Page:Somerville Mechanism of the heavens.djvu/119

fig. 26

are two forces acting at every instant of its motion: the resistance of the air, which is always in a direction contrary to the motion of the particle; and the force of gravitation, which urges it with an accelerated motion, according to the perpendiculars $Ed,Cf$, &c. The resistance of the air may be resolved into three partial forces, in the direction of the three axes $ox,oy,oz$, but gravitation acts on the particle in the direction of $ox$ alone. If $A$ represents the resistance of the air, its component force in the axis $ox$ is evidently $-A{\frac {dx}{ds}}$; for if $Am$ or $ds$ be the space proportional to the resistance, then

$Am:Ec::A:A{\frac {Ec}{Am}}=A{\frac {dx}{ds}}$;

but as this force acts in a direction contrary to the motion of the particle, it must be taken with a negative sign. The resistance in the axes $oy$ and $oz$ are $-A{\frac {dy}{ds}},-A{\frac {dz}{ds}}$; hence if $g$ be the force of gravitation, the forces acting on the particle are

$X=-A{\frac {dx}{ds}};Y=-A{\frac {dy}{ds}};Z=g-A{\frac {dz}{ds}}$.

As the particle is free, each of the virtual velocities is zero; hence we have

${\frac {d^{2}x}{dt^{2}}}=-A{\frac {dx}{ds}};{\frac {d^{2}y}{dt^{2}}}=-A{\frac {dy}{ds}};{\frac {d^{2}z}{dt^{2}}}=g-A{\frac {dz}{ds}}$;

for the determination of the motion of the projectile. If $A$ be eliminated between the two first, it appears that

${\frac {d^{2}x}{dt}}.{\frac {dy}{dt}}={\frac {d^{2}y}{dt}}.{\frac {dx}{dt}}$ or $dlog{\frac {dx}{dt}}=dlog{\frac {dy}{dt}}$;