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P = F/a

  1. Nov 1, 2005 #1
    But what happens when the object is on an inclined surface?

    In this particuliar situation a 30kg block with an area of force of 80cm squared is standing on a surface which 30 degrees inclined relative to the floor. How do you calculate the pressure it exerts?
     
  2. jcsd
  3. Nov 1, 2005 #2

    Pyrrhus

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    You need to find the component of the force that is normal to the surface where it acts, in order to find the pressure.
     
  4. Nov 1, 2005 #3
    Ok so Fn/Fg = cos(30deg) = 0.866
    0.866 * 30kg * 9.8 = 254.6N
    254.6 / 0.008m^2 = 3.2 X 10^4

    Like this?
     
  5. Nov 1, 2005 #4

    Pyrrhus

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    Good work, that's it! :smile:, btw N/m^2 = Pa (Pascal).
     
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