假設,
a block of mass 2kg is placed on a plane inclined at an angle 30° to the
horizontal. the block slides down the plane with an acceleration of 3ms^-2.
a)what is the friction between the block and the plane?
b)what is the normal reaction on the block exerted by the plane?
應如何計算?
力學和摩擦力的數如何計算?
2010-01-04 8:04 pm
回答 (2)
2010-01-04 9:52 pm
✔ 最佳答案
a)Take g = 10 m s^2
The weight of the block = 2(10) = 20 N
The force diagram is shown below:
http://i320.photobucket.com/albums/nn347/old-master/100104_1.jpg?t=1262583827
圖片參考：http://i320.photobucket.com/albums/nn347/old-master/100104_1.jpg?t=1262583827
Along the inclined plane:
F = ma
20•sin30^o - f = 2(3)
f = 4 N
The frictional force between the block and the plane = 4 N
b)
Normal reaction on the block exert by the plane, R
= mg•cos30^o
= 2(10)•cos30^o
= 17.3 N
2010-01-04 9:55 pm
a)By F=ma
a=3ms^-2 (given)
m=2kg (given)
g=10ms^-2 (earth)
f=?
mg sin 30-f=ma
2*10*sin 30-f=2*3
f=4N
b)By F=ma
a=0 (the block does not move perpendicularly to the plane)
m=2kg (given)
g=10ms^-2 (earth)
R=?
mg cos 30-R=ma
2*10*cos30-R=2*0
R=17.3N (corr to 3 sig fig)
a=3ms^-2 (given)
m=2kg (given)
g=10ms^-2 (earth)
f=?
mg sin 30-f=ma
2*10*sin 30-f=2*3
f=4N
b)By F=ma
a=0 (the block does not move perpendicularly to the plane)
m=2kg (given)
g=10ms^-2 (earth)
R=?
mg cos 30-R=ma
2*10*cos30-R=2*0
R=17.3N (corr to 3 sig fig)
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https://hk.answers.yahoo.com/question/index?qid=20100104000051KK00475