x g) that a liquid can exert (on the given liquid section). For if it exerts any other force the equilibrium is disturbed, and our initial assumption fails.

Now image placing a solid cube, of dimensions A x A x A and density P2, gently on the surface. Let it slowly sink. It would displace an equal volume of the liquid, which will pour out of the trough. It should be intuitively obvious to one that this is true. One could collect this overflowing liquid and measure its volume to verify. Assume that the cube just floats, i.e., the upper surface of the cube is along the surface of the liquid. Now there are two forces acting on the cube.

1. Gravity= A^3 x P2 x g, acting downwards and
2. Buoyant force= A^3 x P1 x g, acting upwards.


If gravity is greater the solid cube will sink.
If both are equal it would just float.
If gravity is lesser, then the cube will partially sink.

Now try answering these:

1. For P1 < P2 what percentage of a solid cube will sink?

2. Why does a small piece of solid steel sink while a ship made of tonnes of steel float?


Cheers
 
Last edited:

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