if the coefficient of static friction is 0.4(Us=0.4)find the range of M' for which the system remains at rest.

Asked by abhilipsa satpathy | 14th Jun, 2013, 06:43: PM

Expert Answer:

normal force acting on the inclined plane due to 10 kg block  = 10.g.cos(37)
thus frictional force on the block = (0.4).10.g.cos(37)
hence the mass of 10 kg will start moving after:
1. Either mass m' is heavier enough to overcome this frictional force and pull the block up
    i.e m'>(0.4).10.cos(37)
2. Either the mass block slides down due to its own weight taking m' along with it
    i.e. m'g + 10.g.sin(37) > (0.4).10.g.cos(37)
thus range of m' can be calculated easily now.

Answered by  | 14th Jun, 2013, 10:45: PM

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