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Two blocks and a mass- and frictionless pulley

  1. Nov 1, 2013 #1
    The problem and solutions: http://i.imgur.com/hxlfbmZ.png

    I was able to solve the first two questions, but I guessed part C.

    For the first one, I thought of the two blocks as one unit (since there's no friction or air resistance). If they're both falling with the same acceleration, there shouldn't really be any tension in the rope. So the total work should be:

    [itex]W_{tot}=\int^{0.75}_{0}(32x)dx[/itex] J

    That's for the whole system. For parts A and B I did the above, but for one block a time.

    Part C on the other hand... I can't seem to wrap my head around it. In my opinion, since we have friction, the only force felt by the 20 N block should be the 12 N in the direction of movement, and the friction [itex]f_{k}=20*µ_{k}=6.5[/itex] N in the direction opposite motion. I then want the total work in this case to be:

    [itex]W_{tot}=\int^{0.75}_{0}(5.5x)dx[/itex] J

    This is, apparently, not correct. What am I doing wrong?!
     
    Last edited: Nov 1, 2013
  2. jcsd
  3. Nov 1, 2013 #2

    haruspex

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    No, there will be tension. With no tension the 20N block would not move, and the 12N block would be in free fall. Fortunately your reasoning did not depend on that.
    It will 'feel' the tension, which is less than 12N. (If it were 12N the 12N block would not descend.)
    Btw, you were given a static friction too. Your first step should be to check that the static friction is insufficient to keep things in place.
     
  4. Nov 1, 2013 #3
    Well, the static friction is [itex]f_{s}=0.5*20[/itex] N, which is just 10 N. The small block pulls on the large one with a force of 12 N (am I right in assuming that?), so everything should be set in motion.

    After that, the kinetic friction is all that works against the 12 N from the small block. Shouldn't the total force on the large block then be the 12 N minus the kinetic friction?

    And regarding parts A and B:

    haruspex, you said that:
    "No, there will be tension. With no tension the 20N block would not move, and the 12N block would be in free fall."

    Is that not what is happening, though? The 12 N block is falling freely, just that it has a 20 N block attached to it, so together they are falling freely (no friction, or any type of resistance). Work is then done on the two blocks as a whole, or on the separate blocks if one chooses to look at it that way?
     
  5. Nov 1, 2013 #4

    haruspex

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    No. Some of the 12N is being 'used' accelerating the 12N block. Just write out the free body equations for the two blocks and deduce the tension.
    No. If the 12N block were in free fall its acceleration would be g. But even without friction, the 12N force is having to accelerate (12+20)N/g kg, so the acceleration is only 12g/(12+20).
     
  6. Nov 1, 2013 #5
    So the total force on the small block is [itex]F_{2}=\frac{12}{9.82}*a[/itex], which is the same as [itex](12 - T)[/itex] N.

    This means that the tension in the rope is [itex]T = 12 - \frac{12}{9.82}*a[/itex]

    This force T is the force that sets the large block in motion (?). The total force on the large block is then [itex]F_{1} = T - 20*µ_{k} = T - 6.5 = 12 - \frac{12}{9.82}*a - 6.5 N[/itex]

    And if [itex]F_{1}=\frac{20}{9.82}*a[/itex], then [itex]a*\frac{20}{9.82}=5.5 - \frac{12}{9.82}*a[/itex]

    [itex]a(\frac{20}{9.82} + \frac{12}{9.82}) = 5.5[/itex]

    [itex]a = \frac{5.5}{(\frac{20}{9.82} + \frac{12}{9.82})}[/itex]

    Total force on the large block is then [itex]\frac{20}{9.82}(\frac{5.5}{(\frac{20}{9.82} + \frac{12}{9.82})})[/itex] and the total work is [itex]W_{total}=\int_{0}^{0.75}((\frac{20}{9.82}(\frac{5.5}{(\frac{20}{9.82} + \frac{12}{9.82})}))x)[/itex] dx ?
     
    Last edited: Nov 1, 2013
  7. Nov 1, 2013 #6

    haruspex

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    Yes. That gives the correct answer, right?
     
  8. Nov 2, 2013 #7
    When I calculate what I got there, I get the answer 0.96680 Joule, which is NOT correct. That's what I do not understand! The correct answer is 2.58 Joule. It makes no sense.
     
  9. Nov 2, 2013 #8

    haruspex

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    Sorry, I didn't notice this the first time:
    ##W_{total}=\int_{0}^{0.75}((\frac{20}{9.82}(\frac{5.5}{(\frac{20}{9.82} + \frac{12}{9.82})}))## x ##)dx##
    Where did that x come from?
     
  10. Nov 3, 2013 #9
    Ah! I mixed it up again, I didn't think of dx as a distance, so I added an extra x to get some kind of work to sum up. It all makes sense now, thank you.
     
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