How many people can be on the raft?

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    Buoyancy Physics
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Homework Help Overview

The discussion revolves around a physics problem involving buoyancy and the capacity of a raft based on the volume and mass of logs, as well as the weight of people. Participants are analyzing calculations related to buoyant force and weight equilibrium.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of volume and mass of logs, questioning the use of specific variables and units. There are inquiries about the definition of density and the origin of certain numerical values in the calculations.

Discussion Status

Some participants have offered guidance on the importance of unit consistency and clarity in calculations. There is an acknowledgment of errors in the original calculations, and a few participants suggest a more methodical approach to problem-solving.

Contextual Notes

There are indications of confusion regarding the use of units and the accuracy of numerical values, as well as a suggestion to reconsider the setup of the problem. The original poster's calculations appear to lack clarity and precision, which has led to further questioning and exploration of the topic.

lgeorge
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Homework Statement
A raft is made of 13 logs lashed together. Each is 45 cm in diameter and has a length of 6.0 m. How many people can the raft hold before they start getting their feet wet, assuming the average person has a mass of 68 kg? Do not neglect the weight of the logs. Assume the specific gravity of wood is 0.60.
Relevant Equations
W = mg
Fb=(density)Vg
V = (pi)r^2(h)
density = mV
First I calculated the volume of all of the logs...

V = 13((pi)(.225)^2(g)) = 3.94

Then wanted the mass of the logs
m = density(Volume)
density = .6(997)
m = 2356

Then calculated buoyant force...

Fb = (997)(3.94)(9.8) = 3928

Then set this equal to weight of people and logs

38496 = ((number of people)(68) + 2356)(9.8)

n=23.12

23 people is wrong
 
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Where in your calculation did you use the length of the logs? Also what's the g in the equation for V?
 
Sorry that g is supposed to be a 6
 
It's hard to tell what you're doing when you put down numbers without units.
lgeorge said:
V = 13((pi)(.225)^2(g)) = 3.94
Is this the volume of all the logs together?
lgeorge said:
density = mV
density = .6(997)
What's the definition of density?
If V = 3.94 (units TBA), what does V stand for in the equation above and how did it become 997? Where did this 997 come from anyway?
 
Re check your calculation for V, I think there's an error.
 
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Very true I need to work with my units more because I found my error once I went through the calculation slower with units! Thank you so much!
 
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I strongly suggest never putting in numbers before you arrive at a final expression for what you want to compute. It will make your work clearer as well as make it easier to check for errors and to understand the functional behaviour of your result.

If you insist on putting in numbers, never do so without attaching the appropriate units!
 
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Orodruin said:
I strongly suggest never putting in numbers before you arrive at a final expression for what you want to compute. It will make your work clearer as well as make it easier to check for errors and to understand the functional behaviour of your result.
For a list of the benefits of working algebraically (any I missed?) see https://www.physicsforums.com/insights/frequently-made-errors-equation-handling/
 
haruspex said:
For a list of the benefits of working algebraically (any I missed?) see https://www.physicsforums.com/insights/frequently-made-errors-equation-handling/
I agree with the points, but the precision example is a bit unfortunate. The required ##\Delta T## is part of the input and has two significant digits at best. I’d probably deduct credit for anyone answering with more than two significant digits.
 
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Orodruin said:
I agree with the points, but the precision example is a bit unfortunate. The required ##\Delta T## is part of the input and has two significant digits at best. I’d probably deduct credit for anyone answering with more than two significant digits.
Good point, but of course it was taken from an actual thread. I'll change it to 10.00 seconds.
 
  • #11
haruspex said:
Good point, but of course it was taken from an actual thread. I'll change it to 10.00 seconds.
Then you run into other issues limiting accuracy, such as the Earth not being perfectly spherical, mountains, and how accurately you could actually measure the period. I am of course playing the devil's advocate here ... but I feel there should be possible to define a better example.
 

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