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Dynamics Prob force is a function of time

  1. Feb 25, 2010 #1
    1. The problem statement, all variables and given/known data
    A block is attached to a rope through a single pulley connected to a motor pulling with a force F(t), t in seconds, when the block is falling at a specific velocity > 0 the motors is applied to stop, find the time it takes for v=0.

    neglect mass of pulley & rope.


    2. Relevant equations



    3. The attempt at a solution
    Is this as simple as setting F(t)=m*g? or would I need to set the forces in the y direction = to mass * accel(in y) then take the integral of dv from v to 0 = the integral of a dt from 0 to t. then solve for t?
     
  2. jcsd
  3. Feb 26, 2010 #2

    cepheid

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    The motor isn't turned on until the block is falling at some rate, which we'll call the initial velocity v0. So, given that the block is initally falling at that speed, how long does it take the motor to decelerate it? Let the time the motor is first turned on be t = 0. Let the stopping time, which we are trying to solve for, be ts. Well, the change in velocity in that time interval is given by the integral of the acceleration:

    [tex] v(t_s) = v_0 + \int_0^{t_s} a(t)\, dt [/tex] ​

    Taking downward to be the positive direction, we know that a(t) = Fnet/m, where m is the mass of the block, and Fnet is the net force on it:

    [tex] a(t) = \frac{F_{\textrm{net}}}{m} = \frac{mg - F(t)}{m}[/tex]​

    Since the speed at the stopping time is zero, the integral equation becomes:

    [tex]\Delta v = -v_0 = \int_0^{t_s} g - \frac{F(t)}{m} \, dt [/tex] ​

    Eventually you get:

    [tex] v_0 + gt_s = \frac{1}{m}\int_0^{t_s} F(t)\, dt [/tex] ​

    To be honest, I have no idea what to do with that without knowing the functional form of F(t)
     
  4. Feb 26, 2010 #3
    Nevermind, I think I figured it out.
     
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