Please solve a problem for me
来源:百度知道 编辑:UC知道 时间:2024/05/31 09:10:50
一根木料的左端支在地面上,抬起它的右端时用力60N;把它的右端支在地面上,抬起它的左端时用力0N,则该木料重______N
(请写明过程,Thanks)
左端时用力40N
(请写明过程,Thanks)
左端时用力40N
设木头重力为G,长为L,重心距离左端x,则距右端的距离为L-x.当抬起右端时,左端可看做支点,可列式G·x=F1·L
当抬起左端时,右端为支点,可列式G·(L-x)=F2·L
两式可得G=F1+F2.G=100N
左端时0N???
用平均的思想,两个加起来的一半就是了。
这个没有过程的。
力的方向竖直向上的么?
如果是的话,要用力矩的知识来解答了
设木头重力为G,长为L,重心距离左端x,则距右端的距离为L-x
ps.楼上的909498说的是不对的,x和L并不是力臂
只能说是:G/F1=x/L
同理可得:G/F2=L/(L-x)
解得:GL=(40+60)L,所以G=100N
100N
100N
Please solve a problem for me
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