A block shown in this picture has a weight W=50kN and is acted upon by a force P to start its motion up the inclined plane.Given that the coefficent
of static friction between the block and the plane is u=0.4 Find the minimum value of P
圖片參考：http://i56.photobucket.com/albums/g175/chengqoo2000/111111111.jpg
friction problem
2011-01-15 3:02 am
回答 (2)
2011-01-15 6:04 am
✔ 最佳答案
resolve mg into componentsnormal to inclined plane (Normal Reaction R): mgsin25
along the inclined plane: mgcos25
to move the block, P >= friction + mgcos25 = 0.4(mgsin25) + mgcos25
so P (min) = 5.38*10^4 N
(friction = uR)
2011-01-14 22:06:47 補充：
sorry, the two values have been swapped, actually, it is:
normal to inclined plane (Normal Reaction R): mgcos25
along the inclined plane: mgsin25
to move the block, P >= friction + mgsin25 = 0.4(mgcos25) + mgsin25
so P (min) = 3.93*10^4 N
2011-01-15 4:00 am
Frictional force = 0.4 x (50.cos(25)) kN = 18.126 kNHence, for the block just starts to move,P = (50.sin(25) + 18.126) kN = 39.26 kN
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https://hk.answers.yahoo.com/question/index?qid=20110114000051KK01017