A particle of mass 0.5 kg lies on a rough plane inclined at an angle (alpha) to the horizontal, where sin(alpha) = (5/13).
The particle is acted on by a horizontal force of 8 N and is about to move up a line of greatest slope.
(a) Show that the value of the coefficient of friction between the particle and the plane is 0.71.
ANS: This part is alright. I got [(71/13)/(100/13)] = 0.71.
(b) Determine, with working, whether or not the particle will move when the force of 8 N is removed?????
ANS: I used F=ma. But then, can I allow the acceleration to be ZERO. Only then, I obtained Max Frictional Force = 5(sin alpha) = 5(5/13) = 25/13.
THEN, I see that Max Frictional Force < 0.71(Normal Contact Force = 60/13)
So, the particle does not move!
CAN I ALLOW THE ACCELERATION TO BE ZERO?????????????????????????
THAN KS