Problem: Given an arbitrary number of weights suspended from arbitrary points of a lever whose fulcrum is also given, to find a power which will sustain these weights at a given point.
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Let there be given the weights A, B, and C on the lever DE (with its fulcrum at F), suspended from the points D, C, and H, and a point E at which the power must be applied---.Divide DC at K in such a way that DK is to KC as the weight B is to A; then divide KH at L so that KL is to LH as the weight C is to the weights B and A.As FE is to FL, make the sum of weights A, B, and C be to the power which must be placed at E---.
PROPOSITION XIV
Problem: To make a given power move an arbitrary number of weights at arbitrary places on a given lever.
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Let the given lever be DE, let the given weights be placed as above, and let A be 100, B, 50, and C, 30; and let the given power be 30.Find the point L as before; then divide LE at F in such a way that FE is to FL as 180 is to 30 (that is, as six is to one), and if F is the fulcrum, the power of 30 at E will sustain the weights A, B,- and C.Therefore between L and F take some point such as M, and make this the fulcrum---.
PROPOSITION XV