Density of bar submerged in water

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Homework Help Overview

The problem involves a bar suspended in air and its weight when submerged in water, with the goal of calculating the upthrust and the density of the bar. The subject area relates to fluid mechanics and buoyancy principles.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculation of upthrust based on the difference in weight of the bar in air and water. There are questions regarding the application of Archimedes' Principle and the correct interpretation of weight versus mass in the context of density calculations.

Discussion Status

Some participants have pointed out potential misunderstandings in the calculations, particularly regarding the use of weight in the density formula and the application of buoyant force. There is ongoing exploration of the correct approach to determine the density of the bar and the implications of the calculations presented.

Contextual Notes

Participants note the assumption of the density of water and the need to clarify the relationship between weight and mass in the context of the problem. There is a recognition that the calculations may not align with expected physical behavior, such as the bar's buoyancy.

Roodles01
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Homework Statement


A bar suspended in air weighs 1.75 N
The same bar weigs only 1.4 N when suspended in water.
Calculate;
a. upthrust
b. density of the bar

Homework Equations


D = m/v
F = m*a

The Attempt at a Solution


a[/B].
Upthrust is 1.75 N - 1.4 N = 0.35 N

b.
Assuming density of water is 1x103 kg m-3
and weight of bar displaces that amount of water.
So
volume 1.75 N of water = 1.75 N / (1x103 * g) = 1.78 m3

Density of bar = m/v = 1.75 N / 1.78 m3 = 0.98 N m-3

which has to be wrong, surely, or the bar would float . . .
Help!
Thank you.
 
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Remember the difference between weight and mass.
 
Roodles01 said:

Homework Statement


A bar suspended in air weighs 1.75 N
The same bar weigs only 1.4 N when suspended in water.
Calculate;
a. upthrust
b. density of the bar

Homework Equations


D = m/v
F = m*a

The Attempt at a Solution


a[/B].
Upthrust is 1.75 N - 1.4 N = 0.35 N

b.
Assuming density of water is 1x103 kg m-3
and weight of bar displaces that amount of water.
So
volume 1.75 N of water = 1.75 N / (1x103 * g) = 1.78 m3

It's not clear how you got 1.75 N / (9810) = 1.78 m3

Your calculator must have suffered a nervous breakdown.

Remember, a cubic meter of water weighs a tonne (1000 kg), literally.

Density of bar = m/v = 1.75 N / 1.78 m3 = 0.98 N m-3

which has to be wrong, surely, or the bar would float . . .
Help!
Thank you.

As Dr. Courtney pointed out, density is mass divided by volume, not weight divided by volume. You can't use Newtons and kilograms interchangeably.
 
Roodles01 said:

Homework Statement


A bar suspended in air weighs 1.75 N
The same bar weigs only 1.4 N when suspended in water.
Calculate;
a. upthrust
b. density of the bar

Homework Equations


D = m/v
F = m*a

The Attempt at a Solution


a[/B].
Upthrust is 1.75 N - 1.4 N = 0.35 N
yes, and this called the buoyant force
b.
Assuming density of water is 1x103 kg m-3
and weight of bar displaces that amount of water
So
volume 1.75 N of water = 1.75 N / (1x103 * g) = 1.78 m3
you are not applying Archimedes Principle correctly. The upward thrust or buoyant force is equal to the weight of water displaced. That weight of displaced water is not 1.75 N.
 

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