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Sample text

If the known weight force (6 newtons) is dmwn as a vector we can scale off the magnitude of the unknown support reactions. 17 presents Stevin's solution, virtually identical to the modem anal­ ysis in the preceding diagram. Stevin arrived at this l-esult by deep insight and logical reasoning. All we did was to fOllow a set of rules, step by step. No effurt of imagination and very little thought was required. Moreover; all prob­ lems in statics are attacked in this same way. Does Stevin's little beam-and-pyramid problem seem far removed from practical needs?

This requirement applies only in the special case of a three-force body. The general condition of rotational equi­ librium is simply that the resultant moment of the forces be equal to zero for any point. '1at there is no couple acting on the body and that it is in rotational equilibrium. Consider, for example, the body in Fig. 12. There is a weight force of 6 newtons acting at point T and, together with the loads hanging from it, this is balanced about the fulcrum at point X. - T turn. What we have just found applies to any number of forces.

1 Y(t3 The sum of the moments of the forces acting on a body in equilib­ rium is zero about any point. The weight force of 6 newtons pulling down at point T is not shown. 12 THE FREE-BODY DIAGRAM tant moment about point R is equal to 6 x 2 + 1 X 3 - 20 X 4 The answer to the question posed at the end of the preceding section is noW evident. The horizontal beam with vertical forces acting on it (Fig. 1) is subject to the same conditions of equilibrium as any other body. The sum of the moments about any point (we took moments about point R) must be zero.

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