# A Problem in Dynamics

Dated 19th February, 1854
    AN inextensible heavy chain
    Lies on a smooth horizontal plane,
    An impulsive force is applied at A,
    Required the initial motion of K.
    Let ds be the infinitesimal link,
    Of which for the present we've only to think;
    Let T be the tension and T + dT
    The same for the end that is nearest to B.
    Let a be put, by a common convention
    For the angle at M 'twixt OX and the tension;
    Let V$_t$ and V$_n$ be ds's velocities,
    Of which V$_t$ along and V$_n$ across it is;
    Then ${\frac {V_{t}}{V_{n}}}$ the tangent will equal,
    Of the angle of starting worked out in the sequel.
    In working the problem the first thing of course is
    To equate the impressed and effectual forces.
    K is tugged by two tensions, whose difference dT
    [1] Must equal the element's mass into V$_t$.
    V$_n$ must be due to the force perpendicular
    To ds's direction, which shows the particular
    Advantage of using da to serve at your
    Pleasure to estimate ds's curvature
    For V$_n$ into mass of a unit of chain
    [2] Must equal the curvature into the strain.
    Thus managing cause and effect to discriminate,
    The student must fruitlessly try to eliminate,
    And painfully learn, that in order to do it, he
    Must find the Equation of Continuity.
    The reason is this, that the tough little element,
    Which the force of impulsion to beat to a jelly meant,
    Was endowed with a property incomprehensible,
    And was "given", the the language of Shop, "inextensible."
    It therefore with such pertinacity odd defied
    The force which the length of the chain would have modified,
    That its stubborn example may possibly yet recall
    These overgrown rhymes to their prosody metrical.
    The condition is got by resolving again,
    According to axes assumed in the plane.
    If then you reduce to the tangent and normal,
    [3] You will find the equation more neat tho' less formal.
    [4] The condition thus found after these preparations,
    When duly combined with the former equations,
    Will give you another, in which differential
    [5] (When the chain forms a circle), become in essentials
    No harder than those that we easily solve
    [6] In the time a T totum would take to revolve.
    Now joyfully leaving ds to itself, a—
    Ttend to the values of T and of a.
    The chain undergoes a distorting convulsion,
    Produced first at A by the force of impulsion.
    In magnitude R, in direction tangential,
    [7] Equating this R to the form exponential,
    Obtained for the tension when a is zero,
    It will measure the tug, such a tug as the "hero
    Plume-waving" experienced, tied to the chariot.
    But when dragged by the heels his grim head could not carry aught,
    [8] So give a its due at the end of the chain,
    And the tension ought there to be zero again.
    From these two conditions we get three equations,
    Which serve to determine the proper relations
    Between the impulse and each coefficient
    In the form for the tension, and this is sufficent
    To work out the problem, and then, if you choose,
    You may turn it and twist it the Dons to amuse.
    ${\frac {V_{t}}{V_{n}}}=\tan \beta =-{\frac {e^{(a_{1}-a)}-e^{-(a_{1}-a)}}{e^{(a_{1}-a)}+e^{-(a_{1}-a)}}}$
    1: $dT=mV_{t}ds$
    2: $mV_{n}=T{\frac {da}{ds}}$
    3: ${\frac {dV_{t}}{ds}}=V_{n}{\frac {da}{ds}}$
    4: ${\frac {d^{2}T}{ds^{2}}}=V_{n}{\frac {da}{ds}}=0$
    5: ${\frac {d^{2}T}{da^{2}}}=0$
    6: $T=C_{1}e^{a}+C_{2}e^{-a}$
    7: $R=C_{1}+C_{2}$
    8: $0=C_{1}e^{a_{1}}+C_{2}e^{-a_{1}}$
This work was published before January 1, 1929, and is in the public domain worldwide because the author died at least 100 years ago.
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