Calculating Specific Gravity in Water

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SUMMARY

The specific gravity of an unknown material can be calculated using the weights measured in air and water. In this case, the material weighs 200 N in air and 150 N when submerged, leading to a buoyant force of 50 N. By applying Archimedes' Principle, the density of the material is determined to be approximately 4 times that of water, resulting in a specific gravity of 4. This calculation utilizes the relationship between weight, volume, and density, confirming the material's density as 4000 kg/m³.

PREREQUISITES
  • Understanding of Archimedes' Principle
  • Basic knowledge of weight and buoyancy
  • Familiarity with density calculations
  • Ability to manipulate algebraic equations
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  • Study Archimedes' Principle in detail
  • Learn about buoyancy and its applications in fluid mechanics
  • Explore density calculations for various materials
  • Practice solving problems involving specific gravity and weight measurements
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Students in physics or engineering, educators teaching fluid mechanics, and anyone interested in material properties and buoyancy calculations.

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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