Specific Gravity: Mass 30kg & SG 3.6 in Fluid SG 1.2: True?

[tsecalvink]

An object with mass 30kg and specific gravity 3.6 is placed in a fluid whose specific gravity is 1.2. Neglecting viscosity, which of the following is true?

A. The acceleration of the object is (1/3)g and its apparent weight is 100N.
B. The acceleration of the object is (2/3)g and its apparent weight is 100N.
C. The acceleration of the object is (1/3)g and its apparent weight is 200N.
D. The acceleration of the object is (2/3)g and its apparent weight is 200N.


Equations I used:
g can be rounded to 10
$$\rho$$=m/v
F=mg=$$\rho$$Vg



I know the volume does not change. I know the answer is D but I don't know why.. When I attempted to solve it I attempted to do a proportion and I keep getting an apparent weight of 100N. Please explain why it is D.

 

[gneill]

Perhaps you could show a bit of the calculations you attempted and your reasoning behind them?

 

[tsecalvink]

Volume remains the same so I attempted to do this:

p = m/V, so V = m/p

m1/p1 = m2/p2
30kg/3.6 = m2/1.2
m2=10kg applying w=mg I get an apparent weight of 100N

And I assume gravity is 1/3

 

[gneill]

The mass m2 that you calculated is the mass of the displaced fluid. The buoyant force will equal the weight of the displaced fluid, and this force will decrease the apparent weight of the object. So the object will have apparent weight (30kg - 10kg)*g.

You should be able to express this as a formula involving the specific gravities and mass of the object. Taking the density of water to be 1.0g/cm³, then the specific gravities will be equivalent to densities. Let ρ1 be the specific gravity of the object, ρ2 the specific gravity of the fluid. You've already found that the volume will be given by V = m1/ρ1. So the apparent weight will be:

w = V*(ρ1 - ρ2)*g

You can substitute your expression for V into this, and rearrange to find what the "effective" g is.

 

[rwisz]

First, let's define some terms to make sure we are on the same page. Specific gravity (SG) is a ratio of the density of a substance to the density of a reference substance (usually water). In this case, the specific gravity of the object is given as 3.6, which means it is 3.6 times denser than the reference substance. The specific gravity of the fluid is given as 1.2, so it is 1.2 times denser than the reference substance.

Now, let's look at the equations you provided. The first equation, \rho=m/v, is the equation for density, where \rho is density, m is mass, and v is volume. This equation tells us that density is directly proportional to mass and inversely proportional to volume. This means that if the mass of an object increases, its density also increases, but if the volume stays the same, the density also stays the same.

The second equation, F=mg=\rhoVg, is the equation for weight, where F is weight, m is mass, g is the acceleration due to gravity, \rho is density, and V is volume. This equation tells us that weight is directly proportional to mass and density, and also to the acceleration due to gravity. This means that if the mass and density of an object increase, its weight also increases, but if the acceleration due to gravity decreases, the weight also decreases.

Now, let's apply these equations to the given scenario. The object has a mass of 30kg and a specific gravity of 3.6. This means that its density is 3.6 times greater than the reference substance. Since the volume does not change, the density also does not change. This means that the weight of the object is 3.6 times greater than it would be in the reference substance. In other words, the object is 3.6 times heavier.

Now, let's look at the fluid. Its specific gravity is 1.2, which means that its density is 1.2 times greater than the reference substance. Since the volume does not change, the density also does not change. This means that the weight of the fluid is 1.2 times greater than it would be in the reference substance. In other words, the fluid is 1.2 times heavier.

When the object is placed in the fluid, it displaces some of the fluid,