# Mass of water displaced by a block

1. Jul 20, 2011

### Jgoshorn1

1. The problem statement, all variables and given/known data

A block is suspended from a scale and then lowered into a bucket of water. The density of the water is 1 gm/cm3. The initial reading on the scale is 19N and one complete revolution of the scale is a change of 10N. (The block is completely submerged after one complete revolution)
What is the mass of the the water displaced by the block? What is the volume of the block? What is the density of the block?

2. Relevant equations

Magnitude of buoyant force = weight of displaced fluid.
The mass density is the mas of the substance divided by its volume.

3. The attempt at a solution

I found the mass of the block when completely submerged in water to be .92kg by [(Initial reading-final reading)/gravity]=[(19N-10N)/9.8]
I'm not sure how to use this information to go about finding the mass of the water displaced by the block.

2. Jul 20, 2011

### tiny-tim

Welcome to PF!

Hi Jgoshorn1! Welcome to PF!

You know that the weight of the block is 10 N.

So call the volume of the block V cm3.

What is the weight of the water displaced?

3. Jul 20, 2011

### Jgoshorn1

Hi tiny-tim.
After staring at this problem for a little while long I decided that the mass I had previously calculated for the block when it is completely submerged would be 19N/9.8=1.95 rather than the (19N-10N)/9.8 that I previously thought.
So the apparent weight when submerged is 10N and volume=mass/density. I'm not sure how to find the volume of the block with the information I've been given. I'm super confused.

4. Jul 21, 2011

### tiny-tim

Hi Jgoshorn1!

I'm not really understanding your terminology

the mass of the block is fixed, it's the same whether it's in air water or outer space.

Call the volume V (so the density is m/V) …

then how much would the block weigh (as shown on the scale) when submerged?

5. Jul 21, 2011

### Jgoshorn1

Hi!
So I worked on this problem for a very long time last night and I still wasn't getting an answer provided we can chose from [1.1 kg, 1.6 kg, 2.4 kg, 3.5 kg, 4.8 kg]

The way I tried it was:
Fnet=0 because it's in equilibrium
I made a FBD with Force(gravity) pointing down, Force(buoyancy) and Force(Tension from the scale) pointing up.
(Fb+Ft) - (Fg) = 0
Fb = m(block)g - Ft
Fb = (1.94 kg)( 9.8m/s^2) - 9N
Fb = 10.012
Fb = m(water)g
10.012 = m(water)(9.81)
10.012/9.81 = m(water) = 1.02kg

Does this make sense? I can't figure out any other way to really do it... It is part of an online lab (and we can discuss labs on our discussion posts but nobody seems to be getting an answer provided) and there is a really simple animation that goes with it where the block is lowered into the water attached to the scale.. the scale goes backwards through one complete revolution (started at 19N ending at 9N).. the water level rises in the animation as to be expected and I even tried to calculate the volume of the water displaced from taking measurements on the x-y coordinate system but it doesn't make any sense because if you look at it, when you put the block (aprx 4x1) into the water the water level raises by aprx 10x1. Also there are no given units so that would all just be an assumption that each grid block on the animation plane is 1cm x 1 cm x 1 cm.

Super frustrating.

6. Jul 21, 2011

### Jgoshorn1

I did that:
Magnitude of buoyant force = weight of displace fluid
Fb = m(water)g
10.012 = m(water)(9.81)
10.012/9.81 = m(water) = 1.02kg

The problem is that 1.02kg wasn't a given option and I wanted to know if the way I was doing it was wrong or not.. Or if there is any other way of doing it.

7. Jul 21, 2011

### Aimless

Sorry... somehow I completely missed your post #5 when I was responding.

Yes, that looks correct for the mass of the fluid. As for why it doesn't match up with one of the answers provided, that usually indicates a rounding problem somewhere: either you are rounding somewhere, or the creators of the lab are rounding somewhere that you aren't.

Edit: I caught my error and attempted to delete the offending post, but managed in doing so to cross my deletion with Jgoshorn1's reply. :( Yikes. The full text of the missing post is as he quoted. Sorry for making a mess of this thread.

Last edited: Jul 21, 2011
8. Jul 21, 2011

### Jgoshorn1

It's okay, I wasn't offended. I emailed my prof to figure out whats going on with this problem.. I don't care if I am getting it wrong or miss the credit for the problem.. I just want to know how to do it correctly!

Until then I'm struggling with the second/third question: What is the volume/density of the block.

9. Jul 21, 2011

### Aimless

As you stated in post #1:

"The mass density is the mass of the substance divided by its volume."

What is the mass of the water? What is its density?

Given the volume of the displaced water, what is the volume of the block?

10. Jul 21, 2011

### Jgoshorn1

Mass of water = 1.02kg = 1020 g
Density of water = 1gm/cm^3
v(water)=m/d = 1020g/(1g/cm^3) = 1020 cm^3 = volume of water = volume of block?

The options to chose from are [750 cm^3, 850 cm^3, 1100 cm^3, 1425 cm^3,2400 cm^3]

The volume of the displaced water should = the volume of the block, right? Am I not converting something correctly? Also, on the lab the prof writes "Density of water = 1gm/cm^3" I'm not sure what the unit gm/cm^3 is.. I figured he just meant to write g/cm^3.

11. Jul 21, 2011

### Aimless

The density of water is one gram per cubic centimeter, so "gm" should just mean gram.

Last step. Given the volume of the block, how do you get its density?

12. Jul 21, 2011

### Jgoshorn1

My work looks correct and I don't know any other way to do it but I'm not getting it to match (or even come really close to) any of the other answers.

Mass of block = 1.94kg = 1940g
For density of block:
d=m/v = 1940g/1020cm^3 = 1.90g/cm^3

Answers to chose from: [1.2 gm/cm^3, 1.8 gm/cm^3, 2.4 gm/cm^3, 3.5 gm/cm^3, 4.8 gm/cm^3]

So, given my previous posts; for mass of water displaced would you chose 1.1kg, for volume of block would you chose 1100cm^3, for density of block would you chose 1.8gm/cm^3???

13. Jul 21, 2011

### Aimless

"Coming close" is relative. Like I said above, my suspicion is that you have either a rounding mistake or a measurement mistake somewhere early in your work, which is throwing things off down the line. Given your initial measurements, your math looks correct.

14. Jul 21, 2011

### Jgoshorn1

Thanks!

I went back and re-did the volume and density of the block using 1.1kg as the weight of the water displaced and I got exact answers.