Calculating Apparent Weight of Stone: Specific Gravity 2.50

[akatsafa]

A stone of weight has specific gravity 2.50. (a) what is the apparent weight of the stone when under water? (b) what is its apparent weight in oil (specific gravity=0.90)?

I figured out the density to be 2500, but how do i relate the specific gravity and density to find the apparent weight?

Thanks.

 

[JohnDubYa]

The term appararent weight sucks.

Instead, you should be asked to find what a scale would read when the stone is submerged. Hanging scales (like the kind you find in produce sections) are tension force o'meters, so all you need to find is the tension in the rope and you have your scale reading (and thus your apparent weight).

So, when submerged three forces act on the stone. Gravity acting down. Buoyancy and tension acting up. Newton's second law says that

$$T + F_B - mg = 0$$

Substitute the proper expression for the buoyancy force and solve for T.

 

[M98Ranger]

The specific gravity of a substance is defined as the ratio of its density to the density of water. In this case, the specific gravity of the stone is 2.50, which means that it is 2.50 times as dense as water.

To calculate the apparent weight of the stone when under water, we can use the formula: Apparent weight = Weight - (Volume of stone x Density of water). Since we are given the specific gravity and not the actual weight of the stone, we can substitute the density of the stone with its specific gravity multiplied by the density of water. So, the formula becomes: Apparent weight = Weight - (Volume of stone x Specific gravity x Density of water).

Using this formula, we can calculate the apparent weight of the stone when under water as follows:

(a) Apparent weight = Weight - (Volume of stone x Specific gravity x Density of water)
= Weight - (Volume of stone x 2.50 x Density of water)
= Weight - (Volume of stone x 2.50 x 1000 kg/m^3) [since the density of water is 1000 kg/m^3]

Similarly, to calculate the apparent weight of the stone in oil (specific gravity = 0.90), we can use the same formula:

(b) Apparent weight = Weight - (Volume of stone x Specific gravity x Density of water)
= Weight - (Volume of stone x 0.90 x Density of water)
= Weight - (Volume of stone x 0.90 x 1000 kg/m^3)

So, to relate the specific gravity and density to find the apparent weight, we use the specific gravity as a multiplier for the density of water in the formula.

I hope this helps clarify the concept for you. Let me know if you have any further questions.