Calculating Specific Gravity in Water

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Homework Help Overview

The discussion revolves around calculating the specific gravity of an unknown material based on its weight in air and its weight when immersed in water. The problem involves concepts from fluid mechanics and density relationships.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster expresses uncertainty about how to start the problem and relates specific gravity to density. Some participants suggest using Archimedes' Principle and breaking the problem into manageable parts. Others provide a step-by-step approach to find the volume and density.

Discussion Status

Participants are actively engaging with the problem, offering hints and breaking down the concepts involved. There is a sense of progress as some participants indicate that the problem is becoming clearer, but no consensus has been reached on a final solution.

Contextual Notes

The original poster is working under the constraints of a homework assignment, which may limit the resources they can use to find a solution. There is also a focus on understanding the relationship between weight, volume, and density in the context of buoyancy.

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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