What is the Apparent Weight of an Iron Block Submerged in Water?

[Aoiumi]

Homework Statement

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 apparent weight is...
The answer is 700 N but I don't know how this is determined.

Homework Equations

Apparent weight is equal to mg minus Fb
Fb = ρfgVf

The Attempt at a Solution

Apparent weight = 800N - Fb
Fb = x(9.8)Vf
Density of iron is 8x

 

[Chestermiller]

Homework Statement

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 apparent weight is...
The answer is 700 N but I don't know how this is determined.

Homework Equations

Apparent weight is equal to mg minus Fb
Fb = ρfgVf

The Attempt at a Solution

Apparent weight = 800N - Fb
Fb = x(9.8)Vf
Density of iron is 8x

If the block has a weight of 800 N, what is its mass? From this result, if its density is ρ_B, 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 ρ_B)?

 

[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?

 

[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

 

[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.

 

[Chestermiller]

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?

You just leave it algebraic. v = 81.6/ρ_B
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?

 

[Aoiumi]

That makes sense. Thank you!