Question

A block of mass 2 kg lies on a rough inclined plane making an angle 30 with the horizontal. Coefficient of static friction between the block and the plane is 0.7. The frictional force on the block is

Answer 1

Answer: Frictional force will be mg sin∅

So Frictional force= 2g x $\frac{1}{2}$        {sin∅=30}

                              = g

Explanation: 1. Coefficent of staic friction> tan∅

                       then frictional force = mgsin∅

                       2. Coefficent of staic friction= tan∅

                       then frictional force = mgsin∅= umg cos∅  {u= coefficent of friction}

                      3. Coefficent of staic friction< tan∅

                       then frictional force = uN= umg cos∅     {N=Normal}