1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Trig.-Sand Pile Base

  1. Dec 20, 2013 #1
    1. The problem statement, all variables and given/known data
    A sand pile has an angle of repose of 35 degrees. If the pile is 2.4 meters high, find the area occupied by the base of the pile.

    2. Relevant equations
    Sine=opp/hyp
    Cosine=adj/hyp
    Tan=opp/adj
    a^2+b^2=c^2

    3. The attempt at a solution

    Going off the Pythagorean Theorem, I can find...nothing, right now. I tried that, but realize I need to know two side lengths and I only know one. I'm also not sure what the problem is asking for. Is it asking only for the area of the triangle? I can attach the diagram if needed. I believe the 2.4 meter long side is adjacent to the hypotenuse, correct? I figure I need to find the other two sides, and the other angle. Let's try finding the other angle first. I looked at the sine formula. I can't use that right now since I don't know either side in the equation. I need to use an equation with the adjacent side, since I do know that. 35cosine=2.4m/hypotenuse ≈1.97 m

    Could someone please tell me if that first step is right, and help me figure out the next step? It seems like if that was it, and then calculating the area of the triangle, it would just be too easy.

    Thanks in advance!!! :smile:
     
  2. jcsd
  3. Dec 20, 2013 #2
    A pile of sand is essentially just an upside-down cone. The angle of repose is defined as in the figure here. So basically the problem is just asking us to calculate what the area of the base of the cone is (a circle) since that is the floor area covered up by the sand pile. To do this we will need to find the radius of the base of the sand pile.

    We were given that the angle of repose is 35 degrees and that the height of the pile is 2.4 meters. If we draw a right triangle representing the side view of the sand pile, with one angle marked as 35 degrees, then the hypotenuse represents the slope of the pile, the adjacent leg represents the radius of the pile's base, and the opposite leg represents the height. If we use tan(35) we can write down a relation between the given height of the triangle and the radius of the pile.

    Once we find the radius of the pile it should be rather straight forward to calculate the area of the circle corresponding to this radius.
     
  4. Dec 20, 2013 #3

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    You know one side and all three angles. That's enough to fully describe the triangle.

    No it's the area of the base of the pile which is circular.
     
  5. Dec 20, 2013 #4
    [/QUOTE]To do this we will need to find the radius of the base of the sand pile.

    If we use tan(35) we can write down a relation between the given height of the triangle and the radius of the pile.

    Once we find the radius of the pile it should be rather straight forward to calculate the area of the circle corresponding to this radius.[/QUOTE]


    Okay, I think I see what you're saying. Thank you for being so descriptive. Instead of trying to find ALL the missing side lengths like I had tried above, I only need to find the one adjacent to the 35 degree angle. The picture you described is exactly like the one below my problem.


    CWaters, thanks for your answer! You're right, we know enough to describe my triangle, but my only point from that quote was that we cab't use the classic, straightforward Pythagorean Theorem to find a leg of the triangle at this time. :smile:


    This should prove easier than I thought.

    tan(35)=2.4m/adj
    .7=2.4/adj
    adj≈1.7

    I think I may have done something backwards, but the answer so far seems plausible, so I'll continue with the problem to see if the answer makes sense.


    âˆ= Is this pi? It doesn't look like pi. Oh well, I'll use it anyway.

    âˆ[r][2]=A
    âˆ(1.7)(1.7)=A
    âˆ(2.89)=A
    9.08 m=A


    The answer seems plausible.
     
  6. Dec 20, 2013 #5
    Sorry about the weird quote symbols. I wasn't sure how exactly they were supposed to go. Same with r^2.
     
  7. Dec 20, 2013 #6

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    "This should prove easier than I thought." - Famous last words.

    Are you sure that 2.4 / 1.7 = 0.7? I don't think that's right even in the New Math.
     
  8. Dec 20, 2013 #7
    HA! You're funny. No, that was actually the part I thought I had backwards. Nice sarcasm though. I hate new math. Since when do we round WHOLE numbers and leave them that way, ignoring scientific notation and significant digits and... Anyway, rant over. I think I multiplied instead of dividing, let's try this again.

    tan(35)/2.4=.29
    .29*.29=.0841
    .0841*pi= .264 m^2

    Yeah.....No. Did I divide the wrong thing? Maybe. Here we go.

    2.4/tan(35)=3.43

    3.43^2=11.8pi
    11.8pi= approx. 37 m^2.

    Maybe?
     
  9. Dec 20, 2013 #8

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Looks good.
     
  10. Dec 20, 2013 #9
    The last answer, correct? Thanks!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Trig.-Sand Pile Base
  1. Sand Car (Replies: 2)

  2. Rotating drums and sand (Replies: 21)

Loading...