• Welcome to engineeringclicks.com
  • Lagrange’s Equation with friction that depends on contact force???

    Discussion in 'Calculations' started by mhjones12, May 31, 2012.

    1. mhjones12

      mhjones12 Well-Known Member

      Joined:
      Feb 2012
      Posts:
      78
      Likes Received:
      1
      I am trying to see if I can use Lagrange’s Equations to formulate the equations of motion for my system, but I am running into a problem. All of the examples I have seen where friction is incorporated into Lagrange’s Equations as a generalized force, the friction is something like u*m*g*sign(v). In my case, the m*g is an unknown internal contact force. Lagrange’s Equations do not give the internal forces like the direct method would.

      A simple example that illustrates what I mean is:
      [​IMG]
      The block is allowed to move with friction against ground (in my real system, the friction is between two moveable bodies). If I formulated the generalized force of the example system, it would contain the normal force on the block (an additional unknown), but I would still only have two Lagrange Equations. So I would have too many unknowns (3 unknowns, X, Theta, N, and two Lagrange Equations for X and Theta).

      I came across a method which uses Lagrange Multipliers to enforce constraints, and the physical interpretation of the Multipliers is the force required to enforce the constraint. Would this work on the example system? Would the constraint simply be Y = 0, where Y is the vertical position of the block? Is there another way to do this?, because in my real system, I’m not exactly sure how to formulate the constraint.
       
    2.  
    3. mhjones12

      mhjones12 Well-Known Member

      Joined:
      Feb 2012
      Posts:
      78
      Likes Received:
      1
      Solution

      Well I believe I found my solution; I'll go ahead and post it here for future reference.
      1. You would have to create a new sort of "dummy" variable: the vertical position of the block, Y.
      2. Since the block cannot move in the vertical direction, you also have the constraint equation (which needs to equal zero), that is F=Y=0.
      3. On the left hand side of each Lagrange Equation, you would add a Lagrange Multiplier (lambda) times the partial of the constraint equation (F) with respect to the generalized coordinate of the specific Lagrange Equation.
      4. Now there are three generalized coordinates (X, Y, theta) and thus three Lagrange Equations, plus the constraint makes 4 equations total. There are also 4 unknowns (X, Y, theta, lambda), so the system is solvable.
      5. The point of all of this is that the value of lambda is the force required to maintain the constraint equation, in this case Y=0, which of course is the normal force which I need to calculate the friction.

      phew...
       

    Share This Page