A B is a balance, on which is supposed to hang at one end, B, the scale E, with a man in it, who is counterpoised by the weight W hanging at A, the other end of the balance. I say, that if such a man, with a cane or any rigid straight body, pushes upwards against the beam anywhere between the points C and B (provided he does not push directly against B), he will thereby make himself heavier, or overpoise the weight W, though the stop G G hinders the scale E from being thrust outwards from C towards G G. I say likewise, that if the scale and man should hang from D, the man, by pushing upwards against B, or anywhere between B and D (provided he does not push directly against D), will make himself lighter, or be overpoised by the weight W, which before did only counterpoise the weight of his body and the scale. If the common center of gravity of the scale E, and the man supposed to stand in it, be at _k_, and the man, by thrusting against any part of the beam, cause the scale to move outwards so as to carry the said common center of gravity to _k_ _x_, then, instead of B E, L _l_ will become the line of direction of the compound weight, whose action will be increased in the ratio of L C to B C. This is what has been explained by several writers of mechanics; but no one, that I know of, has considered the case when the scale is kept from flying out, as here by the post G G, which keeps it in its place, as if the strings of the scale were become inflexible. Now, to explain this case, let us suppose the length B D of half of the brachium B C to be equal to 3 feet, the line B E to 4 feet, the line E D of 5 feet to be the direction in which the man pushes, D F and F E to be respectively equal and parallel to B E and B D, and the whole or absolute force with which the man pushes equal to (or able to rise) 10 stone. Let the oblique force E D (= 10 stone) be resolved into the two E F and E B (or its equal F D) whose directions are at right angles to each other, and whose respective quantities (or intensities) are as 6 and 8, because E F and B E are in that proportion to each other and to E D. Now, since E F is parallel to B D C A, the beam, it does no way affect the beam to move it upwards; and therefore there is only the force represented by F D, or 8 stone, to push the beam upwards at D. For the same reason, and because action and reaction are equal, the scale will be pushed down at E with the force of 8 stone also. Now, since the force at E pulls the beam perpendicularly downwards from the point B, distant from C the whole length of the brachium B C, its action downwards will not be diminished, but may be expressed by 8 × BC; whereas the action upwards against D will be half lost, by reason of the diminished distance from the center, and is only to be expressed by 8 × B C/2; and when the action upwards to raise the beam is subtracted from the action downwards to depress it, there will still remain 4 stone to push down