Solving a Raft Submersion Problem - Edge Needs a Second Opinion

[edge]

Okay, I've worked this problem and have what I think is the correct answer...I just want to get a second opinion because I am a little bit unsure.

A raft is constructed of wood having a density of 600 kg/m^3. It's surface area is 5.8 m^2, and its volume is .60 m^3.

- A 80 kg man jumps on the raft. After a while, the man + raft system becomes stable. What is the % of the raft submerged?

My work:

I assume that at equilibrium the weight of the raft + man must equal the buoyant force.

I use mass = (density)*(volume) to get the mass of the raft (360 kg) then add 80 kg to get the new mass, 440 kg. Now I take that mass times gravity to get the weight. And set that equal to the buoyant force of (density of water)*(volume displaced)*(gravity). It looks like this:

(440)g = (1000)(V)(g)
so 440/1000 = v
that gives me .44 as the displaced volume of water.
This is where my question arises. Is 44% the amount of the raft submerged? Or do I then need to divide the displaced volume by the volume of the raft (.44/.6) to get .73 or 73%?

If I've worked it wrong please let me know but I think I've done it correctly...I'm just unsure which of the 2 answers is the correct one.

Thanks!
-edge

 

[NateTG]

It's asking for the percentage of the raft that is submerged, not the volume, so the answer is 73%.

If you keep track of units, you can check because percentages should always be unit-less.

 

[edge]

Awesome, thanks. Silly me, simple errors...they kill me. Thanks for the help!