Basic measurements and conversion not so basic

[Saterial]

Homework Statement

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?

Homework Equations

v = lwh?

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

 

[rickz02]

can I assume that the slope is a perfect vertical(lol) and use that as a length to find volume?

No you can't, this is where you're going to use the dimensions of the mountainside. Also take note that

1.1 m deep slips into a valley in a mud slide.

Can I simply just multiply 1900 kg/m^2 by 2.2 m^2 to obtain 4180 kg?

I believe you simply can't do that, you are given with a volumetric density 1900 kg/m³.

if there is mud stacked on top itself at a higher height, would that not change the mass?

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

What is the mass of the mud sitting above a 2.2 m2 area of the valley floor?

Where exactly is 2.2 m² area of the valley floor, is it near the slope?

 

[rickz02]

can I assume that the slope is a perfect vertical(lol) and use that as a length to find volume?

No you can't, this is where you're going to use the dimensions of the mountainside. Also take note that

1.1 m deep slips into a valley in a mud slide.

Can I simply just multiply 1900 kg/m^2 by 2.2 m^2 to obtain 4180 kg?

I believe you simply can't do that, you are given with a volumetric density 1900 kg/m³.

if there is mud stacked on top itself at a higher height, would that not change the mass?

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

What is the mass of the mud sitting above a 2.2 m2 area of the valley floor?

Where exactly is 2.2 m² area of the valley floor, is it near the slope? I guess it is...

 

[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?