geos100spr2012_isostasy

geos100spr2012_isostasy - 1 Consider a wooden block...

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Unformatted text preview: 1 Consider a wooden block floating in water, as depicted below: According to Archimedes, any floating object displaces its own weight of fluid. That is to say, the weight of the floating object and the weight of the fluid displaced by the object are equal. In the case of our floating wooden block, we can summarize this principle as follows: 2 1 W W (1) where W 1 is the weight of the wood and W 2 is the weight of the displaced water. Weight is defined as gravitational force acting on the mass of an object: Mg W (2) where m is the objects mass and g is the acceleration due to gravity (9.81 m/s 2 ), which is a constant. Substituting equation 2 into equation 1 gives us: g M g M 2 1 (3) where M 1 is the mass of the wood and M 2 is the mass of the displaced water. Since g is a constant and is on both sides of the equation, we can ignore it and simplify equation 3 as: 2 1 M M (4) Density ( ) is mass per unit volume ( V ), and can be expressed as: V M (5) Rearranging equation 5 to solve for mass, we get: V M (6) Substituting equation 6 into equation 4, we get: 2 2 1 1 V V (7) where 1 and V 1 are the density and volume of the wood, and 2 and V 2 are the density and volume of the displaced water. Volume is the product of the length ( l ), width ( w ), and depth ( d ) of an object: lwd V (8) Substituting equation 8 into equation 7, we get: 2 2 2 2 1 1 1 1 d w l d w l (9) where l 1 , w 1 , and d 1 , are the length, width, and depth of the wood, and...
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This note was uploaded on 03/08/2012 for the course GEOS 100 taught by Professor Sammatson during the Spring '12 term at Boise State.

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geos100spr2012_isostasy - 1 Consider a wooden block...

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