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Let an upward force $P$ act along $x=0$ and a downward parallel force $Q$ along $x=d>0$. An equivalent resultant force must preserve both total force and torque. Its signed vertical component is $R=P-Q$, and taking torques about the first line gives $R{x}_R=-Qd$. ThusIts magnitude is $|P-Q|$ and its direction agrees with the larger force. Its line lies outside the two original lines on the side of that larger force; for example, $P>Q$ gives $x_R<0$. The products of the original forces and their perpendicular distances from this line are equal.

If $P=Q\ne 0$ and $d\ne 0$, the total force vanishes but the torque is $-Pd$. No single force is equivalent to this system: it is a force couple of magnitude $Pd$. Its torque is independent of the reference point because translating that point changes the torque by the cross product of the shift and the zero resultant force. Equal opposite forces on the same line instead cancel completely.

The eye is rotated by six extraocular muscles. The rectus muscles approach it from behind; the oblique muscles have oblique lines of pull, with the superior oblique muscle tendon redirected by an orbital pulley. Attachments on the sclera give lever arms about the rotation centre. In the straight-ahead position their actions are as follows; vertical and torsional contributions vary with gaze direction.^[ref]

Coordinated contraction selects the desired rotation and balances unwanted components. This illustrates torque control; the full ocular geometry is not simply the parallel-force pair of the first calculation.

Take downward displacement as positive and gravitational acceleration $g$ as constant. In a vacuum, $mdv/dt=mg$, so with initial downward velocity $u$,For release from rest these become $v=gt$ and $s=g{t}^2/2$.

A raindrop encounters an increasing upward drag force as its speed grows. At terminal velocity, gravitational weight is balanced by drag force and the small buoyancy contribution:The drop remains in motion: zero acceleration does not imply zero velocity. In a quadratic drag force model, $F_{\mathrm{drag}}={\rho }_{\mathrm{air}}{C}_{D}A{v}^2/2$, giving $v_t=\sqrt{2(m-{\rho }_{\mathrm{air}}V)g/({\rho }_{\mathrm{air}}{C}_{D}A)}$. That formula assumes a fixed effective area and drag coefficient; sufficiently small drops or deforming large drops need an appropriate model. The question's “effective force” at settled terminal velocity is the approximately zero resultant force, not its nonzero gravitational weight.