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Simple kinetics problem / Dynamics /

  1. Apr 22, 2012 #1
    Simple kinetics problem / Dynamics / Simultaneous equations

    I've been trying to solve this problem for hours, not finding the solution. Therefore this question to PhysicsForums homeworks section. Question is just part of the problem, but in order to fully understand issue I must have these joint forces to be clear for myself.

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

    Problem is to solve reaction forces in the joint B when the moment is starting to affect.

    Known variables are;

    L1 = 0.36 m
    L2 = 0.40 m
    m1 = 3.2 kg
    m2 = 3.2 kg
    θ = 26°
    M = 30 Nm

    roller C is expected to be weightless.


    2. Relevant equations

    I do have created equations for this. From the F = ma i got all in all six equations. (Described in the attachment.)

    3. The attempt at a solution

    I've been trying to solve equations based on the fact that number of equations = Unsolved variables, but finding the needed 3 kinematics equations I am not so sure.

    How to find Bx and By with the given data?


    EDIT;

    - Added picture "equations". - 22.4
    - Title corrected.
     

    Attached Files:

    Last edited: Apr 22, 2012
  2. jcsd
  3. Apr 22, 2012 #2
    Re: Simple kinetics problem / Dynamics / Simultaneous equations

    Problem is likely to be that I am not able to compose these kinetics equations. Any hints for learning how to create these needed equations?
     
  4. Apr 23, 2012 #3
    I found some similar looking example from the internet. (See attachment.)

    But still I do not know Angular velocity (ω) for this. I must derive it somehow? Any assistance for this, or hints where to start?
     

    Attached Files:

  5. Apr 23, 2012 #4
    This file might help to learn this issue also. Just don't get it. (See attachment in case interested in understanding this prob.) Similar looking examples with some variation.
     

    Attached Files:

  6. Apr 23, 2012 #5

    OldEngr63

    User Avatar
    Gold Member

    You have just stumbled upon a great fundamental truth! The hardest part of almost all dynamics problems is the kinematics portion. There is a great tendency for most folks to want to simply leap directly over the kinematics, saying, "Oh, that is only kinematics, but that is most often the key to the problem!

    The length of the inclined bar is

    L3 = √((L1)2+(L2)2)

    For your problem, let us let your frame move slightly, so that the upright arm, L1, is at an angle θ with the vertical, and then let the inclined arm, L3 make an angle [itex]\phi[/itex]. Then we can write two geometric relations that must apply for all possible positions:

    L1 sinθ + L3 cos [itex]\phi[/itex] - X = 0
    L1 cosθ - L3 sin [itex]\phi[/itex] = 0

    where x is the new length along bottom edge after the structure has moved. These two equations are obtained by simply summing displacements around the structure, first horizontally, then vertically, and in each case insisting that the summation must close, i.e., that we must return to the starting point.

    Now, if you treat θ as independent variable, then [itex]\phi[/itex] and x each depend upon θ
    so that you can write all of the equations of motion in terms of this one variable. To do this, you need to know how the velocities and accelerations are related, so first solve the previous equations for [itex]\phi[/itex] and x.

    Looking at the second equation,
    L3 sin [itex]\phi[/itex] = L1 cosθ
    sin [itex]\phi[/itex] = (L1 / L3) cosθ

    d[itex]\phi[/itex]/dt cos [itex]\phi[/itex] = - (L1 / L3) sinθ
    from which d[itex]\phi[/itex]/dt is easily evaluated. The other equation can be used to evaluate dx/dt in terms of dθ/dt. They can both be differentiated again to get the accelerations. See if this will get you going.
     
  7. May 4, 2012 #6
    Thanks for your efforts, but it still not helped me much. So the problem remains unsolved for me. :)
     
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