# Finding density of item in a liquid

• PascalPanther
In summary: The difference between the density of the blob and the density of water is the ratio of the weight of the blob to the weight of the water it displaces.In summary, the conversation discusses the problem of finding the density of an unknown thick syrup-like liquid without knowing its mass, volume, or density. The approach is to use Newton's second law and Archimedes principle to calculate the density in terms of the density of water. The resulting equation involves the mass of the unknown liquid and the mass of an equal volume of water, which can be solved for the density of the unknown liquid. The final density is very close to that of water, indicating that the blob is sinking slowly.
PascalPanther
I am suppose to find the density of a thick syrup-like liquid (does not disperse).
The only information I have is that when a small portion is poured into a cup of unknown volume, it takes 5 seconds for it to reach the bottom. The cup has a height of 12 cm.

I am not sure how to approach this problem. When I was doing a FBD, the only thing I can see to put on it is weight, and buoyancy, but I don't have a mass, volume, or density of the unknown heavy liquid.
The only equation I have that has enough information to fill out is:

pressure = (initial pressure) + (density)(gravity)(height)

That would give me the pressure on the unknown liquid at the bottom, which I can't see how that helps me.

With kinematics, I can find the deceleration necessary, but I can't find the buoyancy force because I don't have the mass.
How should I start this problem?

I assume the cup is full of water before you pour in the unknown liquid, and I assume also you are ignoring any drag effects. So what you have is a constant buoyant force and gravity acting on the blob of liquid. The bouyant force is related to pressure, but by Archemedes principle it is equal to the weight of a volume of water equal to the volume of the unknown liquid.

Write a Newton's second law equation for the blob

ma = F

where m is the mass of the blob, a is the constant acceleration that you calculate from the information given in the problem (5 seconds to fall 12cm starting from rest) and F is the difference between the weight of the object and the weight of the water it displaces.

Your equation will involve the mass of the blob and the mass of an equal volume of water. You do not need to know the mass of the blob or its volume numerically. If you divide each mass by the volume of the blob, the equation will become an equation relating the densities (m/V). You can solve for the density of the unknown liquid in terms of the density of water. Then use the known density of water to complete the calculation.

I'm still trying to get my head around this.

So it sounds like:
F = ma = F
(mass of unknown)*(acceleration) = ((mass of unknown)-(mass of displaced water))*g

a = 0.0096 m/s^2 (1/2at^2 = .12 m)

(density of unknown) = (mass unknown)/(volume of unknown)
(density of water) = (mass of displaced water)/(volume of unknown)
(volume of unknown) = (mass of displaced water)/(density of water)

d(water) = 1.0g/cm^3
I seem to be stuck at this point, since I don't have the mass of the displaced water. I think I did something wrong somewhere.

----
I tried just putting in numbers to see what would happen:
1.0g/cm^3 = 1.0g/1 cm^3
volume of unknown = 1 cm^3
mass of displaced water = 1.0g

(mass of unknown)*(acceleration) = ((mass of unknown)-(mass of displaced water))*g
(mass of unknown)(0.96 cm/s^2) = ((mass of unknown)-(1.0g))9.8
(0.0096 m/s^2) = (1 - (1.0g/mass of unknown))(9.8m/s^2)
(0.0096 m/s^2)/(9.8m/s^2) = (1 - (1.0g/mass of unknown))
9.8E-4 = 1 - (1.0g/(mass of unknown))
-.999 = -1.0g/(mass of unknown)
(mass of unknown) = 1.0g
Which means I just ended up with basically the exact same density as water. Which wouldn't make sense at all.
-----

The volume of the unkown is the same as the volume of the water it displaces. Call that simply volume. Divide your equation

(mass of unknown)*(acceleration) = ((mass of unknown)-(mass of displaced water))*g

by that common volume and replace the ratios (mass/volume) by densities. Using the acceleration you calculated and the density of water you will have an equation for the density of the unknown.

So when I'm dividing by the volume, this isn't a normal math operation where I have to apply it to all terms right? It only applies to the mass? I am kind of just substituting it looks like.

p(u)*a = (p(u)-p(w))*g

So I guess if that was the case, the answer is real near the same density of water. I got 1.001 kg/m^3 vs. 1.000kg/m^3 for water. Actually I think the way where I just started with the density of water to fill in mass and volume yields the exact same answer I just need to keep a lot of significant figures.

PascalPanther said:
So when I'm dividing by the volume, this isn't a normal math operation where I have to apply it to all terms right? It only applies to the mass? I am kind of just substituting it looks like.

p(u)*a = (p(u)-p(w))*g

So I guess if that was the case, the answer is real near the same density of water. I got 1.001 kg/m^3 vs. 1.000kg/m^3 for water. Actually I think the way where I just started with the density of water to fill in mass and volume yields the exact same answer I just need to keep a lot of significant figures.
It absolutely is a normal math operation. When you divide a product of factors by something, you divide the whole product once and only once. Equivalently, you can divide any one of the factors in the product before you multiply.

The blob is sinking quite slowly, so the density is very nearly the same as that of water.

## What is the formula for finding the density of an item in a liquid?

The formula for finding density is D = m/v, where D stands for density, m stands for mass, and v stands for volume.

## What units are commonly used for density measurements?

Density can be measured in a variety of units, but the most commonly used units are grams per milliliter (g/mL) for liquids and grams per cubic centimeter (g/cm3) for solids.

## How do you measure the volume of an irregularly shaped object?

To measure the volume of an irregularly shaped object, you can use the water displacement method. Submerge the object in a known volume of water and measure the change in water level. This change in volume is equal to the volume of the object.

## How does the density of a liquid affect the density of an object in that liquid?

The density of a liquid can affect the density of an object in that liquid. If the density of the liquid is greater than the density of the object, the object will float. If the density of the liquid is less than the density of the object, the object will sink.

## What factors can affect the accuracy of density measurements?

There are several factors that can affect the accuracy of density measurements, including temperature, pressure, and impurities in the liquid. It is important to control for these factors when conducting density experiments to ensure accurate results.

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