• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Density and Apparent Weight

  • Thread starter Aoiumi
  • Start date
23
0
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:
18,890
3,667
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
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)?
 
23
0
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?
 
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
 
3,722
405
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.
 
18,890
3,667
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/ρB
This is the volume of the block, so the buoyant force exerted by the water on the block is:
[tex]F=9.8 ρ_Wv=(81.6)(9.8)\frac{ρ_W}{ρ_B}=800\frac{ρ_W}{ρ_B}[/tex]
Does that make sense?
 
23
0
That makes sense. Thank you!
 

Want to reply to this thread?

"Density and Apparent Weight" You must log in or register to reply here.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top