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Dynamics involving dependent motion

  1. Jan 26, 2006 #1
    If cylinder B has a downward velocity in feet per second given by
    vB = t2/2 + t3/6 , where t is in seconds. Calculate the acceleration of A
    when t = 2 seconds.

    How does the velocities of A & B relate...Seems to be at C but I am not clear how to state it.

    Attached Files:

  2. jcsd
  3. Jan 28, 2006 #2


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    If you just look at A and C, you should see that vA=-vC. In other words, A and C move at the same speed and when A moves down, C moves up (and vice versa).

    You might be able to see the relationship between B and see, but let's try it a little more analytically: Let's call yC the position of C and yB the position of B. Now let's write yB as a function of yC:

    [tex]y_B=y_c - x[/tex]

    where x is a variable I just made up to represent the length of rope between the pulley and weight B. But the length of the rope (let's call it L) is constant and equal to L=yC-x

    Plug this into the equation for yB:

    [tex]y_B=y_C - (L - y_C) [/tex]
    [tex] y_B = 2y_C - L [/tex]

    Take the derivative to find the relationship between velocities (remember L is constant):

    [tex] v_B=2v_C [/tex]




    [tex] v_B=-2v_A [/tex]
  4. Jan 28, 2006 #3
    Thanks, I finally arrived at this answer Thursday.
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