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Constrained motion problem

  1. Aug 13, 2013 #1
    1. The problem statement, all variables and given/known data
    Calculate the acceleration of the block B in the figure, assuming the surfaces and the pulleys ##P_1## and ##P_2## are all smooth and pulleys and string are light.

    (The mass of block C is m)

    Ans: F/(7m)

    2. Relevant equations



    3. The attempt at a solution
    I measured distances from a fixed wall on the right of B. The distances are shown in the second attachment.
    Writing down the expression for length of string, (##R## is the radius of pulleys)
    $$L=x_{P1}-x_B+\pi R+x_{P1}-x_{P2}+\pi R+x_{C}-x_{P2}$$
    Differentiating twice with respect to time,
    $$0=2\ddot{x_{P1}}-\ddot{x_B}-2\ddot{x_{P2}}+\ddot{x_{C}}$$
    ##\because \ddot{x_{P1}}=\ddot{x_A}## and ##\ddot{x_{P2}}=\ddot{x_{B}}##
    $$2\ddot{x_A}=3\ddot{x_B}-\ddot{x_C} (*)$$
    Assume that the tension in the string is T. Applying Newton's second law for A,
    $$F-2T=2m\ddot{x_A} (**)$$
    For B,
    $$3T=4m\ddot{x_B} (***)$$
    For C,
    $$T=m\ddot{x_C} (****)$$
    Solving the four equations,
    [tex]\ddot{x_B}=\frac{3F}{13m}[/tex]

    Any help is appreciated. Thanks!
     

    Attached Files:

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  2. jcsd
  3. Aug 13, 2013 #2
    I would suggest to keep things simple.

    Let the lengths be x1(top),x2(middle),x3(bottom).

    A and B move to the left,C move to the right.

    We have, [itex] \ddot{x_1}+\ddot{x_2}+\ddot{x_3}=0 [/itex]

    [itex] \ddot{x_1}=a_A-a_B[/itex]

    [itex]\ddot{x_2}=a_A-a_B[/itex]

    [itex]\ddot{x_3}=-a_B-a_C[/itex]

    where aA,aB,aC are magnitude of accelerations of blocks A,B,C.

    aA = (F-2T)/(2m)

    aB = 3T/(4m)

    aC= T/m
     
  4. Aug 13, 2013 #3
    That does gives the answer but how do you get the second derivatives of lengths in terms of ##a_A##, ##a_B## and ##a_C##?
    Using your method, I get ##2a_A=3a_B+a_C## where as from my method, I get ##2a_A=3a_B-a_C##. Where did I go wrong? :confused:
     
  5. Aug 13, 2013 #4
    Lets consider case of x1:

    What causes a change in x1?

    Block A moving to the left causes x1 to increase and B moving to the left causes the string length x1 to decrease.So, [itex]\dot{x_1}={v_A}-{v_B}[/itex],where [itex]{v_A}[/itex] and[itex] {v_B}[/itex] are speeds of a and B respectively.

    Hence [itex]\ddot{x_1}=\ddot{a_A}-\ddot{a_B}[/itex].
     
  6. Aug 13, 2013 #5
    Thanks! :)

    Any idea what's wrong with my method?
     
  7. Aug 13, 2013 #6
    It should be $$-T=m\ddot{x_C} (****)$$

    You are considering left direction to be positive .:smile:
     
  8. Aug 13, 2013 #7
    Great! Thanks a lot Tanya! :)
     
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