Floating Copper Ball
Date: 02/26/2002 at 10:27:04 From: Amber Subject: Floating Copper Ball Find the wall thickness of a hollow copper ball (sphere) with an outside radius of 50.0 cm that "just floats" in water.
Date: 02/26/2002 at 10:47:13 From: Doctor Ian Subject: Re: Floating Copper Ball Hi Amber, By 'just floats' I assume you mean that the top of the sphere is even with the top of the water. Is that correct? The key principle here is that if you submerge something in water, the water will provide a buoyant force equal to the weight of the water displaced. (This is known as Archimedes' principle, because according to legend, he came up with as a solution to the problem of figuring out whether a crown was made from pure gold or an alloy.) Since you know the outside radius of the sphere, you can compute the volume of the sphere. That will tell the volume of the water displaced. If you look up the density of water, you can use that to compute the weight of the water displaced. And that is the buoyant force. Now, if the sphere 'just floats', it means that the buoyant force (i.e., the water pushing up) exactly cancels out the gravitational force (i.e., gravity pulling down). So what you need to do is figure out how thick the sphere has to be so that the weight of the copper is equal to the weight of the water displaced. If the outside radius of the sphere is R, and the inside radius is r, then the volume of copper used will be V = volume of outside sphere - volume of inside sphere = (4/3) pi R^3 - (4/3) pi r^3 To find the weight of the copper, you'll need to look up the density of copper. Can you take it from here? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum