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Pulley and Spring Problem

  1. Mar 20, 2016 #1
    1. The problem statement, all variables and given/known data
    The mass of block A is 10kg. The block is at rest and the spring has a K = 25N/m is unstretched when the block is in the position as shown. One second after the block is released, find the acceleration and velocity of the block and the tension in the cable. My teacher gave us the answers of T= 89N, a= 9.81, and v= .064, but I'm still stuck. He wants us also to write a constrained equation for the total length.

    2. Relevant equations


    3. The attempt at a solution
    2T-mg= ma
    L= 2ya + yk Pulley Problem.jpg
     
  2. jcsd
  3. Mar 21, 2016 #2

    Orodruin

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    You need to describe your attempt much better. It is not at all clear what you are trying to do and impossible to understand if your thoughts are correct or not.
     
  4. Mar 22, 2016 #3
    Asked the professor again and not much help. Did get one piece of info that was helpful. He said T=kx

    My length equation and derivatives: L= 2sa + sk
    0= 2va + vk
    0= 2aa + ak
    Solve for aa: aa = -.5ak

    For my first equation: 2T - mg = ma

    If I substitute kx for T I get:

    2(kx) - mg = ma

    If I then substitute ak for aa:

    2(kx) - mg = -.5mak

    When I plug in mass, g, k, the equation comes to:

    2(25x) -98.1 = -5ak

    Simplify:

    50x -98.1 = .5x( double dot)

    So what I'm trying to figure out is do I have to solve this differential equation?:

    .5x(dd) -50x + 98.1 =0
     
  5. Mar 22, 2016 #4

    BvU

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    Hello Bama, :welcome:

    Unfortunately you forgot
    in the problem statement. Our telepathic capabilities are very limited. Could you help us understand what all these names stand for ? And their units ?

    Then: If teacher expects an acceleration of 9.81 m/s2 after a full second, doesn't that mean that the spring still doesn't do anything (cf your relevant equation) ?
     
  6. Mar 22, 2016 #5
    I am having a little confuse, why acceleration nearly equaled gravity ? If it happens then why we need tension ?, the spring had to prevent movement of block A, right ?
    Oh and I assume that block A is moving down
     
  7. Mar 22, 2016 #6

    haruspex

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    It is the acceleration at a particular point in time. What might the state of the spring be at that time?
     
  8. Mar 22, 2016 #7

    haruspex

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    Yes, the working is unintelligible without being told what all these variables represent. I assume that in many or most cases the second of the pair of letters is intended as a subscript. @Bama_Brian please use the subscript button at the top of the editing window.
    Still? It may have been doing something during the second.
     
  9. Mar 22, 2016 #8

    TSny

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    The answer of 9.81 m/s2 for the acceleration doesn't necessarily imply that the spring is (approximately) unstretched after 1 second. The direction of the acceleration is not specified. Note that there is a significant amount of tension at that time. (I believe there is a misprint in the value given for the tension.)
     
  10. Mar 23, 2016 #9
    Sorry for the confusion guys. Instead of using x let me switch to the variable y becuase it makes more sense because the motion is up and down. I wish there was a way to write y double dot in this text editor but I guess not. What I mean by (dd) is y double dot or the second derivative of y with respect to time. Also, I thought the same thing when I saw the acceleration or y (dd) as 9.81 exactly after one second. The only thing I can think is acceleration is changing which always makes these problems more fun. In this dynamics class I keep trying to use the 3 Kinematics equations for physics class which gets me in trouble because most of the time he gives us problems with a changing acceleration.

    5y(dd) +2y = 98.1

    With this equation I look at it two ways. First it looks like a differential equation because you have a second derivative and a non derivative of the same variable. If I take my math hat off and put on my dynamics hat I look at it, it looks like acceleration is a function of displacement.

    y(dd) = (-2y + 98.1)/ 5 is the same thing as:

    a= (-2y +98.1)/5

    I talked to my differential equations instructor yesterday and he was like yea thats a differential equation and you should be able to solve it because these type were on the last test lol! I'm still very shaky with differential equations because I've just been exposed to them the past couple of months and I'm also not doing too great in the class.

    He showed me that it should be solved somewhat like this:

    5y'' + 2y = 98.1
    5r2 +2 =0
    r = sqrt(-2/5) or r= sqrt(2/5)i

    Since there are imaginary roots, the trial solution should be:

    yp = Acos(sqrt(2/5))t + Bsin(sqrt(2/5))t

    From here I don't know how to solve for the coefficients A and B because on the right side of the original equation there are not any cos or sin to set up your system of equations. I'll check back with him tomorrow and in the mean time try to work on solving a = (-2y + 98.1)/5 with the equations we learned in dynamics class like ads=vdv
     
  11. Mar 23, 2016 #10
    My apologies again: the equation in the last message should be:

    ay = -10y + 9.81
     
  12. Mar 23, 2016 #11

    haruspex

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    Assuming y is the displacement downwards of the mass from the relaxed spring position, I think your first version was correct. (Functionally, you can get rid of the upper pulley and take the spring as anchored to the ceiling instead.)
    Yes, that is the general form of the solution to this (simple harmonic motion) differential equation. To find the coefficients, just plug it into your differential equation and see what happens.
    To find this solution from scratch, you can get rid of the g constant by a simple change of variable (equivalent to measuring y from the equilibrium position instead). Then multiply through the equation by ##\dot y##. You can now integrate all terms to obtain a first order ODE.

    If you are going to be posting more ODE questions on this forum, take the trouble to learn a bit of LaTeX. Type a double hash (#) to start it. To put one dot over a variable prefix it with \dot. Leave a space between that and the variable. For double dot, \ddot. End the LaTeX with another double hash. Check the result using the PREVIEW button.
     
  13. May 4, 2016 #12
    Ok just want to wrap up this post. Wound up getting close with the differential equation path to this problem, but the teacher finally gave us the calculus way to solve the problem. Here's the solution for anyone else that might be struggling with this type problem:

    Length equations:

    First:
    L= 2b + c - lo

    b= length from roof to pulley with weight attached
    c= distance from roof to floor
    lo= length of spring unstretched

    Second:

    L= 2(b+y) + (c-lo -x)

    y= distance from the weight hanging from the movable pulley to the ground
    x= distance the spring will stretch

    Subtract first length equation from the second and you're left with:

    x=2y

    Now, sum of forces from FBDs:

    ΣFy= may

    10(9.81) -2T = 10a

    T= 25x (Fs= kx)
    T= 50y (x = 2y)

    98.1 -100y = 10a

    But, a =v dv/dy

    Finally,

    98.1 - 100y = 10v dv/dy

    Integrate the above equation with respect to y:

    5v2 = 98.1y - 50y2 +C1 (C1 =0 because starts from rest when y=0)

    v= sqrt ( 19.62y -10y2)

    Because v = dy/dt:

    Now integrate again with respect to time:

    ∫dy/ sqrt(19.62y -10y2) = ∫dt

    This integral can be gotten from an interation table:

    t + C2= -1/√10 sin-1 (1- 20/19.62 y)

    1.019y - 1 = sin (√10t - π/2)

    When t=1:

    v= .0642 m/s
    a= 9.81 m/s2
    T= 98.1 N

    Alright guys that's it. Hope this helps anyone else who might be struggling as much as I was :)
     
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