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!

Cross Sectional Area to Newtons

  1. Dec 9, 2006 #1
    1. When the hot water in a certain upstairs bathroom is turned on, a series of 18 "ticks" is heard as the copper hot-water pipe slowly heats up and increases in length. The pipe runs vertically from the hot-water heater in the basement, through a hole in the floor 5 m above the water heater. The "ticks" are caused by the pipe sticking in the hole in the floor until the tension in the expanding pipe is great enough to unstick the pipe, enabling it to jump a short distance through the hole. If the hot-water temperature is 50°C and room temperature is 20°C

    This is a two part problem:
    1st is a) the distance the pipe moves with each "tick"
    Coefficient of Expansion is 17 x 10 ^-6 [(°C)^-1]
    So to solve this all I did is following:
    17 x 10 ^-6 [(°C)^-1] by 5 by 50°C = a
    17 x 10 ^-6 [(°C)^-1] by 5 by 20°C = b
    and then (a - b)/18 = 0.000146 meters

    So here comes my question:
    b) the force required to unstick the pipe if the cross-sectional area of the copper in the pipe is 3 x 10-5 m^2

    I don't have an idea how to even approach this problem. If anyone can give me any clue it would be great



    Thank you
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 9, 2006 #2

    OlderDan

    User Avatar
    Science Advisor
    Homework Helper

    For part b you need to know something about the force required to compress the pipe by the length of one tick movement. Note that the area given is the cross-sectional area of the copper in the pipe, not the cross-sectional area of the pipe.
     
    Last edited: Dec 9, 2006
  4. Dec 9, 2006 #3
    Well I have found that the typical value for the elastic modulus of copper is 1.1 x 10^11 Pa. Now I have length of the pipe which is 5 meters, Cross-sectional area of the copper in the pipe in m^2, how much it moves with each tick (18 ticks as well) and typical value of elasticity of copper in N/m^2. I am still a little confused on how to solve this problem.
     
  5. Dec 9, 2006 #4

    OlderDan

    User Avatar
    Science Advisor
    Homework Helper

    Assume that after each tick, the pipe achieves a relaxed state (no compression) As it heats up, its natual length increases by thermal expansion, but it is physically constrained to stay at one length. This requires a force, and the amount of force increases with temperature because the amount of compression from natural length is steadily increasing. When the force is no longer strong enough to keep the pipe compressed, it suddenly moves one tick to a new relaxed position.

    The assumption of achieving a relaxed postion at each movement may not be completely valid, but you have not been given enough information to know exactly how much force is already acting after each move.
     
  6. Dec 10, 2006 #5
    what else do you think I would need to solve this?
     
  7. Dec 10, 2006 #6

    OlderDan

    User Avatar
    Science Advisor
    Homework Helper

    Nothing. You have all that is needed. Find out how much force it would take to compress the pipe by the length of one tick.
     
  8. Dec 10, 2006 #7
    Should I multiply
    elasticity of copper (N/m^2) by
    cross-sectional area of the copper in the pipe (3 x 10-5 m^2)
    multiple by the distance each tick the pipe moves
    and divide by the distance of the pipe (5m)?
     
  9. Dec 10, 2006 #8
    I would like to thank you for your help.

    Thank you
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Cross Sectional Area to Newtons
Loading...