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例1 如图1,质量为m,长为l的均匀杆AB由系于杆两端的长也为l的两细线悬于O点,在B点挂质量为M的重物,求平衡时杆与水平面夹角θ为多少? 解 因为AO=BO=AB=l,所以 △ABO为等边三角形. 设C是杆AB重心,则有 AC=BC=l/2,且 OC⊥AB,OC=(3~(1/2))/2l.又过O的竖直线与过B的水平线交于K,所以 点I、C、K、B四点共圆,所以 ∠COK=∠ABK=θ.又将m与M看作整体,其重心必在过O的竖直线
Example 1 As shown in Fig.1, a uniform rod AB with a mass of m and a length of l is suspended at point O by two fine lines of l long at both ends of the rod, and weights of weight M are hung at point B. What is the angle θ between the rod and the horizontal plane? Solution Because AO = BO = AB = l, so △ ABO is an equilateral triangle. Let C be the rod AB center of gravity, then there is AC = BC = l/2, and OC ⊥ AB, OC = (3~(1/2))/2l. The vertical line passing through O crosses with the horizontal line passing through B at K, so the four points I, C, K, and B are co-circular, so ∠COK=∠ABK= θ. Also consider m and M as a whole, and the center of gravity must be in the vertical line of O