Question

A rectangular object has a width of 40 meters, height of 15 meters, and length of 2 meters. It floats consistently when 3 meters of its height is below the surface of the water.

1. Find the volume of the displaced water.
2. How much is the buoyant force on the object?
3. What is the object's weight?
4. What is the mass density of the block?

Answer 1

Is the volume of the displaced water equal to 3*40*2, or 240?

Answer 2

And I got 240*1000*9.8 = 2352000 N for the buoyant force

Answer 3

I'm not sure how to get the third and fourth parts.

Answer 4

Let me ask you a silly question, what is 2352000 N ?

Answer 5

2352000 Newtons, the buoyant force

Answer 6

Density = mass/volume, right? But how do I get the mass with the given information @ganeshie8 ?

Answer 7

use this

weight = mass*g

Answer 8

I made a silly mistake in part 3. Let's start over..

According to archimedes principle, that must equal the weight of the `COMPLETE object`, yes ?

Answer 9

But that calculation was just for the part that was submerged...what about the rest of the object?

Answer 10

No, we were wrong about our inerpretation of the archimedes principle earlier. Here is the correct interpretation :

The weight of the water displaced equals the weight of the COMPLETE object.

Answer 11

Let me delete our earlier incorrect discussion, its so embarassing lol

Answer 12

So...what do we do next :)

Answer 13

So... what's the answer for part 3 ?

Answer 14

2352000 N?

Answer 15

yes, its same as the part 2 :

`Weight of displaced fluid = buoyant force on the object = Weight of the COMPLETE object`

Answer 16

I'm going to use that number in the mass density calculation.

mass = weight/g
mass = 2352000/9.8
m = 240000

density = mass/volume
density = 240000/1200
d = 200

Answer 17

That looks good to me..

Answer 18

Is it right that I divided by 1200 instead of 240?

Answer 19

240000 is the mass of the complete object
1200 is the volume of the complete object

To get the density, we use the number 1200 yea

Answer 20

http://physics.weber.edu/carroll/archimedes/principle.htm

Answer 21

Ok...one last question

5. How would the amount of the object under the surface change if it were floating in glycerin?

I'm thinking that I should use the density of glycerin instead of water in the calculation for amount of displaced water. Is this right?

Answer 22

Yes, density of glycerin is greater than the density of water. So I'd guess the height of the submerged part of the object decreases ?

Answer 23

Thank you very much for all the help!

Answer 24

Np :)