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Inclined plane and pulley: how to know the acceleration's direction

  1. Feb 9, 2015 #1
    1. The problem statement, all variables and given/known data
    A block of mass m1= 3.70 kg on a frictionless plane inclined at angle θ=30.0° is connected by a cord over a massless, frictionless pulley to a second block of mass m2=2.30 kg. What are (a) the magnitude of the acceleration of each block, (b) the direction of the acceleration of the hanging block, and (c) the tension in the cord?

    2. Relevant equations
    F=ma

    3. The attempt at a solution
    Block 1's acceleration's coordinates are:
    (vector)a (sinθ*g+T/m1, cosθ*g +N/m1)

    Block 2's acceleration's coordinates are:
    (vector)a' (0,T/m2-g)

    a and T are the unknowns.
    However, how can I "gather" ax and a'y if I don't know their signs?
    Indeed, I don't know if block 1 is sliding towards the bottom of the plane or if block 2 is falling. One of the accelerations has to be given a minus sign (since one of them will be pointing in the opposite direction of the unit vectors), but which one is it?
    I don't know if I am clear enough; I hope you understood my problem though.


    Thank you!! :)
     
  2. jcsd
  3. Feb 9, 2015 #2

    DEvens

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    You seem to have selected the coordinates for block 1 such that the "x" direction is parallel to the inclined surface. Good. Be sure you have all the signs correct so that forces that oppose actually do. You seem to have negative g meaning gravity opposes the tension for block 2. Make sure you do the correct thing for block 1.

    How much motion is there perpendicular to the inclined surface? What does that tell you about the net force in that direction?

    Conserve string. So something about block 1 has to be the same as something about block 2.
     
  4. Feb 9, 2015 #3
    Thank you for your answer! :)

    Oops, I did forget the signs indeed. Here's something better:
    Block 1's acceleration's coordinates are:
    (vector)a (-sinθ*g+T/m1, -cosθ*g +N/m1)
    Block 2's acceleration's coordinates are:
    (vector)a' (0,T/m2-g)

    Since the rope is non-stretchable, ax and a'y have the same magnitude, which is why I was talking about "gathering" them. But one of them is negative, and I don't know which one.
    Nevertheless, I've just understood that, mathematically speaking, it doesn't matter whether I use -sinθ*g+T/m1= -(T/m2-g) or -(-sinθ*g+T/m1)= T/m2-g. However, is it the right way to explain it?
     
    Last edited: Feb 9, 2015
  5. Feb 9, 2015 #4

    lightgrav

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    the hanging block has the same acceleration as the rope. Does the rope run parallel to the incline, or along the horizontal?
    (set these situations up with components || the acceleration, and perp the acceleration).
    sum the Force components first, individually.
     
  6. Feb 10, 2015 #5
    Thanks for you help lightgrav ;)
    The rope runs parallel to the incline.
    I don't understand your next question, however. I mentioned the acceleration's components above; is that what you are talking about?
     
  7. Feb 10, 2015 #6

    lightgrav

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    since the velocity and acceleration are parallel the rope, set "x" along the rope
    ... that coordinate will bend at the pulley, like the rope does.
    Don't skip so many steps ... add the Force components along the rope
    ... it is ∑F that equals ∑ma
     
  8. Feb 10, 2015 #7
    Are you suggesting using F=ma for both blocks at once?
    What do you mean by "the coordinate will bend at the pulley"? Is it mathematically consistent? I mean, I've never seen a coordinate system bending along the way.
     
  9. Feb 10, 2015 #8

    lightgrav

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    yeah, it is one system, with 2 masses that have the same speed and acceleration at every instant.
    Surely you've seen inclined coordinates along a ramp before (|| a) ... that's the natural way to do things.
     
  10. Feb 11, 2015 #9
    I am indeed familiar with titled coordinate systems. But what you mention would rather look like this, wouldn't it?
    http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0014/tablewpulleyFBD.gif

    This way of dealing with the problem doesn't seem very intuitive to me. I'm used to using one coordinate system per system and to project the vectors on this single system. I understand that using two systems here works and gives the right answer, but it doesn't feel "right". Could you help me get the intuition here?
     
    Last edited: Feb 11, 2015
  11. Feb 13, 2015 #10

    lightgrav

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    it is not 2 coordinate systems, it is actually only 1 coordinate (x), along the rope length ... it is really a 1-dimensional situation.
    you don't need a coordinate perpendicular to the rope (except to obtain Normal Force, to find friction Force if it is non-zero)
    since there is no motion nor acceleration in that direction.
    If you LIKE to solve multiple coupled simultaneous equations, keep using vertical and horizontal components in every situation;
    if you want straight-forward simple (clear) math, select a coordinate parallel to the acceleration (or || velocity, if a's direction is unknown)
    Atwood's machine (2 blocks connected over a pulley - balanced until a small extra mass added to the right side):
    do you really want to treat that as 2 sub-systems with opposite accelerations? What about the pulley's acceleration? (around!)
    it is clearly the small UNbalanced mg that drives this system to start moving ... it all moves together, in the same sense (clockwise).
    Intuition corresponds closely to natural language ("toward the heavy side", or "down the ramp") - you can write physics that way, too.
    And it will make more sense then, and probably be simpler also.
     
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