How much ice to float a 80 Kg person

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Homework Help Overview

The problem involves determining the volume of ice required to float an 80 kg person in salt water, taking into account the densities of salt water and ice. The original poster presents an equation related to buoyancy and attempts to calculate the necessary volume of ice.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the balance of forces, including the weight of the person and the buoyant force provided by the displaced water. There are attempts to incorporate the weight of the ice into the calculations, with some questioning how to correctly account for it in the equations.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and the equations involved. Some have provided insights into the necessity of including the weight of the ice, while others are questioning the assumptions made regarding the densities and the setup of the equations.

Contextual Notes

There is a mention of the specific densities of salt water and ice, and participants are considering whether the ice should be treated as salty, which introduces additional complexity to the problem. The original poster's approach raises questions about the correct application of buoyancy principles.

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Homework Statement



Find the volume of the smallest block of ice which when floating on salt water will carry a person of mass 80 Kg

Densities:

Salt Water = 1025
Ice = 917

Homework Equations



F_b=P_f * V_f * G

The Attempt at a Solution



I figured F_b would have to be equal to the force of the person down which is equal to 784 N. Then plugging that into the equation and solving for V_f I get .078. I used the Salt Water for P_f. Is there somewhere that I need to plug in Ice?
 
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Don't forget the weight of the ice...:wink:
 
ok so I tried to add the weight of ice in here is what I get.

like before I have the equation.

W_p * G = 784

This is the force of the pearson. Next I have

784 = Volume of the ice * Density of water * Gravity

Then to add in the weight of ice I do this.

Volume of the ice * Density of Ice * Gravity + 784 = Volume of ice * density of water * gravity

This makes an equation has everything cancel out.
What am I doing wrong?
 
iceman_ch said:
ok so I tried to add the weight of ice in here is what I get.

like before I have the equation.

W_p * G = 784

This is the force of the pearson. Next I have

784 = Volume of the ice * Density of water * Gravity

Then to add in the weight of ice I do this.

Volume of the ice * Density of Ice * Gravity + 784 = Volume of ice * density of water * gravity

This makes an equation has everything cancel out.
What am I doing wrong?

(vol ice*(density ice)+m)g=volume ice(density SALT water)*g Now the question is ice salty? I think not. On the left hand side we have the total weight of the ice plus man, on the right the weight of water displaced.
 

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