How Many People Can a Raft Hold While Keeping Feet Dry?

[chantalprince]

Homework Statement

A raft is made of 11 logs lashed together. They are 38.0 cm in diameter and 6.10 m long and has a density of 700 kg/m^3. How many 60 kg persons can the raft hold in freshwater while keeping everybody's feet dry?

Homework Equations

Volume immersed/Volume total = density obj/ density fluid
% immersed= density obj/ density fluid
buoyant force = roh x V x g
P = F/A (?)
m = roh x V

The Attempt at a Solution

I have found the volume of the raft: 2.42 cubic meters

Here are my thoughts: To keep everybody's feet dry, the raft can be immersed up to 99.9%. Not sure how to use the buoyant above to my advantage... Also, I can find the buoyant force of the raft. Now, I'm stuck!

Thanks in advance to all who respond :)

 

[LowlyPion]

Think of the volume of the logs as your budget. Figure the weight of the water displaced by that volume.

The first outlay from the budget is for the weight of the wood.

What's left over - volume times the density of that volume of water - is what you have in the bank to pay out then for additional weight.

So ... how many 60 kgs can you afford from your budget?

 

[chantalprince]

I can't quite wrap my mind around what your saying. However, if it's of any help to get me started, I think I found the weight of water displaced by the raft. I used: m = roh x V then used W = mg. Is this correct?

 

[LowlyPion]

I can't quite wrap my mind around what your saying. However, if it's of any help to get me started, I think I found the weight of water displaced by the raft. I used: m = roh x V then used W = mg. Is this correct?

That's the idea. Now you know the total of how much the logs and the people can weigh.

 

[chantalprince]

But I thought I just found the weight of the raft with the equations in my last reply... How do I find out the TOTAL weight I'm allowed??

 

[LowlyPion]

But I thought I just found the weight of the raft with the equations in my last reply... How do I find out the TOTAL weight I'm allowed??

That's the whole point. The weight of the displaced water is what the logs will support. (After allowing for the weight of the logs themselves of course.)

 

[LowlyPion]

Maybe this will help?
http://hyperphysics.phy-astr.gsu.edu/hbase/pbuoy.html#buoy

 

[chantalprince]

Alright- so, I found the buoyant force, and then the weight of the raft, so the rest of the weight can go to people. So, what does the whole, "and not get their feet wet" bit have to do with this? Does it just mean that if we go over our "account balance" that the raft will sink? So, the raft will float up until the next person gets on?

 

[LowlyPion]

Yes. They don't want you getting into the density of the people and the volume they displace.

 

[chantalprince]

Ok :) Now another issue I am having. I took the force the raft can support, which is in Newtons, as is the weight of the raft. So, how do I figure how many 60 kg persons this thing will hold? I thought it would be a simple conversion, but I am missing something somewhere. I did this:

(1 lb/4.448 N) (1kg / 2.2lb)

But- I still have Newtons on top.

(Can you tell I SUCK at this physics stuff?)

 

[LowlyPion]

How many Newtons is a 60 kg person?

Hmmmm... F = m*g = 60*9.8 = 588 N

 

[chantalprince]

Thank you so much for all of your help, LP