A block is initially at rest on a rough inclined plane which makes an angle of 10' to the horizontal. When a force F of 25N parallel to the plane is applied on the block, the block moves at a constant speed.
Suppose the weight of the block is 60N, find the frictional force acting on the block by the inclined plane.
我個solution 就係60sin 10' = 10.4N
但個soultion 錯o左....而係25+60sin 10'
點解要加埋25N 個force...而唔係整係用sin 搵個frictional force....???
thx!!!
FRICTION 問題
2006-10-22 8:30 am
回答 (2)
2006-10-22 12:42 pm
✔ 最佳答案
Because fritional force always has the same magnitude as the net force, that is where you did wrong. In here, the net force is not 60N,
it is (weight)sin10 + 25N (you did not add this 25N force)
=> 60sin10 + 25N
2006-10-22 8:53 am
因為俾一個25N 既力個block
個平面係rough 既, 所以就會有個25N既friction
唔知岩唔岩
個平面係rough 既, 所以就會有個25N既friction
唔知岩唔岩
收錄日期: 2021-04-23 15:46:23
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061022000051KK00177