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% David R. Wilkins
% School of Mathematics, Trinity College, Dublin 2, Ireland
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% Trinity College, 2000.
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\centerline{\Largebf ON THE COMPOSITION OF FORCES}
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\centerline{\Largebf By}
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\centerline{\Largebf William Rowan Hamilton}
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\centerline{\largerm (Proceedings of the Royal Irish Academy,
2 (1844), pp.\ 166--168.)}
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\centerline{\largerm Edited by David R. Wilkins}
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\centerline{\largerm 2000}
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\centerline{\largeit On the Composition of Forces.}
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\centerline{{\largeit By\/}
{\largerm Sir} {\largesc William R. Hamilton.}}
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\centerline{Communicated November~8, 1841.}
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\centerline{[{\it Proceedings of the Royal Irish Academy},
vol.~2 (1844), pp.\ 166--168.]}
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The Chair having been taken, {\it pro tempore}, by the Rev.\ J.H.
Todd, D.D., V.P., the President communicated the following proof
of the known law of Composition of Forces.
Two rectangular forces, $x$ and $y$, being supposed to be
equivalent to a single resultant force~$p$, inclined at an
angle~$v$ to the force~$x$, it is required to determine the law
of the dependence of this angle on the ratio of the two component
forces $x$ and $y$.
Denoting by $p'$ any other single force, intermediate between $x$
and $y$, and inclined to $x$ at an angle~$v'$, which we shall
suppose to be greater than $v$; and denoting by $x'$ and $y'$ the
rectangular components of this new force~$p'$, in the directions
of $x$ and $y$, we may, by easy decompositions and
recompositions, obtain a new pair of rectangular forces, $x''$
and $y''$, which are together equivalent to $p'$, and have for
components
$$\eqalign{
x'' &= {x \over p} x' + {y \over p} y';\cr
y'' &= {x \over p} y' - {y \over p} x';\cr}$$
the direction of $x''$ coinciding with that of $p'$, but the
direction of $y''$ being perpendicular thereto. Hence,
$${y'' \over x''} = {x y' - y x' \over x x' + y y'};$$
that is,
$$\tan^{-1} {y'' \over x''}
= \tan^{-1} {y' \over x'} - \tan^{-1} {y \over x};$$
or finally,
$$f(v' - v) = f(v') - f(v),
\eqno ({\sc a})$$
at least for values of $v$,~$v'$, and $v' - v$, which are each
greater than $0$, and less than
$\displaystyle {\pi \over 2}$;
if $f$ be a function so chosen that the equation
$${y \over x} = \tan f(v)$$
expresses the sought law of connexion between the ratio
$\displaystyle {y \over x}$
and the angle~$v$. The functional equation~({\sc a}) gives
$$f(mv) = m f(v) = {m \over n} f(nv),$$
$m$ and $n$ being any whole numbers; and the case of equal
components gives evidently
$$f \left( {\pi \over 4} \right) = {\pi \over 4};$$
hence
$$f \left( {m \over n} {\pi \over 4} \right)
= {m \over n} {\pi \over 4},$$
and ultimately,
$$f(v) = v,
\eqno ({\sc b})$$
because it is evident, by the nature of the question, that while
$v$ increases from $0$ to
$\displaystyle {\pi \over 2}$,
the function~$f(v)$ increases therewith, and therefore could not
be equal thereto for all values of $v$ commensurate with
$\displaystyle {\pi \over 4}$,
unless it had the same property also for all intermediate
incommensurable values. We find, therefore, that for all values
of the component forces $x$ and $y$, the equation
$${y \over x} = \tan v
\eqno ({\sc c})$$
holds good; that is, the resultant force coincides {\it in
direction\/} with the diagonal of the rectangle constructed with
lines representing $x$ and $y$ as sides.
The other part of the known law of the composition of forces,
namely, that this resultant is represented also {\it in
magnitude\/} by the same diagonal, may easily be proved by the
process of the M\'{e}canique C\'{e}leste, which, in the present
notation, corresponds to making
$$x' = x,\quad
y' = y,\quad
x'' = p,$$
and therefore gives
$$p = {x^2 + y^2 \over p},\quad
p^2 = x^2 + y^2.$$
But the demonstration above assigned for the law of the
{\it direction\/} of the resultant, appears to Sir William
Hamilton to be new.
\bye