By J. Duncan
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These are lettered Fig. 40 (b\ and the polygon of forces in Fig. 40 (c). to correspond with those for the derrick crane, and will be followed readily. is In Fig. 41 Forces acting at a point but not in the same plane. shown in outline a pair of sheer legs such as is used for moving heavy loads. Two legs AB and and are hinged at the ground rotating as a whole about the BC are jointed together at the top B, at A and C so as to be capable of The legs are line AC in the plan. supported in any given position by means of a back leg DB, which is FORCES ACTING AT A POINT 35 D capable of BD in the plan.
Balance, provided or, Thus, the forces P 15 P 2 P 3 , , etc. (Fig. 17), will P1 + Pf -Pt -P4 -Pi*a 2P = o, the interpretation being that the algebraic sum of all the forces of which one only is given as a type immediately after the symbol 2 must be equal to zero. Suppose in a given case it is found that the algebraic sum of the zero. is not forces may infer from this that a single force given may be substituted for the given forces without altering the effect. towards B positive, Thus, in Fig. 18, calling forces of sense from We A we have 2 + 3 + 5 _8-i=+i.
6()). Find the reactions of its supports. Taking moments about B we have EXAMPLE and is P = 2-765 Taking moments about Qx A tons. we have i6 = (2X2)+(ix5)+(fx Q = 1-484 tons.