How to find the surface density for a given linear density

  • #1
Given a square with a linear mass density of:

λ(x) = a * x (see image below where black is high density and white is low density)

200px-XO-4-test-display-5-black-vs-white-horizontal-gradient.png

how would you deduce what the surface mass density is?

I get confused for the following reason:

To me it seems that the surface mass density should depend on x, because the further left or right you go the linear density changes, so why wouldn't the surface density change as well. A surface mass density must have units of $$\frac{m}{L^2}$$ so since λ(x) has units of $$\frac{m}{L}$$ I would think I need to divide λ(x) by some length. I don't think you would divide by y (the height coordinate) because the density is translationally invariant in y. But that means you have to divide by x to get the right units, and that would mean that the surface mass density is a constant, which doesn't make physical sense to me.
 

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  • #2
jbriggs444
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I don't think you would divide by y (the height coordinate) because the density is translationally invariant in y.
Seems to me that's pretty much why you should divide by the linear mass density by y. The linear mass element (a point) is getting smeared up and down a surface mass element (a vertical line) of length y.
 
  • #3
Ibix
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If I told you that the total mass of a square sheet was m and that it had uniform areal density, how would you get that density?
 
  • #4
If I told you that the total mass of a square sheet was m and that it had uniform areal density, how would you get that density?
I would divide the total mass m by the total area A:
$$\sigma = \frac{m_{tot}}{A}$$
 
  • #5
Ibix
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I would divide the total mass m by the total area A:
$$\sigma = \frac{m_{tot}}{A}$$
And the area is the square of the side length, ##A=L^2##. But now you've got a 1d density variation. What's the appropriate length you were looking for?
 
  • #6
And the area is the square of the side length. But now you've got a 1d density variation. What's the appropriate length you were looking for?
Is this what you mean:

$$\sigma = \frac{m_{tot}}{L^2}$$

but the total mass is equal to the masses of the infinitely thin lines that you stack up together to create a square:

$$\frac{\int_0^L {\lambda(x) dx} * L}{L^2} = \frac{\int_0^L {a*x dx} * L}{L^2} = \frac{a *\frac{L^2}{2}* L}{L^2} = \frac{aL}{2} $$
 
  • #7
Ibix
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That seems over-complex. My point was that you were correct to say you need to find a length to divide by.

In the simple uniform density case the linear density is ##m/L##. The areal density is ##m/L^2##. What is the length I divided by to get the areal density? Given that your density is uniform in the y direction, should your approach be different?

Another way to look at it is that the total mass is ##m=\iint\sigma(x,y)dxdy##. But you know that ##\sigma(x,y)=\lambda(x)/l##, where ##l## is this length that's confusing you. So you have ##m=\iint(\lambda(x)/l)dxdy##. But from the definition of linear density you know that ##m=\int\lambda(x)dx##. What's ##l##?
 
  • #8
That seems over-complex. My point was that you were correct to say you need to find a length to divide by.

In the simple uniform density case the linear density is ##m/L##. The areal density is ##m/L^2##. What is the length I divided by to get the areal density? Given that your density is uniform in the y direction, should your approach be different?

Another way to look at it is that the total mass is ##m=\iint\sigma(x,y)dxdy##. But you know that ##\sigma(x,y)=\lambda(x)/l##, where ##l## is this length that's confusing you. So you have ##m=\iint(\lambda(x)/l)dxdy##. But from the definition of linear density you know that ##m=\int\lambda(x)dx##. What's ##l##?
In the uniform density case you had $$\lambda = \frac{m}{L}$$ where in this case L is the distance in x. Then you divide by the y coordinate to get the area density.

So then it looks like it should be l = y (or maybe l = dy because you have $$\iint (\lambda(x) / l) dxdy = \int \lambda(x) dx$$)
 

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