Density of bar submerged in water

  • #1
128
0

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.
 

Answers and Replies

  • #3
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,798
1,670

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.
 
  • #4
PhanthomJay
Science Advisor
Homework Helper
Gold Member
7,167
507

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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