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If the coefficient of friction between M and the inclined surfaces is mu = 1//sqrt(3) find the minimum mass m of the rod so that the of mass M = 10 kg remain stationary on the inclined plane |
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Answer» SOLUTION :Angle of repose ` phi = tan^(-1)((1)/(sqrt(3)))= 30^(@)` Here ` theta gt phi` HENCE , the block has a tendency to slide down  `Mg cos theta+ N' , N = mg cos theta` `N' = Mg cos theta + mg cos theta`….(i) For the block to remain stationary . `f = Mg sin theta`...(ii) If f is static as nature `f LT f_(max)` `Mg sin theta lt mu N'` `Mg sin theta lt mu(Mg cos theta + mg cos theta)` `Mg sin theta - mu mg cos theta lt mu mg cos thetan` `m gt M((tan theta)/(mu) -1) rArr m gt 10((sqrt(3))/(1//sqrt(3)) - 1) rArr m gt 20 kg` 
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