Density of bar submerged in water

In summary: It is equal to the weight of the bar in water - 1.4 N. Now you can use that weight of water to find the volume of water that weight represents using ρ = m/V. Once you know the volume of water displaced, then you can use the definition of density to find the density of the bar.
  • #1
Roodles01
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.
 
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  • #2
Remember the difference between weight and mass.
 
  • #3
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.
 
  • #4
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.
 

What is the definition of density?

Density is a measure of the amount of mass per unit volume of a substance. It is typically expressed in units of grams per cubic centimeter (g/cm^3) or kilograms per cubic meter (kg/m^3).

How is density calculated?

Density is calculated by dividing the mass of an object by its volume. The formula for density is D = m/V, where D is density, m is mass, and V is volume.

Why is the density of a bar submerged in water important?

The density of a bar submerged in water is important because it can help determine the material of the bar. If the bar is less dense than water, it will float. If it is more dense, it will sink. This can also be used to determine the purity of a substance, as impurities will affect the density.

What is the relationship between density and buoyancy?

The density of an object determines its buoyancy, or ability to float in a liquid. Objects with a higher density than the liquid they are in will sink, while objects with a lower density will float. This is why some objects, such as wood, can float on water while others, like rocks, sink.

How does temperature affect the density of a bar submerged in water?

In most cases, as the temperature of water increases, its density decreases. This means that a bar submerged in warmer water will experience less buoyant force and may sink. However, there are some exceptions to this, such as when water reaches its maximum density at around 4°C. In this case, a bar may float in colder water and sink in warmer water.

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