Calculating Specific Gravity in Water

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To calculate the specific gravity of an unknown material weighing 200 N in air and 150 N in water, Archimedes' Principle is applied. The buoyant force, which is the difference in weight (50 N), helps determine the volume of the material. Using the equation for buoyancy, the volume is found to be approximately 5.102 liters. The density is then calculated using the relationship between weight and volume, resulting in a density of about 4 times that of water. This leads to a specific gravity of 4 for the material.
davidatwayne
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Homework Statement



A sample of an unkown material weighs 200 N in air and 150 N when immersed in water. What is the specific gravity of this material?

Homework Equations


specific gravity = (density/density of water)
density of water is 1000 kg/m^3


The Attempt at a Solution



I don't even know where to start. All my textbook says is the specific gravity of a substance is the ratio of its density to the density of the water at 4 degrees C. I don't know how to relate this to an equation or begin to solve it.
 
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Hint: use Archimedes' Principle (or the Law of Buoyancy - a different name for the same thing)
 
davidatwayne;1191839 A sample of an unkown material weighs 200 N in air and 150 N when immersed in water. What is the specific gravity of this material?[/QUOTE said:
The key is to solve one thing at a time and not try to make one big impossible equation.

The volume of this thing is related to 50N of water (200 - 150)
50 = (v)(d)(g)
50 = (v)(1)(9.8)
v = 5.102 litres i think

200 = (5.102)(d)(9.8)
d = ~4

You can also do it as an expression in one shot. The in-water weight is 150, which is related to (d-1), and your normal weight affected by d alone is 200. Put the stand alone conditions as the numerator on each side and the in-water as the denom.:
200/150 = d/(d-1)
d = 4
 
Thanks a lot guys, it makes a lot more sense now.
 
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