Density and Apparent Weight

1. Nov 20, 2013

Aoiumi

1. The problem statement, all variables and given/known data
The density of iron is 8 times larger than the density of water. An iron block has weight 800 N when it is weighed in air. When the block is weighed when completely submerged in water, the apparrent weight is....
The answer is 700 N but I don't know how this is determined.

2. Relevant equations
Apparent weight is equal to mg minus Fb
Fb = ρfgVf

3. The attempt at a solution
Apparent weight = 800N - Fb
Fb = x(9.8)Vf
Density of iron is 8x

Last edited: Nov 20, 2013
2. Nov 20, 2013

Staff: Mentor

If the block has a weight of 800 N, what is its mass? From this result, if its density is ρB, what is its volume? If it is totally submerged in water, with this volume, what is the buoyant force on the block (given that the density of water is 1/8 of ρB)?

3. Nov 20, 2013

Aoiumi

800N = m (9.8 m/s^2)
m = 81.6 kg

If Density of iron: ρ = m/v
ρ = 81.6 / v

How do I express ρ?

4. Nov 20, 2013

gabriel.dac

Water is more dense than air, so it will push the block up. It will seems the block will weigh less. But I think that how deep the block is in the water matters too. It's apparent weight will be different in the bottom of a pool and in the bottom of the ocean. I'm not really sure though

5. Nov 20, 2013

nasu

You don't need to find the mass explicitly.
You wrote a formula for Fb.
Write a similar one for the weight of the block (W).
And compare the two. You know W. You can easily find Fb and then the apparent weight.

6. Nov 20, 2013

Staff: Mentor

You just leave it algebraic. v = 81.6/ρB
This is the volume of the block, so the buoyant force exerted by the water on the block is:
$$F=9.8 ρ_Wv=(81.6)(9.8)\frac{ρ_W}{ρ_B}=800\frac{ρ_W}{ρ_B}$$
Does that make sense?

7. Nov 20, 2013

Aoiumi

That makes sense. Thank you!