Density of bar submerged in water

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SUMMARY

The discussion centers on calculating the upthrust and density of a bar submerged in water, weighing 1.75 N in air and 1.4 N in water. The upthrust is correctly calculated as 0.35 N, but the density calculation is flawed due to incorrect application of Archimedes' Principle. The density of the bar cannot be derived from weight alone; mass must be used instead. The correct approach involves understanding that the buoyant force equals the weight of the water displaced, not the weight of the bar.

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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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