(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A raft is made of 11 logs lashed together. Each is 35 cm in diameter and has a length of 5.6 m. How many people can the raft hold before they start getting their feet wet, assuming the average person has a mass of 70 kg? Do not neglect the weight of the logs. Assume the density of wood is 600kg/m^{3}

Just want to make sure that I did my work correctly because this looks like a really big raft and I am not sure if only ten people would actually stand on such a big raft.

2. Relevant equations

F_{b}- m_{total}g = 0

Thus

F_{B}= m_{total}g

where m_{total}= mass of 11 logs + mass of unknown number of people

3. The attempt at a solution

[tex]\rho[/tex]_{water}* Volume occupied by the logs * number of logs * g = m_{people}* number of people + [tex]\rho[/tex]_{logs}* Volume occupied by the logs * number of logs

Subtracting the desnity of logs times the volume from the right hand side and moving it to the left, then factoring out the volume and the number of logs, and then dividing the whole thing on the left by the mass of the people I get the following.

[number of logs * Volume occupied by logs ([tex]\rho[/tex]_{water}- [tex]\rho[/tex]_{logs})] / mass of people = number of people

Plugging in the numbers I get

[11* ((35/100)/2)^2 * 5.6 * (1000 - 600)] / 70 = 10.78 People which means 10 people can stand on the boat.

Am I being paranoid for thinking this is wrong?

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# Homework Help: Fluid Buoyancy Check

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