Theory question 115
StaticsA uniform beam, XY, 4 m long and weighing 350 N rests on two pivots P and Q. It is kept in equilibrium by weights of 80 N attached at X and 1000 N attached at a point between P and Q such that it is 0.6 m from Q. If |XP| = 0.8 m and |PQ| = 2.2 m: (a) calculate the reactions at P and Q. (b) if the 1000 N weight is replaced with a 1200 N weight, at what point from Q should it be placed in order to maintain the equilibrium.
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Distances from X: P 0.8 m, centre 2 m, 1000 N at 2.4 m, Q 3.0 m.
(a) Moments about P: 2.2R_Q + 80(0.8) = 350(1.2) + 1000(1.6), so 2.2R_Q = 1956 and R_Q ≈ 889.09 N.
R_P = 80 + 350 + 1000 − 889.09 ≈ 540.91 N.
(b) Keeping the reaction at P the same (moments about Q): 1200d = 1000(0.6), so d = 0.5 m from Q (between P and Q). Then R_Q = 1630 − 540.91 ≈ 1089.09 N.
Take moments about one support to find the other reaction.