作者haowei37 (Hao)
看板NTUMEB98
标题Re: [问题] A班静力学的题目
时间Mon Nov 2 17:29:51 2009
5.142
r=CB(0.3125,0,-1)
R=CA(0.5,0,-1.6)
F1=Fd/2.18(-1.5625,0.875,1.25)
F2=Fe/2.18(-1.5625,-0.875,1.25)
F3=(0,67.5,42.75)
M1=r X F1=(0.3125,0,-1)XF1
=Fd/2.18(0.875,1.1718,0.273)
M2=r X F2=Fe/2.18(-0.875,1.1718,-0.273)
M3=R X F3=(108,-21.37,33.75)
M1+M2+M3=0
Fd/2.18*1.1718+Fe/2.18*1.1718+(-21.37)=0
Fd+Fe=39.76
Fd/2.18*0.875+Fe/2.18*(-0.875)+108=0
Fd-Fe=-269.51
Fd=-114.88
Fe=154.64
感谢B班林政贤提供XD
--
※ 发信站: 批踢踢实业坊(ptt.cc)
◆ From: 140.112.7.59
1F：推 sirius733:漂亮喔=D 11/02 18:13
2F：推 physicsdk:水啦 推一个! 11/02 19:31
3F：推 thatismyway:我感动到哭了 11/02 21:13