the scale; because 8 × B C - 8 × B C/2 = 4BC. Consequently, a weight of 4 stone must be added at the end A to restore the æquilibrium. Therefore a man, &c., pushing upwards under the beam between B and D, becomes heavier. Q. E. D. On the contrary, if the scale should hang at F, from the point D, only 3 feet from the center of motion C, and a post G G hinders the scale from being pushed inwards towards C, then, if a man in this scale F pushes obliquely against B with the oblique force above mentioned, the whole force, for the reasons before given (in resolving the oblique force into two others acting in lines perpendicular to each other) will be reduced to 8 stone, which pushes the beam directly upwards at B, while the same force of 8 stone draws it directly down at D towards F. But as C D is only equal to half of C B, the force at D, compared with that at B, loses half its action, and therefore can only take off the force of 4 stone from the push upwards at B; and consequently the weight W at A will preponderate, unless an additional weight of 4 stone be hanged at B. Therefore, a man, &c., pushing upwards under the beam between B and D, becomes lighter. The other problem presented by Rev. Desagulier is denominated by him "An Experiment explaining a Mechanical Paradox, that two bodies of equal weight suspended on a certain sort of balance do not lose their equilibrium by being removed, one farther from, the other nearer to, the center." The article concerning this problem is as follows: If the two weights P W hangs at the ends of the balance A B, whose center of motion is C, those weights will act against each other (because their directions are contrary) with forces made up of the quantity of matter in each multiplied by its velocity; that is, by the velocity which the motion of the balance turning about C will give to the body suspended. Now, the velocity of a heavy body is its perpendicular ascent or descent, as will appear by moving the balance into the position _a b_, which shews the velocity of P to be the perpendicular line _e a_, and the velocity of B will be the perpendicular line _b g_; for if the weights P and W are equal, and also the lines _e a_ and _b g_, their momenta, made up of _e a_ multiplied into W, and _b g_ multiplied into P, will be equal, as will appear by their destroying one another in making an equilibrium. But if the body W was removed to M, and suspended at the point D, then, its velocity being only _f d_, it would be overbalanced by the body P, because _f d_ multiplied into M would produce a less momentum than P multiplied into _b g_. [Illustration]