Mass in Given Area: 820m x 820m

[Ivan92]

Homework Statement

During heavy rain, a section of a mountainside measuring 2900 m horizontally, 1000 m up along the slope, and 2.6 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 820 m x 820 m and that the mass of a cubic meter of mud is 1900 kg. What is the mass (in kg) of the mud sitting above a 2.2 m^2 area of the valley floor?

2. Relevant material
I am assuming this is a triangular prism, so I am going to use the formula for volume:
V=Al=1/2(b)(h)(l)

I also got a hint: Convert km to m (all ready did that). Find the volume in m^3. Spread the volume over the given area. Find the depth. Find the volume over the given area. What is the corresponding mass?

The Attempt at a Solution

I don't even know if I am doing the volume right. The wording in the problem is confusing me. So I let b = 2900 m, h = 1000 m, and l = 2.6. Next part is to find m^3. So:

V= (1/2)(2900)(1000)(2.6) = (3.77*10^6) m^3

Next thing it says to do is to find the depth using given area. Given area = 6724 m^2 so:

(3.77*10^6) = (6724) (l); l = 560 m

Then to use that depth and the 2nd area to solve find m^3

V= (2.2) (560) = 1232 m^3.

Then to find the mass corresponding to the volume. I am kind of stumped at this step. I am not even sure if the work I did is even right. Can someone guide me in the right direction? Thanks in advance.

 

[cepheid]

Then to find the mass corresponding to the volume. I am kind of stumped at this step.

Why? You have been given the density of mud:

Assume [...]that the mass of a cubic meter of mud is 1900 kg.

I am not even sure if the work I did is even right. Can someone guide me in the right direction? Thanks in advance.

Yeah, I didn't check your geometry or arithmetic, sorry.

 

[Ivan92]

Oh I didn't notice that, now I am a bit more lost. The answer is to find the mass corresponding to the given area. I just need to know if I am going in the right direction because I don't even know how to get the answer.

 

[cepheid]

Oh I didn't notice that, now I am a bit more lost. The answer is to find the mass corresponding to the given area. I just need to know if I am going in the right direction because I don't even know how to get the answer.

Assuming that your computation of the total volume of mud is correct, (which you need to do more work to verify) then your last few steps are correct. Because after the landslide, the mud is just in the form of a rectangular slab or box, and its volume = length*width*thickness = area*thickness. Hence, thickness = volume/area (which is exactly what you did). Then you found the volume contained below the sub-area of only 2.2 m^2 by applying volume = area*thickness again.

I was merely attempting to point out that you can easily find the mass of this small sub-volume, because you have been given the density of the mud. Recall that density = mass/volume (it's just a measure of how tightly-packed the matter in a substance is). So, given volume and density, how would solve for mass? It's a simple rearrangement of the equation.

 

[cepheid]

Okay, I thought about it some more, and this is my interpretation of the problem (NOT TO SCALE):

http://img31.imageshack.us/img31/9039/triangularprism.png

If this is correct, then I agree that the object is a triangular prism, whose bases are right triangles. I disagree a little bit over which side is b, which is h, and which is l, since l is supposed to be the distance between the triangular faces. In the end, it doesn't matter too much, since you can exchange the order of operands in multiplication without affecting the result.

One things that does matter, however, is that for a right triangle, the base and the height are the two perpendicular edges. However, I don't think you've been given the height (vertical edge) in this problem. Instead, I think you've been given the hypotenuse of the right triangle (since the problem states that the 1000 m is the distance along the slope)

DISCLAIMER: My interpretation of the geometry could be wrong. There is room for ambiguity in the wording of the problem.

 

[Ivan92]

Oddly enough, I did not get a notification to this answer...huh...Well I thank you for the diagram and your assistance. I actually got the answer though I forgot how. It has been a week or more, since I finished this. Thank you for your time! :)