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

AI Thread Summary
To determine how much of the granite rock ball must be submerged to float, the weight of the ball must equal the buoyant force from the water pressure. The granite ball has a diameter of 1.3m and a density of 2691kg/m^3, leading to a calculated area of 136.2 in^2 that needs to be submerged under a pressure of 50 lbs/in^2. The weight of the ball can be calculated using its density and volume, while the buoyant force is derived from the pressure and submerged area. The discussion highlights the importance of equating the forces to find the necessary submerged area for flotation. Understanding the relationship between pressure, area, and weight is crucial for solving this problem.
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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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