Object with mass 30kg and specific gravity 3.6 in fluid of specific gravity 1.2

[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.