Past exam of the medicine course of the University of Oxford / 1867 / First examination / Physics / 1

If two parallel forces act upon a rigid body in opposite directions, state how to determine the magnitude and position of their single resultant.

Discuss the case in which the two forces become equal in magnitude.

Describe the arrangement and action of the muscles by which the motion of the eye-ball is produced.

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.