# Density of an Object

1. Feb 9, 2008

### 7starmantis

An object weighs 15N in air and 13N when submerged in water. determine the density of the object.

Am I setting this up correctly, I'm confused about the two masses.

Volume = W(air) - W(water) / pg
15-13 / (1000)(9.8) = 2.0x10^4
p(density) = m/V
What mass do I use here though, the one in air or water?

Last edited: Feb 9, 2008
2. Feb 9, 2008

### BishopUser

I don't know whether your solution is correct, but I can tell you that mass and weight are not the same thing. The object's mass is the same no matter where it is

3. Feb 9, 2008

### 7starmantis

How do I go about finding the density? Don't I need to find volume first then use volume in the density equation?

4. Feb 9, 2008

### John Creighto

Weight in air is:

m*g - volume of the object* density of the air* g

weight in watter is

m*g - volume of the object* density of water*g

You already solved for the volume now solve for the mas and then apply your density formula.

5. Feb 9, 2008

### 7starmantis

I'm confused, that still leaves me with two separate values for mass. Which value would I use in the density equation?

6. Feb 9, 2008

### HallsofIvy

Staff Emeritus
How does that leave you with "two separate values for mass"? John Creighto told you to set the two quantities equal and solve for mass. There will only be one solution.

7. Feb 9, 2008

### 7starmantis

I didn't get that he said to set them equal to each other, but I also don't know the density of the air.

So:
m*g - volume of the object* density of the air* g = m*g - volume of the object* density of water*g

I have everything except density of air.

8. Feb 9, 2008

### mgb_phys

Generally it is small enough to ignore (it is roughly 1/1000 the density of water )
If you look at the equation it is an additive term so calculate your answer ignoring it and then calculate the answer again using 1.3kg/m^3 and you will see that the difference in the mass of the object is small.

In general you can ignore small terms when they are added/subtracted but yuo need them when they are multiplied/divided - can you see why?

9. Feb 9, 2008

### 7starmantis

I'm so frustrated by this problem I can't even think straight anymore.
I don't understand what you mean by it being an additive term, isn't it multiplied in the equation? But at this point I can't even figure out how to solve the equation with or without it.

10. Feb 9, 2008

### 7starmantis

We did a lab like this, but my teacher hasn't returned our lab reports yet and it has all my data in it. I just can't for the life of me understand how to setup the problem.

11. Feb 9, 2008

### mgb_phys

Not quite set them equal, you are given two differnet weights
m*g - volume of the object* density of the air* g = 15N
m*g - volume of the object* density of water*g = 13N

Lets just look at the equation:
m*g - V*air*g = 15N
m*g - V* water*g = 13N

Since density of air is small then anything multiplied by it is small, so the term V*air*g is small. This is added to the m*g, this is what is meant by an additive term. Adding a small number to a big one doesn't do much so you can ignore it.

m*g = 15N
m*g - V* water*g = 13N

or 15N - V*water*g = 13N, so V*water*g = 2N
You know density of water so can work out the volume.
Once you have the volume and you know it's weight in air you can work out it's density, once again ignoring the density of air.

It's easy to get confused once you start writing the equations, it's important to have a clear picture of what is happening first.
Draw a diagram showing all the forces. And finally do a quick check that the units balance,
V*water*g = 2N is m^3 * kg/m^3 * m/s^2 = kg m/s^2 = N

Last edited: Feb 9, 2008
12. Feb 9, 2008

### Kurdt

Staff Emeritus
Considering you're looking for the density it might be better to use that term in your equations.

Weight in air:

$$V_o (\rho_o-\rho_a)g = 15N$$

Weight in water:

$$V_o(\rho_o - \rho_w)g = 13N$$

Then one has two variables and two simultaneous equations. You can solve for Vo from one of the equations and substitute back into the second to find the final answer. As others have said you can probably ignore the density of air.

13. Feb 9, 2008

### 7starmantis

So basically since$$\rho$$gv = 2N
Volume = 2N / $$\rho$$g
Volume = 2.0X10^-4

Then the weight in air is 15N / 9.8 = 1.53kg

$$\rho$$ object = m/v
= 1.53kg/2.0X10^-4
= 7650

That make sense? I hope so, cause it does to me.... :)

14. Feb 9, 2008

### Kurdt

Staff Emeritus