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Homework Help: Force required to lift a bar pivoting on one end

  1. Sep 17, 2011 #1
    1. The problem statement, all variables and given/known data

    This is actually for a "hobby" project, not at all academically related, but I'm posting it here because I know it's very elementary. I'm just having more trouble with it than I should but it's been five years since my last Physics class so I felt like I partly have an excuse. ;-)

    I have sliding glass doors at home and use a "charley" (security bar) to secure them. For reasons that would waste time to get into, I want to make it so that I can move this security bar into and out of place from outside my home using an RFID tag in my wallet. As shown in the attached drawing, the security bar is a piece of square tubular steel that is attached by a pivot to the right edge of the sliding door. When someone wants to open the door, they pivot the security bar from the horizontal to the vertical position. I want to mimic this action with a linear actuator.

    2. Relevant equations

    In the drawing I have given the total length of the security bar (34"), a proposed attachment point of the linear actuator (12"), and the weight of the security bar (about 1lb).

    3. The attempt at a solution

    I know that when the actuator starts to move (pick up the bar from its resting position) it must overcome gravity. This problem has been partially already answered here but I don't fully understand what is being said.


    After spending a while trying to find the right way to calculate this, I actually decided to try and figure it out experimentally. I attached a piece of rope (12" from the pivot of the bar) and ran it through a pulley above the bar. Then I hung a jug of water on the other end and filled it until the bar moved upwards. After doing the calculation for the weight of water, I found that a counterweight of about 6lbs is needed for the bar to move. Does this sound right?

    Attached Files:

    Last edited by a moderator: Apr 26, 2017
  2. jcsd
  3. Sep 18, 2011 #2
    What is important is torque, force times a perpendicular distance, see:



    In your case I took measurements off your diagram and assuming the drawing is roughly to scale we have the following torques about the bar pivot,

    1 pound X 17 inches - force of actuator X 8 inches = 0

    So with the bar in a horizontal position a force of about 17/8 pounds just keeps the bar in equilibrium. As the bar rises these numbers will change, that is another problem.
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