# Thread: linear mass density?

1. ## linear mass density?

If I need to find the work in lifting a chain and I am given linear mass density, how would I go about it? I am used to being given simply the density.

I have a feeling I would need to integrate the linear mass density to find *just* the mass? And then integrate just 'y' with limits to account for how high I would be lifting a chain off the ground multiplied by mass*acceleration due to gravity? But I'm not sure. If someone could help explain what linear mass density is and how I would use it to solve for the the work in lifting the chain? Hopefully I'm on the correct track?

Sorry it's vague, I don't want the problem solved for me so I left it out. I am hoping someone could simply help in explaining linear mass density and give me an idea as to if I am headed in the right direction?

2. ## Re: linear mass density?

Originally Posted by cathrinbleu
If I need to find the work in lifting a chain and I am given linear mass density, how would I go about it? I am used to being given simply the density.

I have a feeling I would need to integrate the linear mass density to find *just* the mass? And then integrate just 'y' with limits to account for how high I would be lifting a chain off the ground multiplied by mass*acceleration due to gravity? But I'm not sure. If someone could help explain what linear mass density is and how I would use it to solve for the the work in lifting the chain? Hopefully I'm on the correct track?

Sorry it's vague, I don't want the problem solved for me so I left it out. I am hoping someone could simply help in explaining linear mass density and give me an idea as to if I am headed in the right direction?
Linear mass density is similar to "volume mass density" which is what we usually simply call "density." Linear mass density is the same, only over a length. The relevant formulas are:
$\rho = \frac{M}{V}$ <--volume mass density

$\lambda = \frac{M}{L}$ <--linear mass density, where L = length

As far as the chain is concerned consider how much mass is currently being lifted and use that to find the mass lifted (as opposed to the mass lying on the floor) as a function of time.

-Dan

3. ## Re: linear mass density?

Originally Posted by cathrinbleu
If I need to find the work in lifting a chain and I am given linear mass density, how would I go about it? I am used to being given simply the density.

I have a feeling I would need to integrate the linear mass density to find *just* the mass? And then integrate just 'y' with limits to account for how high I would be lifting a chain off the ground multiplied by mass*acceleration due to gravity? But I'm not sure. If someone could help explain what linear mass density is and how I would use it to solve for the the work in lifting the chain? Hopefully I'm on the correct track?

Sorry it's vague, I don't want the problem solved for me so I left it out. I am hoping someone could simply help in explaining linear mass density and give me an idea as to if I am headed in the right direction?
"Linear density" is just the one dimensional analog of the physicist's density. If a material object has constant density, you find its mass by multiplying density by volume. With variable density, integrate the density function over the volume. In "one dimension", where "length" is the only dimension, we find the "mass" by multiplying the linear density by the length. (The corresponding idea in two dimensions is "areal density".)
But finding the work done in lifting a chain is a little more complicated than just finding the mass because you don't have to lift the entire chain the same distance. Instead imagine a small "piece" of the chain, of length " $\Delta x$" at height x above the floor. Taking $\gmma(x)$ to be the density function, it has mass $\gamma(x)\Delta X$, so weight $g\gamma\Delta x$ and, lifting it to height h above the floor, it must be lifted a distance h- x so that work $(h- x)g\gamma(x)\Delta x$. The total work would be, approximately, the sum of the work done in lifting all the different "pieces" of the chain: $\sum (h- x)g\gamma(x)\Delta x$.

That is a "Riemann sum" and we can make it exact by converting to the corresponding integral: $\int_0^h (h- x)g\gamma(x)dx$.