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Density: Example

by Jared Rovny
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    00:01 Let’s do a quick example before we move any forward. This discusses the density of a person as well as the specific gravity. So, suppose a person has 50 kilograms of mass with a specific gravity of 1, 15 kilograms of bone mass with a specific gravity of 1.5, a little heavier, and then 10 kilograms of mass with a specific gravity of 2.0, much heavier. The question might be, what is the total specific gravity and then what is their density. Using the definitions of density and specific gravity that we’ve already talked about, go ahead and give this a shot. See if you can do it using the definitions we’ve introduced so far. Then we’ll try it here. So, for this problem, we have three different kinds of mass that the person has. We have 50 kilograms and a specific gravity for this is 1.0. We have 15 kilograms with a specific gravity of 1.5. Then we have 10 kilograms with a specific gravity of 2. The question is what is the total specific gravity first? So, just calling this total specific gravity, S. If we’re going to try to solve for it, we know from our definition that this is equal to the density of the person divided by the density of water. This is the written way of writing the letter ρ (rho), the Greek letter which stands for density in our equations. This is equal to 1 over the density of water times the density of the person.

    01:25 So, what is that density? Well, density is defined as mass divided by volume; so, the mass of the person divided by the volume of the person. The mass of the person, we know from this problem because what we have is m1 + m2 + m3, these three different kinds of masses given here, here, and here on the left. The total volume of the person is also something that we could figure out.

    01:53 But it’s not given to us in this problem. So, this is where the real trick of the problem is.

    01:58 What are these volumes? So, let’s see if we can find those volumes. Let’s resort to the definition of the specific gravity. The specific gravity is defined to be the density of an object over the density of water. So, for the density of an object, we have that as the mass per unit volume of that object divided by the density of water. So, if we know the specific gravity and we also know the density of water and the masses of our objects, we can solve for the volume.

    02:28 So doing that, we have the volume is equal to the mass divided by specific gravity of the object times the density of water. So, this expression for the volume will be something that we can use over here where we’re not sure what the volumes are. So, let’s go ahead and do that.

    02:45 In this problem, we have 1 over the density of water times these three masses, m1, m2, and m3.

    02:52 Now, we are going to divide by these new expressions that we found for volume.

    02:59 So, this will be m1 over the specific gravity of the first kind of substance times the density of water, plus m2 times its specific gravity times the density of water, plus m3 divided by its specific gravity also times the density of water. So, in this case, it’s not so tricky. All we can do is maybe pull out this density of water. Let’s first factor that. If we factor out this density of water, notice that it is in the denominator of a denominator. When that’s true, it’s in the numerator.

    03:31 So, if we factor that out, it would in fact go like this. Density of water divided by the density of water upfront where this top term density of water came from these lower terms, again, being in the denominators of the denominators. Then we just have the sum of the masses, m1 + m2 + m3, which is something we know, 15 + 15 + 10 divided by the ratios of each of the masses to their specific gravities. So, we have 50 over its specific gravity + 15 over its specific gravity + 10 divided by its specific gravity of 2. Nicely, this cancel and we’re almost done.

    04:12 We have 50 + 15 + 10 is 75, divided by 50 divided by 1 is 50, plus 15 divided by 1.5 is 10, plus 10 divided by 2 is 5. So, we have an answer of 75 divided by 65. This is equal to about 1.15.

    04:37 This is our specific gravity. The second part of this question asks about the density of the person.

    04:43 But since we have the specific gravity, this is quite easy. The density will be equal to the specific gravity times the density of water. This comes just from the definition of the specific gravity as the density of the object, in this case a person, divided by the density of water.

    04:57 We can always multiple both sides by the density of water to find the density of our object.

    05:01 So, doing that, we know that the specific gravity of our object first is 1.15 and then the density of water is something that we would know and that is 1000 kilograms per cubic meter. This comes up quite often, so I would be aware of this quantity because we use this number quite frequently.

    05:23 The other number that you see for this is measured in grams per cubic centimeter. So, let’s very quickly see how that works since we’ll be using it over and over again in the future. We can change the units of these by saying 1 kilogram is 1000 grams. Then saying there are 100 centimeters in 1 meter. Since this meter is cubed, we actually have to cube this quantity to make sure the units cancel up properly. So, in the numerator, we will have 1000. In the denominator, we will have two zeroes to the third power which is six zeroes.

    05:58 So, this gives us six zeroes in the denominator. So, we have 1000. Kilograms cancel with kilograms Meters cubed cancel with meters cubed. Then we have grams per cubic centimeter. The numbers, as I said, that we’re multiplying are 1000 grams times 1X10 to the two which is this, 10 to the two.

    06:20 So, the third power will be 10 to the 6th. So, we have 10 to the 6th down here. So now, we have one, two, three zeroes, one, two, three zeroes, six zeroes competing against six zeroes.

    06:30 Our answer is 1 gram per cubic centimeter. So, this is another expression for the density of water.

    06:37 So, we’ll also be using this one as well as this value of 1000 kilograms per cubic meter. So in final, the density of our person is just 1.15 times 1000, so 1150 kilograms per cubic meter, where if they were giving you options in terms of grams per cubic centimeter, which you’ll have to be careful to look out for, we might instead get an answer of 1.15 times 1 gram per cubic centimeter, in which case we can measure the density of the person in this way instead, 1.15 grams per cubic centimeter.

    07:14 So, be careful about these options because sometimes you’ll be given the mass in terms of one quantity. Sometimes you’ll be given the mass with the density, in this case in terms of another quantity. So, be careful to look at those options and look at the units that comes with each option to make sure you properly assess which kind of density you’re referring to.


    About the Lecture

    The lecture Density: Example by Jared Rovny is from the course Fluids.


    Included Quiz Questions

    1. Mass per unit volume.
    2. Amount of matter per unit volume.
    3. Mass / Strength of the Gravitational Field.
    4. Weight per unit volume.
    5. Mass times volume.
    1. N/dm3.
    2. g/cm3.
    3. g/ml.
    4. kg/m3.
    5. kg/L.
    1. 75,600 gms.
    2. 75,600 kg.
    3. 7.556 kg.
    4. 75,600 mg.
    5. 75,600 Newtons.
    1. 1.1gm/cm3, No it would not float.
    2. 1.1gm/cm3, Yes it would float.
    3. 100 gm/cm3,Yes it would float.
    4. 100 gm/cm3, No it would not float.
    5. 1.1gm/ml, No it would not float.

    Author of lecture Density: Example

     Jared Rovny

    Jared Rovny


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