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    Particle A moves on a horizontal table and is joined to a particle B of equal mass, by a light inextensible string passing over a frictionless pulley on the edge of the table. Find the accelerations of A and B:
    (i) assuming that the table top is frictionless, and
    (ii) assuming it has a coefficient of friction u.
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    It would be nice if you asked politely for help, I'm not getting paid for this!

    Basically the resultant force pulling down = mg and the total mass of the system is 2m therefore the vertical acceleration = mg/2m = g/2

    as the string is inextensible this will also be the horizontal acceleration of B
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    By the way, this is a 2D problem (horizontal and vertical directions are involved in the problem).
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    (Original post by div curl F = 0)
    By the way, this is a 2D problem (horizontal and vertical directions are involved in the problem).
    I disagree. The problem is a single variable problem. There exists a co-ordinate system where the problem can be described only with one variable.
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    (Original post by Mehh)
    I disagree. The problem is a single variable problem. There exists a co-ordinate system where the problem can be described only with one variable.

    Whilst there may be a connection between the vertical coordinate of one mass with the horizontal coordinate of the other (effectively eliminating one of them), two variables are needed to completely describe the position of each of the two masses, i.e. the (x,y) coordinates, although the second variable may not be as interesting as the first since the horizontal surface will most probably be y = 0.
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    I think he was getting at that you could use a 1D co-ordinate system where the sole axis was one that ran along the table and then off the edge (i.e. always parallel to the string)
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    (Original post by div curl F = 0)
    Whilst there may be a connection between the vertical coordinate of one mass with the horizontal coordinate of the other (effectively eliminating one of them), two variables are needed to completely describe the position of each of the two masses, i.e. the (x,y) coordinates, although the second variable may not be as interesting as the first since the horizontal surface will most probably be y = 0.
    Concider the Euler Lagrande equations associated with the problem.
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    While the idea of using two variables and Euler Lagrande equations to calculate the problem is interesting, the title 1D Dynamics should indicate that I needed an explanation using 1D dynamics. The mathematics course that I am doing assumes no knowledge of multivariable calculus.
    For the imput that you have all put in, I do truely thank you for your time that you have spent in this debate.
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    (Original post by Mehh)
    Concider the Euler Lagrande equations associated with the problem.

    Theres no need to be considering lagrangians for such a simple dynamics question, constraints remove coordinate dependence but they don't remove the coordinate system. As I stated in my previous post, you need two coordinates to completely specify the position of any one mass in this problem; if the mass that starts on the horizontal table begins at (0,0) then its position sometime later will be (x,0) and if the string is of length l then the other mass starts at (l,0) and at some time later will be at (l,-y) (if you like your positive vertical axis to go upwards and positive horizontal axis to go to the right).
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    (Original post by div curl F = 0)
    Theres no need to be considering lagrangians for such a simple dynamics question, constraints remove coordinate dependence but they don't remove the coordinate system. As I stated in my previous post, you need two coordinates to completely specify the position of any one mass in this problem; if the mass that starts on the horizontal table begins at (0,0) then its position sometime later will be (x,0) and if the string is of length l then the other mass starts at (l,0) and at some time later will be at (l,-y) (if you like your positive vertical axis to go upwards and positive horizontal axis to go to the right).
    If you can remove a coordinate's dependence then surely you can find a different set of co-ordinates where only 1 will be needed (i.e. the one I suggested before)?

    For e.g, with those coordinates (take origin at the pulley) you would have one mass at +x and one at -x'
 
 
 
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