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!

Pressure Problem! help

  1. Jan 16, 2005 #1
    Hey,

    Ok here's the problem.

    A house has a roof with the dimensions in drawing. Determine the magnitude and direction of the net force that the atmosphere applies to the roof when the outside pressure rises suddenly by 10.0mm of mercury, before the pressure in the attic can adjust.

    Ok here's what I got so far...

    Ok so there are three forces in the vertical direction. One upwards on the roof and the the other two downwards, namely atmospheric pressure (P1) and weight (W). The other being the opposite force P2, acting upwards.
    So the Fnet is really just the sum of these forces right? So to find force for P1 and P2, I just used the Pressure x Area = force formula for both, except instead of atmospheric pressure, I added 10 to 760 and converted back into Pa. Then I got the forces and subtracted, and I also found the weight to be 2816N, I also subtract this number. I get something like 1.81 x 10^6 N in the downward direction. Something's wrong...

    Any ideas?
     

    Attached Files:

  2. jcsd
  3. Jan 16, 2005 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Okay,what weight??Weight of the roof??

    Daniel.
     
  4. Jan 16, 2005 #3
    Yes, the weight of the roof. I found that by using the rho (p) of air as 1.29 and volume of the triangle or roof to get m. p=m/V.
     
  5. Jan 16, 2005 #4

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    1.Is the roof made up of air???
    2.What is the formula which gives the volume of a triangle???
    3.What is the problem's question???


    Daniel.
     
  6. Jan 16, 2005 #5
    No, the roof's inside is air though.
    volume of a triangle : area of triangle, so 1/2bxh x the length.
    Determine the net force, which I really think is acting downwards. Am I at least on the right track with the vertical forces?
     
  7. Jan 16, 2005 #6

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    There is no such thing as the volume of a triangle...Not even area...In your case,that would be a rectangular triangular prism...It has a volume,the problem is that u don't need that.
    Apply the second law of dynmics for the roof...What is the total force acting on the roof??

    How many forces act on the roof??

    Daniel.
     
  8. Jan 16, 2005 #7
    ok let's see. I'm pretty sure there are 3. The one acting downwards on the roof, the roof's weight acting down, and the reaction force acting upwards. So Fnet=P2A - P1A. So, (760 x (7.29x14.5)) - (770x((14.5x 4.21m)x2)=0. I tried this before, didn't work. I figure the forces acting upwards P2A acts on the rectangle surface, so I need the area for that. And the other P1A acts on both rectangles on the roof downwards so I need to multiply by two.
     
  9. Jan 16, 2005 #8

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    The weight of the roof has nothing to do with this problem.It's totally another interaction.There are only 2 forces that act on the roof which are interesting:The force from above and the one from below.They are pressure forces,having to do with the fact that air molecules transmit constantly momentum to the roof by their collisions with the roof.

    Daniel.

    PS.The net force is the vector sum of the 2.
     
  10. Jan 16, 2005 #9
    Vector sum? I don't understand this, am I suppose to break it into components? I just figure there are 2 forces up/down, pressure difference of 10 between outside/inside, and just solve for Fnet by adding the P1A=F1 and the P2A=F2 together.
     
  11. Jan 16, 2005 #10

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper


    Yes.
    There's not too much "breaking",the 2 forces have only one component each.

    Okay,it's good you got the picture.Now deliver the results...


    Daniel.
     
  12. Jan 16, 2005 #11
    ok here's a question: the force outside acting on the roof...It acts on both rectangular areas, so for A I have two double it? The other pressure acts only on the base rectangle. I tried both ways, but still not correct.

    Fnet= P2A-P1A=0
    0=(1.013 x 10^5 Pa)(14.5m x (2 x (4.21cos30))) - (1.022 x 10^5 Pa x 2(14.5m x 4.21m))
    =1.07x 10 ^7 N - 1.25 x 10^7 N
    =1.8 x 10 ^6 N acting downwards.??

    That's not right.
     
  13. Jan 18, 2005 #12
    So...Any ideas where I went wrong?
     
  14. Jan 18, 2005 #13

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    How come u used the angle of inclination only at one force??Both forces are inclined...


    Daniel.
     
  15. Jan 18, 2005 #14
    Even the one acting upwards under the roof? I figure there are two forces one acting on the outside on both sides (hence multiply area by 2) and one reacting back at the outisde presssure. So the first pressure acting down doesn't use the angle for the calculation, I think only the one reacting back does, because you need to find the length across the house.
     
  16. Jan 18, 2005 #15
    Someone, anyone...at least check if my reasoning is correct.
     
  17. Jan 19, 2005 #16
    I'll be patient lol....
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Pressure Problem! help
Loading...