How Much of the Granite Rock Ball Must Be Submerged to Float?

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SUMMARY

The discussion focuses on calculating the submerged area of a granite rock ball with a diameter of 1.3 meters, which is necessary for it to float under a water pressure of 50 lbs/in². The granite density is specified as 2691 kg/m³. A calculation provided indicates that an area of 136.2 in² must be submerged to achieve buoyancy, derived from the relationship between the weight of the ball and the pressure exerted by the water. The fundamental principle discussed is that the buoyant force must equal the weight of the ball for it to float.

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1. At Lagoon, there is a large granite rock ball that is supported by water pressure, so people can spin the rock. The diameter of the rock is 1.3m. Granite has a density of 2691kg/m^3. Let’s assume a water pressure if 50 lbs/in^2. Calculate the area of the ball that must be in the water under pressure in order for the ball to “float”.
2. I was given A=2691kg/m^3 * 2.2 lbs/kg * 4/3 x pi * (.65m)^2 * in^2/50lbs = 136.2 in^2 though I am not sure what full relevancy this has to the problem. Is this the answer to the equation or a beneficial equation?
3. I am attempting to help my niece as I am the only one in my family who went beyond pre-algebra in school but this is beyond me. Any help (if possible, I know I am breaking the rules with not having attempts to solve it) would be appreciated.
 
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The equation looks like an attempt at an approximate solution.
The idea is that the force from the water is equal to the weight of the ball.

The weight of the ball is mg, get the mass from the density and the volume ... you should know how to find the volume of a sphere.

The force from the pressure is the pressure times the area that it acts.
On the ball, the pressure on different parts results in a force in different directions - but that may not be important if the area is very small compared with the overall surface area of the ball.
 

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