Specific Gravity

[SunChip]

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EK Physics 1001, page 56, #550.

An object completed submerged in a fluid with a specific gravity of 3 has an apparent loss of weight of 40 N. If the mass of the object is 12 kg, what is its specific gravity?

Answer: 9

Can someone explain this to me?

 

[SunChip]

Ok, here's the explanation, but where did the density of "3000" come from?

If you have a certain object with a certain weight and you want to lift it up, you have exert a force equal to its weight to lift it.

If you take the object (like a big rock) under water (or some other fluid) with you, and you again want to lift it up, you no longer now have to exert a force on it equal to its total weight. Why? The water that it's submerged in is trying to "lift" some of it for you. That is, when the rock is submerged, there is an upward bouyant force on it, that opposed the weight of the rock. So now when you go to lift it, YOU only have to lift the weight of the rock that the water is NOT lifting for you. This is the apparent loss of weight. Objects feel lighter under water. In fact, a lot of weightless training for astronauts is done in large "swimming" pools. The astronauts can wear weight belts and suits filled with various amounts of air to keep the total bouyant force on them equal to their own weight. In that case, what would be the apparent weight of the astronaut be?

0. If you wanted to lift him/her, you wouldn't have to exert any upward force, since the bouyant force is doing all of it for you.

So the apparent weight of something is:

Weight(apparent) = Weight(actual) - Bouyant force

or Bouyant force = apparent LOSS of weight

So the apparent loss of weight is 40 N = the bouyant force is 40 N

Fb = rho*g*V

Fb is Bouyant force, rho = density of the fluid, g = 10, V = the volume of fluid displaced = the volume of the object (since the object is completely submerged). If the specific gravity of the fluid is 3, it's density is 3000.

So we have:

40 = 3000*10*V

V = 4/3000

So we have the volume of the object, and we want to find its density, given that its mass is 12 kg.

rho = 12 / (4/3000) = 12 * 3000 / 4 = 9000

A density of 9000 (kg / m^3) is a specific gravity of 9.

 

[Rabolisk]

An object is completely submerged, so the force that pulls it down is its weight, which is mg = (12kg)(10m/s^2) = 120N. That would be the weight in vacuum, or on the ground. However, its apparent weight is less, because the buoyant force acts upwards, lifting the object. We know that this force is 40 N, because the object has an "apparent loss" of 40 N. That means that ρVg = 40. Since the object is fully submerged, the volume of the object, and the volume of the water displaced are the same. That means that mg = 3*ρVg. Mass is equal to density times volume, and since volumes are equal, that means that the density of the object is three times the density of the fluid. Specific gravity is just a ratio of density to that of water, so 3 * 3 = 9.

 

[SunChip]

Thanks! 🙂

 

[livinitup]

Why was 3000 used instead of 1000? I thought the buoyant force equation was (density of fluid)(V)(g), not the density of the object.