# Calculating the mass of a beam of non uniform density.

1. Nov 7, 2012

### Craptola

Been doing some physics problems from my mechanics class. This is the first time I've attempted a problem like this and am not so confident with solving it, I could be correct and just being paranoid but I just have a feeling I've made a mistake somewhere, would appreciate if someone who is comfortable with this kind of thing would tell me if/where I'm going wrong.

1. The problem statement, all variables and given/known data
Consider a thin beam of length L=3 m, and height and width equal to w= 2 cm and h=2 cm respectively. Assume that the composure of the beam is from a mixture of materials, that have given it a non-uniform mass density, which varies continuously along x [where x represents the coordinate for the direction along the length of the beam, measured from the left edge of the beam]. Assume that such mass density as a function of x is given by:
$$\rho = \rho _{0}e^{\alpha x}$$
where ρo= 9.10^3 kg/m3 and α=1/L
Find the total mass for the beam.

2. Relevant equations

$$M=\int_{0}^{3}dm$$

$$dm=\rho dv$$

3. The attempt at a solution

$$dv=0.0004dx$$
$$\therefore dm=0.0004\rho dx = 3.6e^{-\frac{1}{3}x}dx$$
$$M=\int_{0}^{3}dm = 3.6\int_{0}^{3}e^{-\frac{1}{3}x}dx$$
$$= -10.8\left ( e^{-1} - 1 \right ) = 6.83kg$$

2. Nov 7, 2012

### haruspex

It is a good idea to work with the algebra until you have the final formula and only then substitute in the values of the known constants. You're less likely to make a mistake, it's easier to spot where a mistake has been made (e.g. by considering dimensionality) and it's easier for others to follow.
Where did the minus sign come from? Did you omit that in the formula for density?

Otherwise, looks fine.

3. Nov 7, 2012

### Craptola

Yeah, that's a typo, thanks for the help.