A box of mass 2 ~kg is placed on a inclined plane that makes 30^ with the horizontal. The coefficient of friction between the box and inclined plane is 0.2 . A force F is applied on the box perpendicular to incline to prevent the box from sliding down. The minimum value of F is (acceleration due to gravity =10 ~ms ⁻² )

Step-by-step solution

As, block has a tendency to slide downwards, friction acts in upward direction. Box will not slide if friction balances component of weight acting in downward direction. f m g Here, f= friction = N or f= [(m g )+F] As, applied force is perpendicular to surface of incline. Normal reaction, N= net perpendicular to surface force =m g +F Hence, by Eqs. (i) and (ii)we have aligned & (m g )+ F m g & or F m g - m g aligned here, aligned & m=2 ~kg & g=10 ~m / s ^2 aligned gathered = 30^ = 1 2 =0.2 gathered So, F ( 2 10 1 2