如题所述
. ΣF =0, 即:F4-40N-20N+27N=0, 得:F4=33N ( ↑ ),
. 设F4作用线到O点距离为x,
. ΣMo =(F4)(x)+(27N)(7.5m)-(20N)(2.5m)-(40N)(0m)=300Nm,
. 将F4=33N代入上式得:x ≈4.47 (m)