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Forces on a Mechnical System (Spring, Damper etc )

  1. Sep 13, 2013 #1
    1. The problem statement, all variables and given/known data

    Write equations that could be used to solve for, [tex]V_{1}(s) \quad , \quad V_{2}(s)[/tex] in the Laplace domain for the mechanical system shown in the figure attached.

    2. Relevant equations



    3. The attempt at a solution

    I think I understand most of the problem, but I think I am confused about the direction of the forces.

    I am looking to write two equations as follows,

    [tex]\sum \text{Forces on M1} = 0 \quad , \quad \sum \text{Forces on M2} = 0 [/tex]

    First,

    [tex]\sum \text{Forces on M1} = 0[/tex]

    [tex]b_{2}(v_{1}(t) - 0) + b_{1}(v_{1}(t) - v_{2}(t)) + M_{1}\frac{dv_{1}(t)}{dt} = r(t)[/tex]

    Second,

    [tex]\sum \text{Forces on M2} = 0[/tex]

    [tex]b_{1}(v_{2}(t) - v_{1}(t)) + k\int v_{2}(t) dt + M_{2}\frac{dv_{2}(t)}{dt} = 0[/tex]

    Here are the things I am confused about,

    1. For the forces on the dampers, how do I figure out whether it is (v2-v1) or (v1-v2) in each of the two cases?
    2. Why is the force from the spring not considered in the first equation, i.e. summation of forces on M1?

    I am capable of completing the rest of the problem without any issues, but I just want to clarify my understanding with regards to those two questions.
     

    Attached Files:

  2. jcsd
  3. Sep 13, 2013 #2
    First question 2: If you isolate M1 as a free body, the spring is not connected to M1 and is not exerting a force on it.

    Question 1: If the velocity difference makes the dashpot get bigger, then the force is tensile. If the velocity difference makes the dashpot smaller, then the force is compressive.
     
  4. Sep 13, 2013 #3
    Hi Chestermiller! :smile:

    The answer to question 2 is clear, but I am still confused about question 1.

    Let's take the damper b2 as an example.


    There are two possible velocity differences

    [tex]v_{1}(t) - 0, \quad \text{or} \quad 0 - v_{1}(t)[/tex]

    I'm confused as to how to interpret these two velocity differences. How are you supposed to know which difference makes the dashpot bigger or smaller without knowing whether the velocity is positive or negative?

    If you assume its positive, then how would we interpret the first velocity difference? Are we saying that the bottom portion of the dashpot is moving down faster than the top portion of the dashpot effectively making the dashpot bigger, thus tencile force upwards?

    I think I am understanding something backwards.
     
  5. Sep 13, 2013 #4
    If the bottom portion of the dashpot is moving down faster than the top portion of the dashpot, then the dashpot is in tension, and wires or bars connecting it to the ceiling and to mass M1 are in tension. The wire connected to the ceiling is pulling downward with a force of b2v1, and the wire connected to the mass is pulling upward with a force of b2v1.
     
  6. Sep 14, 2013 #5
    If this is the case, and,

    [tex]M_{1}\frac{dv_{1}(t)}{dt}[/tex]

    is a force acting on the mass in the downwards direction, then how are the forces on the same side of the equation with the same sign?

    i.e. [tex]b_{2}(v_{1}(t) - 0) + b_{1}(v_{1}(t) - v_{2}(t)) + M_{1}\frac{dv_{1}(t)}{dt} = r(t)[/tex]
     
  7. Sep 14, 2013 #6
    Still looking for clarification on this!
     
  8. Sep 14, 2013 #7
    [itex]M_{1}\frac{dv_{1}(t)}{dt}[/itex] is not an actual force acting on the mass. Mathematically, according to Newton's second law, [itex]M_{1}\frac{dv_{1}(t)}{dt}[/itex] is the difference between the actual forces acting in the downward direction and the actual forces acting in the upward direction. Draw a free body diagram showing M1. There are two forces acting in the upward direction, and one force acting in the downward direction.

    Forces acting in downward direction = r(t)

    Sum of forces acting in upward direction=[itex]b_{2}(v_{1}(t) - 0) + b_{1}(v_{1}(t) - v_{2}(t))[/itex]

    Maybe it would help if the force balance equation were expressed in the following equivalent form:
    [tex]M_{1}\frac{dv_{1}(t)}{dt} = r(t)-(b_{2}(v_{1}(t) - 0) + b_{1}(v_{1}(t) - v_{2}(t))) [/tex]
     
  9. Sep 15, 2013 #8
    Hi again Chestermiller! :smile:

    Thank you for your explaination, things are much more clear now!

    I tried to apply my new knowledge to a seperate problem just to make sure its clear in my head! (See figure attached)

    So the sum of the forces on the cart are,

    [tex]m\frac{dv_{c}(t)}{d_{t}} = k \int v_{c}(t)dt + b(v_{t}(t) - v_{c}(t)) + k \int v_{t}(t)[/tex]

    The part I am confused about here is how to use the relative velocities and determine the direction of the force on the cart due to the spring.

    You had explained how to reason out the force direction for the damper based on the relative velocity in a previous post i.e.
    but what about for the spring?

    Thanks again for all your help!
     

    Attached Files:

  10. Sep 15, 2013 #9
    It's the same concept for the spring as for the dashpot. There should be a minus sign in front of the integral for the cart velocity. The tension is the spring is determined by the displacement of one end relative to the other end.

    Chet
     
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