口例题 §5静力学基本概念 例题3 H: M(F, F2)=MA(ED=MC(F2) AC×F1=CA×F2 C(O0) AC=-ai+ck B(0,b,0 F=Fsin ay-Fcos ak sin a (F,F2) cos C 5 A(a,0,0 X M(F2F2)=AC×F1=F-a0 C 0 sin a -cos al =(-241-39j-40kkNm( , ) M F1 F2 例 题 3 § 5 静力学基本概念 例题 解: 1 2 1 2 1 2 ( , ) ( ) ( ) AC F CA F M F F M A F MC F = = = = AC ai ck = − + F F j F k 1 = sin − cos 0 sin cos ( , ) 0 1 2 1 − = = − a c i j k M F F AC F F = (−24i −39 j − 40k )k Nm 5 3 cos 5 4 sin = = O x y z C(0,0,c) B(0,b,0) A(a,0,0) F1 F2 5 4 3