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Pulleys and Mass becoming infinitely large

  1. Jul 5, 2012 #1
    This is 2 problems that continue one another; I solved the first one but do not understand the second.
    1. The problem statement, all variables and given/known data
    A 100kg man dangles a 50kg mass from the end of a rope. If he stands on a frictionless surface and hangs the mass over a cliff with a pulley, the tension in the rope will be: (Note: Ignore any frictional forces) [SOLVED]

    In the question above, as the mass becomes infinitely large, the man's acceleration becomes:
    A. 5m/s2
    B. 10m/s2
    C. 20m/s2
    D. infinite

    2. Relevant equations
    F = ma


    3. The attempt at a solution
    The solution to the first problem:
    50g = (50+100)a
    a = g/3 ~ 3.33
    T = mass of man (a) = 100(3.33) = 333N

    For the second problem:
    Now the acceleration is directly tied to the mass of the block, so when it becomes infinite, I think the acceleration should be infinite.

    However, in the answers it says: "B is correct. No matter how large the mass gets, its acceleration can never be greater than g, because it is gravity that is acting on it. Of course, the man will not be accelerated faster than the mass."

    They go on to repeat this method again in the next problem. I don't really understand how the man's acceleration would still be finite with a block of infinite mass.
     
  2. jcsd
  3. Jul 5, 2012 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Good.
    Not good. Redo this.

    Forget the rope and the 50 kg mass for a second. If the man just dropped, what would be his acceleration? If his mass doubled, what would change? What if he were 1000 kg, not 100 kg?

    Instead of trying to do it 'in your head', just redo your first calculation with increasing mass and see what happens. Make his mass first 100 kg, then 1000 kg, then 10000 kg. See the pattern.
     
  4. Jul 5, 2012 #3
    If the man just dropped, then mg = ma and a = g. I see how acceleration will approach g as the mass goes to infinity now.

    What do you mean redo this? If you do it from the perspective of the block, then
    mg - T = ma
    T = mg - ma = 50(10 - 3.33) = 333N
     
  5. Jul 5, 2012 #4

    Doc Al

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    Staff: Mentor

    Cool.
    As applied to the block, that equation has a sign error. The tension and acceleration are both upward.
     
  6. Jul 5, 2012 #5
    Mg-T=Ma
    T=ma

    [itex]a=(\frac {M}{M+m}) g[/itex]

    M>>m, M+m≈M
     
  7. Jul 6, 2012 #6

    Doc Al

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    Staff: Mentor

    Your second and third equations are incorrect.

    But I agree that the best way to see the answer is to solve it symbolically and let M → ∞.
    Why not let the OP solve it?
     
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