# Specific Gravity

1. Jan 24, 2009

### jan2905

An object with mass 30kg and specific gravity 3.6 is placed in a fluid whose specific gravity is 1.2. Neglecting viscosity, what is the objects acceleration and weight?

I guessed on this one. Not sure how to make things come together.

I said that a=2/3(g) and F=200N

2. Jan 24, 2009

### chrisk

Refer to Archimedes' Principle then show your approach if you do not get the desired result.

3. Jan 24, 2009

### jan2905

is specific gravity (rho)?

4. Jan 24, 2009

### nvn

jan2905: No. Density is mass per unit volume, and is denoted by the symbol rho. Density (rho) has SI units of kg/m^3. Specific gravity is the density of a substance divided by the density of water, and is therefore a dimensionless ratio. Specific gravity is sometimes denoted by the symbol SG, or maybe G (?), but never rho. Thus, specific gravity SG = rho/rhow, where rhow = density of water.

Regarding your acceleration answer, excellent work! That is correct. From the definition of specific gravity, and from the definition of density, you can solve for volume of the object. Your answer for the apparent weight of the submerged object is currently incorrect. However, remember Newton's second law? Try it again.

5. Jan 25, 2009

### jan2905

how can that be? F=mg=30kg(2/3)(9.81)=196...

6. Jan 25, 2009

### nvn

That is correct; F = m*a = m*(2/3)g = (30 kg)(2/3)(9.807 m/s^2) = 196.1 N.

How can that be? The water pressure on the bottom of the object is pushing upward harder than the water pressure on the top of the object is pushing downward. This creates a net upward force, called buoyancy force, which reduces the magnitude of the downward acceleration of the object.

7. Jan 26, 2009

### jan2905

you said it wasn't 200N... because it's 196N? ... sorry I rounded.

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