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A cart and pulley problem with a twist

  1. Oct 23, 2009 #1
    A friend and I were imagining a physics problem for fun:

    A 3kg rectangular cart has a pulley system on its front edge. 1kg weight suspended is down the front, and it's attached by a rope through a mass-less pulley to a 1kg mini-cart on top of the cart. There is no friction between the ground and the cart, or the cart and the mini cart, the cart and the front block, or on the pulley.

    http://img98.imageshack.us/img98/7416/cartproblem.png [Broken]

    The question is, how hard do you have to push the cart so the little carts don't roll off? In other words, with some level of F_p, the acceleration of the top mini-cart relative to the ground is equivalent to the acceleration of the big cart relative to the ground.

    I said the answer was simple: 5kg * g, or 49N.

    My argument: Part 1: The acceleration of the system must be g, or 9.8m/s/s. That way, the yellow front box is held up by the pull of the blue box backwards. Or in other words, if the cart had dice in its rearview mirror, they would make a 45deg angle when the cart on the windshield is no longer falling, and the cart on the roof is no longer slipping forwards. To make these forces equal, you'd have to put the pedal to the metal and travel at 9.8m/s/s (which is a 2.7s "zero to sixty" time if you want to imagine that).

    Part 2: The mass under acceleration is 5kg, the same as it would be if the mini-carts were bolted down.

    Part 3: F = m * a = (1+3+1)kg * g = 49N

    My friend (call him Bob) said the answer is 4kg * g/2, or 2kg * g, or 19.8N

    His argument: Part 1:
    The equations of motion for block 1 on top:
    T = 1kg*a
    For block 3 (small one on the side's) vertical direction:
    1kg*g - T = 1kg*a
    -> 1kg*g - T + T = (1kg + 1kg)*a
    -> a = 1kg*g/ (1kg + 1kg) = g/2

    Part 2: "Without friction there is no way for the Blue cart to "know" that the Red cart is being pushed. It is absolutely impossible. There is nothing about Blue's horizontal motion that has anything to do with Red. There is simply no coupling between the horizontal motion of Red and the horizontal motion of Yellow." Therefor he only uses the mass of the Red cart and the Yellow cart, which is being pushed by the normal force of "Red,Yellow."

    Part 3: F = m*a = (1kg + 3kg)*(g/2) = 2g = 19.6N

    I replied that (1) T = 1kg*a only works if "a" encompasses the relative acceleration of Blue over Red AND that of Red over the ground. And (2) the horizontal force between Blue and Red comes by virtue of the normal force on the pulley.

    Who is right?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Oct 23, 2009 #2


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    I think the argument goes this way.
    To keep the Yellow cart suspended, you need to accelerate it vertically by g.
    This means that the Blue cart must provide its own 'weight' of force on the string. You can achieve this by accelerating the big cart forwards at g; the tension will be just enough to achieve this and, at the same time, keep the yellow cart suspended. Of course, you are also accelerating the dangling cart forwards at g but that is achieved by pushing the big cart. The total force needed must be the total weight of the three carts, because they are all accelerating at g.
    I think that makes you right and your friend wrong. Which is nice!
  4. Oct 23, 2009 #3
    Thanks, Sophie. I've talked to my two engineering roommates about it, and they agree that I'm right, but they emphasize that I'm making intuitive jumps that will hurt me if I use them for less clear problems.

    For the side cart, this is correct:
    T = 1kg*g - 1kg*a

    But for the top minicart, the horizontal acceleration is actually the acceleration relative to the cart plus the acceleration of the cart relative to the ground, which I'll call X.
    T = 1kg*(a + X)

    Also, we need to include the tension of the rope in the equation for F_p:
    F_p = (M_Blue + M_Yellow) * X + T

    ***The equation modifications result from the facts that the relative accelerations of Blue are additive, and that Red must have a normal force against the rope.***

    From here, the problem solves itself. The question simply asks, "What is F_p when a = 0?"
    That's easy:

    Find the acceleration:
    1kg*g - 1kg*a = 1kg*(a + X)
    1kg*g - 0 = 1kg*(0 + X)
    g = X

    Find the tension:
    T = 1kg (a + X)
    T = 1kg (0 + g)
    T = 1kg * g

    Plug it in:
    F_p = (M_Blue + M_Yellow) * X + T
    F_p = (3kg + 1kg) * g + 1kg * g
    F_p = 5kg * g = 49N
  5. Oct 23, 2009 #4


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    Sometimes, one man's intuitive jump is another man 's way into the problem.
    I think you are now making this particular problem more complex than necessary. As you are dealing with a stiuation of balance, the simple logic of tension in the rope produces equal moduli of acceleration on two equal small masses and the relative accerations of the top and large carts have to be equal because the initial problem demands it.

    In other situations, the full monty may be required but, here, you can afford to do the thinking before deciding which equations you need to write down.
    Maths is your slave and not your master.
  6. Oct 23, 2009 #5
    I dove into the math because my friend failed to see my intuitive claims, so I just decided to play ball on his court, so to speak.

    I did come up with a very interesting side conclusion:

    The acceleration of Red from force F_p N is not (1/5)(F_p)m/s/s.
    It's actually [(2/9)F_p - (49/45)]m/s/s.
    That means for the box just to stay in a neutral position as the mini-carts fall, you have to hold it at 49/10 newtons, or "half a g".
  7. Oct 23, 2009 #6


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    That's nice.
  8. Oct 23, 2009 #7
    Thanks. Glad you like it.
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