6
Yes the correct ans is mg.
But according to the question "the total force the inclined plane exerts on the block" should be the reaction force i.e is N and since it is at rest hence N=mgcos@.
So,i am confused how the answer is mg.
plz help.
A block of mass m is placed at rest on an inclined plane of inclination @ to the horizontal.If the coefficient of friction between the block and the plane is U then the total force the inclined plane exerts on the block is :
(a)mg
(b)U mgcos@
(c)mg sin@
(d)U mgtan@
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UP 0 DOWN 0 0 4
4 Answers
1mg
as the force acting downwards is gravity . the frictional and normal forces will add up to this in opposite direction
6
1the force of friction is also applied by incline on the block
so net force on block will be:-
F=[(MG COS@)2 + (MG SIN@)2]1/2=MG
HERE 2 MEANS SQUARE AND 1/2 MEANS SQUARE ROOT