F23=F32= 9293 4T8a2 =67.5N 9193 F3=F31=4元ea2 =67.5N 9x109x1义1×10 4×10-4 =22.5N F=J(Fg)2+(F232-2F3Fc0s60°=59.5N F2=(F22+(F2)2-2F12F23c0s60°=67.5N F3=J(F)2+(F2-2F1F2Cos60°=81.1N q受力最大 2)F2=67.5N 0= =60° 2 题号结束π a 4ε 2 0 F23 = F32 = q2 q3 =67.5N π a 4ε 2 0 F13 = F31 = q1 q3 =67.5N = 9×109× =22.5N 1×1×10-12 4×10-4 2 60 59.5 N 2 0 F1 = (F13) + F13 F23 = 2 (F ) 23 cos q3受力最大 (2)F2 =67.5 N 2 60 81.1 N 2 0 F3 = (F31) + F31 F32 = 2 (F ) 32 cos 2 60 67.5 N 2 0 F2 = (F12) + F12 F23 = 2 (F ) 23 cos a q = = 2 600 题号 结束