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Particle dynamics, incline

  1. Oct 20, 2011 #1
    1. The problem statement, all variables and given/known data

    Two blocks with equal masses m1=m2=5kg are connected via a pulley, as shown. Take [itex]\mu[/itex] = 0.25. Find the acceleration given that a) m1 is moving down, b)m1 is moving up. If m2=6 kg for what values of m1 will the pair move att constant speed?

    2. Relevant equations

    F=ma, f=[itex]\mu[/itex]N

    3. The attempt at a solution

    I've managed to solve a) and c) but not b).
    I don't understand how m1 can be moving up, since m1=m2 and the force m2 affects m1 with is sin(37)*m2g < m1g.
    If m1 is moving up doesn't that mean that it will not affect the other mass? I mean shouldn't the tension on m2 be zero? If so then we have [itex]\Sigma[/itex]F=5*9.8*sin(37)-0.25*9.8*cos(37)*5=19.7
    Which gives a=19.7/5=3.94 m/s^2 However it should be 2.94 m/s^2 according to the answer sheet.

    Attached Files:

  2. jcsd
  3. Oct 20, 2011 #2


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    No. I highly recommend that you draw a free body diagram for m2, and then another free body diagram for m1. Apply Newton's 2nd law, namely that ƩF = ma. You should be able to solve for a. It doesn't matter that it is impossible for m2 to pull m1 up by gravity alone. All that matters is that you're told that right now m2 happens to be moving down the incline, while m1 is moving upward. These are the initial conditions you've been given. (Maybe somebody applied a force to make this happen. Who cares? It doesn't matter, just accept this as the initial condition).

    Indeed, if you choose the "down the incline" direction to be positive in your force balance equations for m2, you'll find that you get a negative value for a, meaning that although m2 is presently moving down the incline, its acceleration points up the incline, meaning that it is slowing down and will eventually reverse direction.

    No, the fact that m1 is moving up doesn't mean that the rope is going slack. As I said before, it means that m2 is moving down the incline while m1 moves upward (and the rope remains taut).
  4. Oct 20, 2011 #3
    Experience tells me that if I would give m1 a push the rope would go slack. Or is that because i accelerate it? Also wouldn't the rope go slack if v1 > v2 because s=vt and if the rope is taut then s1=s2?

    Anyways, new try

    The relevant forces acting on m2 is

    friction, f=0.25*cos(37)*5*9.8
    Tension T=5*9.8
    Gravitational component, Fx=sin(37)*5*9.8

    And so
    a=(0.25*cos(37)*5*9.8-sin(37)*5*9.8+5*9.8)/10=2.93 m/s^2
  5. Oct 21, 2011 #4


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    That's true. An example of the adage, "you can't push on a rope." However, nobody is saying that you (or anyone else) is pushing on m1. All that is being said is that, presently, m1 is moving upwards, whilst m2 is simultaneously moving down the ramp. That's it. Do you understand?

    Yeah, that's right. I busted out a "whilst." :rofl:

    Once again, you're not wrong. Any upward advancement of m1 must be accompanied by an advancement of m2 down the ramp by an equal distance. That, in fact, is precisely what is happening (i.e. this is the situation you are asked to consider). It's just the exact reverse of what happens when m1 falls down, and m2 consequently gets pulled up the ramp. The difference is that the latter situation can happen spontaneously, while the former cannot.

    This part in red is not correct. If the tension were just equal to the weight, then the net force on m1 would be zero, and it would have zero acceleration. However, if m1 had 0 acceleration, then the whole system would have zero acceleration. It doesn't.

    To get T, you need to do the force balance for m1, which tells you that T - mg = ma.

    It's a fluke that you got the right answer here. You made a second mistake that negated the first one that I mentioned above (in red). Your second mistake was to assume that the 'm' in the ma term was the total system mass. However, since the force balance equation you wrote above was for m2 only, it should be that m = 5 kg, not 10 kg.

    Without this second mistake, you see that your answer is too large by a factor of 2.

    Also, I'd really recommend keeping everything algebraic, until the last possible step, and not plug in numbers until you've simplified things as much as possible. If you'd done that, you'd have seen that 'm' cancelled from both sides of the equation, making it unnecessary to plug in the 5 kg. Furthermore, you'd see that g factors out of every term on the left-hand side.

    Anyway, if you correct your first mistake (the one in red) then you'll have this problem nailed.
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