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!

Force on the top screw in a hanging basket scenario

  1. Nov 23, 2009 #1
    1. The problem statement, all variables and given/known data

    Hi and Thanks to anyone that takes time to help!

    This is not a homework thing but a 'discussion' with friends.
    Please excuse my crappy drawings...
    5lcojk.jpg

    The measurments are not important only that it is the same frame either with strut or reversed for tie.

    The difference in opinion is simply that one says there is more force pulling the top screw from the wall in the tie senarios of each design, while the other says there is no difference in force on the top screw either way round.

    (We are both in our late 30's so it was some time since working out any actual equations- let alone remembering which ones are relevant)


    2. Relevant equations

    Errr...vectors, moments..??

    3. The attempt at a solution

    It's been too long but only exact maths is gonna conclude this little discussion!

    Please, please can you prove the error in one of our heads and make the other safe in the knowledge they are not a complete idiot!

    EDIT:Have I put this in teh wrong section? It just seemed the best place fer an exact answer.
     
    Last edited: Nov 23, 2009
  2. jcsd
  3. Nov 23, 2009 #2

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    If we assume the lower screw as the pivot, then the moment around it and the pull on the upper screw are the same in each case.
     
  4. Nov 24, 2009 #3
    Thanks A.T.,

    Would the pivot not be the bottom end of the bar against the wall and does that make any difference?

    What equations need to be used to work out exact numbers for the force acting on the top screw in both senarios for each design?
     
  5. Nov 24, 2009 #4

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    That might be more sensible. Then both screws are pulled.
    Not between the designs, if the distance of the screws to the bottom end of the bar is the same for all of them
    http://en.wikipedia.org/wiki/Torque
    You compute the torque created by the given force (M = F1 * r1) and then you use this torque to compute the pull force on the screw (F2 = M / r2). Where r is the shortest distance between the pivot and the line of action of F.
     
  6. Nov 24, 2009 #5
    Thanks again A.T., I just need to try to work it out when I have time and will maybe show my workings here for probable corrections!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Force on the top screw in a hanging basket scenario
  1. Hanging sign forces? (Replies: 4)

Loading...