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Pulley Problem for Engineering Dynamics Course

  1. Jun 2, 2006 #1
    The cable at C is being pulled down at a velocity of [tex]4 \frac{ft}{s}[/tex]. The cable at C is also decelerating at a rate of [tex]2 \frac{ft}{s^2}[/tex]. At what velocity and acceleration is the block at A moving?

    [​IMG]

    I have the following equations for the problem worked out:

    [tex]\ell\,=\,2\,S_A\,+\,\left(h\,-\,S_B\right)[/tex]

    [tex]\frac{d}{dt}\,\left[\ell\,=\,2\,S_A\,+\,\left(h\,-\,S_B\right)\right][/tex]

    [tex]0\,=\,2\,V_A\,-\,V_B[/tex]

    [tex]V_A\,=\,-\frac{V_C}{2}[/tex]

    [tex]V_A\,=\,-\frac{-4}{2}\,=\,2\,\frac{ft}{s}[/tex]

    This answer is NOT correct, it is off by a factor of two. Can anyone tell why?
     

    Attached Files:

  2. jcsd
  3. Jun 2, 2006 #2

    Andrew Mason

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    Homework Helper

    Pulley C moves at half the speed of the cable around it. So C is moving down at 2 ft/sec and the cable attached to C is also moving at 2 ft/sec, which is the cable around A. Pulley A must be moving at 1/2 the rate at which the cable around it is moving.

    AM
     
  4. Jun 6, 2006 #3
    Thank you very much!

    I guess I was just not thinking of a simple solution because I thought that I would be complicated to figure. The obvious 1/2 velocity result of the right hand pulley is what threw me!

    Anyways, the correct answer is:

    [tex]V_C\,=\,\frac{1}{2}\,(-4\,\frac{ft}{s})\,=\,-2\,\frac{ft}{s}[/tex]

    [tex]V_A\,=\,-\frac{-2}{2}\,=\,1\,\frac{ft}{s}[/tex]
     
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