# Basic measurements and conversion not so basic

1. Sep 18, 2011

### Saterial

1. The problem statement, all variables and given/known data
During heavy rain, a section of a mountainside measuring 3.6 km horizontally, 0.53 km up along the slope, and 1.1 m deep slips into a valley in a mud slide. Assume that the mud ends up uniformly distributed over a surface area of the valley measuring 1.2 km x 1.2 km and that the mass of a cubic meter of mud is 1900 kg. What is the mass of the mud sitting above a 2.2 m2 area of the valley floor?

2. Relevant equations
v = lwh?

3. The attempt at a solution

To be honest, I can't necessarily figure out where to start. Looking at the question, I have a length, width and height... can I assume that the slope is a perfect vertical(lol) and use that as a length to find volume? This question seems really basic to me, the answer is looking for the mass of the mud sitting above a specific height, however, it states that a cubic meter of mud is 1900 kg.

Can I simply just multiply 1900 kg/m^2 by 2.2 m^2 to obtain 4180 kg ? I can't figure out why the other supplied information would be of any help at all. If I think about this logically, if there is mud stacked on top itself at a higher height, would that not change the mass ? This is where I thought maybe the dimensions of the mountain side were necessary.

Can anyone shed some light on this problem?

Thanks

2. Sep 19, 2011

### rickz02

No you can't, this is where you're gonna use the dimesions of the mountainside. Also take note that
I believe you simply can't do that, you are given with a volumetric density 1900 kg/m3.

Obviously the mass would. Aside from the fact you are adding mud, the mountainside is also sloping so the mass from a lower layer is lesser than that of the above.

I suggest you draw a diagram so you can easily see the whole problem.

Honestly the question of this problem is quite tricky. It says
Where exactly is 2.2 m2 area of the valley floor, is it near the slope?

3. Sep 19, 2011

### rickz02

No you can't, this is where you're gonna use the dimesions of the mountainside. Also take note that
I believe you simply can't do that, you are given with a volumetric density 1900 kg/m3.

Obviously the mass would. Aside from the fact you are adding mud, the mountainside is also sloping so the mass from a lower layer is lesser than that of the above.

I suggest you draw a diagram so you can easily see the whole problem.

Honestly the question of this problem is quite tricky. It says
Where exactly is 2.2 m2 area of the valley floor, is it near the slope? I guess it is...

4. Sep 19, 2011

### Saterial

I believe that the 2.2 m^2 is after the mudslide and off the slope, so it becomes level area in the valley.

How can the dimensions of the mountain be used? If the slope does contain a curvature, how can the 0.53 km even be accounted for?

The mud is uniformly distributed over a 1.44 km^2 (or 1440 m^2) area. Would that mean I need to determine the volume of mud in a 2.2 m^2 area and multiply that by 1900 kg? If so, can I assume no curvature in the dimensions and do the following?

V = (3.6 km)(0.53 km)(0.0011 km)
= 2.10 x 10^-3 km^3
= 2100000 m^3

Take that volume and use it as the total volume of the mudslide, determine how much of that volume is enclosed in a 2.2 m^2 area, and multiply it by 1900 kg?