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Simple problem for anyone who knows what he is doing

  1. Jul 11, 2003 #1
    I’m stuck on a problem involving a cord that has mass. I know how to do problems where the cord's mass is 0 but ones the factor of the cords mass is introduced I’m lost.

    I’m stuck on the following problem:

    A block (mass m1) on a smooth horizontal surface, connected by a cord that passes over a pulley to a second block (mass m2), which hangs vertically.

    Determine a formula for the acceleration of the system if the cord has non-negligible mass mc specify in terms of L1 and L2 the lengths of cord from the respective masses to the pulley (total cord length L =L1+L2)

    How do I approach this problem?

    The answer (if it helps) is a=(m2+mc(L2/L))g/(m1+m2+mc)

    Any help will be highly appreciated.
  2. jcsd
  3. Jul 11, 2003 #2


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    I think what you need to do is note that the cord has total length L= L1+ L2 and so has "linear density" mc/L= mc/(L1+ L2). That means the mass of the cord hanging vertically is
    mc(L2/(L1+ L2))= mc(L2/L). That gives precisely the answer given.

    From that you can calculate the total mass hanging down and total of all mass.
  4. Jul 11, 2003 #3
    I already understood that the mass of the cord hanging down in L2/L.

    How do you deal with the fact that the horizontal part of the cord is being pulled down and what about the tension in the cord?

    I don’t understand how all the forces involved effect the system and therefore I don’t know how to set up an expression for the acceleration.

    For the hanging down part of the system Net force = (m2+mc*L2/L)*g- Ft
    Where Ft is the force of tension.

    a=((m2+mc*L2/L)*g- Ft)/( m2+mc*L2/L)

    I think I’m correct on that, but how do I find Ft.
  5. Jul 11, 2003 #4


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    I thought you said you could do it if the cord was massless- once you know how to calculate the mass of each part of the cord, include that into the masses m1 and m2 and the rest of the problem is exactly as if the cord was massless.

    The total mass "hanging down" is m2+ mc(L2/L). The downward pull is it's weight: (m2+ mc(L2/L)g. But that force (weight) has to accelerate the entire mass: m1+ m2+ mc. Using F= ma we have

    (m2+ mc(L2/L)g= (m1+ m2+ mc)a.

    Solve for a.
  6. Jul 11, 2003 #5
    Thanks, the explanation helped a lot .
    I think I understand now, but there is still one thing, the force of tension.
    Last edited: Jul 11, 2003
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