Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Active force of spring on mass (D'alemberts)

  1. Apr 22, 2015 #1
    I am trying to understand the effect that eliminating the mass m1 (m1 = 0) has on the active forces (Fa). I have gone through a scenario where m1 is taken into consideration (refer to uploaded images). The active force on m1 is from the spring and does not have m1 in the expression (Fs1 = (l-r)k), so I did not think that there would be any change. However the set of D'alemberts reduces if m1 is eliminated, so how does active force from the spring effect the remaining equation of motion.

    If more info or clearing up is required please do not hesitate to ask. Any help would be much appreciated.

    Cheers.
     

    Attached Files:

  2. jcsd
  3. Apr 25, 2015 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Can you write it up there? The attachments are hard to read and their wrong orientation does not help either.

    I don't understand what you want to test/show.
    The force of the spring at a specific location does not depend on the mass, but the equations of motion certainly do because acceleration is force divided by mass. This also means you cannot expect a physical answer if you set the mass to zero. If you have a spring without mass attached, you still have the mass of the spring (which is often neglected).
     
  4. Apr 26, 2015 #3
    Figure1.png
    figure1.

    In figure1, the generalised variables are q and θ, and there are two masses (mass of the spring is neglected). The active forces are Fa1 = Fspring1 (horizontal direction) and Fa2 = Fweight (vertical direction). Therefore it follows that there should be four parts for the equations of motion:
    (Fa1 - m1a)∂r1/∂q + (Fa1 - m1a)∂r1/∂θ + (Fa2 - m2a)∂r2/∂q + (Fa2 - m2a)∂r2/∂θ = 0 [eq1]

    this is simplified to:
    (Fa1 - m1a)∂r1/∂q + (Fa2 - m2a)∂r2/∂q + (Fa2 - m2a)∂r2/∂θ = 0 [eq2]

    as ∂r1/∂θ=0

    My query is what happens when m1 is eliminated from the system (m1 = 0)
    The is now two generalised variable and only one mass:
    (Fa2 - m2a)∂r2/∂q + (Fa2 - m2a)∂r2/∂θ = 0
    [eq3]


    figure2.png
    figure 2.

    Now consider figure2 (Spring is removed and m1=0)
    The equation of motion of system in figure2 is the same as that for eq3, where the spring was included. This intuitively does not appear correct, and I wish to understand how to take into account the effect of spring on the remaining system when m1 =0;
     
  5. Apr 26, 2015 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    If you set m1 to zero, you cannot ignore the forces on r1 - they are still part of the mechanics. The only difference is the exact cancellation of the force from the spring and the horizontal component of the force in the beam (the experiment does not work properly with a string of course)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook