Page:Somerville Mechanism of the heavens.djvu/115

Chap II.]

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 is the direction of $$oz$$ 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 $$dt$$ 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{ds}{ds}; Y = \frac{dy}{ds}; Z = g-A\frac{dz}{ds} $

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

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

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

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