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Acceleration of a Block on pulley

  1. Oct 13, 2009 #1
    1. The problem statement, all variables and given/known data

    Determine the acceleration of block A when the system is released. The coefficient of kinetic friction and the weight of each block are indicated. Neglect the mass of the pulley and cord.

    azcsw3.png

    2. Relevant equations

    [tex]\sum F = ma[/tex]

    3. The attempt at a solution

    [tex]w_{B} = 20 lb[/tex]

    [tex]w_{A} = 80 lb[/tex]

    [tex]\mu_{k} = 0.2[/tex]

    [tex]\theta = 60 deg[/tex]

    2m4f987.jpg

    Block A

    [tex]\leftarrow\sum F_{x} = ma_{x}[/tex]

    [tex]-F_{f}cos60 + Nsin60 - 2Tcos60 = ma_{x}[/tex]

    [tex]\downarrow\sum F_{y} = ma_{y}[/tex]

    [tex]-F_{f}sin60 + w - Ncos60 - 2Tsin60 = ma_{y}[/tex]

    Block B

    [tex]\downarrow\sum F_{y} = ma_{y}[/tex]

    [tex] w - T = ma_{y}[/tex]

    Kinematics

    [tex]2S_{A} + S_{B}[/tex]

    [tex]2a_{A} + a_{B}[/tex]

    [tex]a_{A} = -\frac{a_{B}}{2}[/tex]

    I am confused do I just say [tex]N = ma = (\frac{80lb}{32.2ft/s^{2}})(32.2ft/s^{2})[/tex] or do I have to solve N as an unknown?
     
    Last edited: Oct 13, 2009
  2. jcsd
  3. Oct 13, 2009 #2

    rl.bhat

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    What is the value of Ff and N?
     
  4. Oct 13, 2009 #3
    Wouldn't they be:

    [tex]N = ma = (\frac{80lb}{32.2ft/s^{2}})(32.2ft/s^{2}) = 80lb[/tex]

    and

    [tex]F_{f} = N(0.2) = (80lb)(0.2) = 16lb[/tex]
     
  5. Oct 13, 2009 #4

    rl.bhat

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    No.
    N is the normal reaction of the surface on the sliding body.
    You have to resolve the weight of the body ( mg) in to two components. One pqarallel to the inclined plane and the other along the inclined plane.
    Now redraw the FBD.
    You have written 2SA = SB. Can you explain why it so?
     
  6. Oct 13, 2009 #5
    Block B FBD is same and block A:

    29l22kg.png

    I modified my positive directions a little:

    Block A:

    [tex]\sum F_{x} = ma_{x}[/tex]

    [tex]-2T + wsin60 -F_{f} = ma_{x}[/tex]

    [tex]\sum F_{y} = ma_{y}[/tex]

    [tex]N - wcos60 = ma_{y}[/tex]

    Block B:

    [tex]\downarrow\sum F_{y} = ma_{y}[/tex]

    [tex] w - T = ma_{y}[/tex]

    I have said 2SA + SB because:

    34e886x.png
     
  7. Oct 13, 2009 #6

    rl.bhat

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    When the block A moves down through a distance x, two segments of the ropes attached to A will also lengthen by x each. Since the total length of the string is constant, B must move up by 2x. Is it not so?
     
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