As shown in the figure a block of mass 10 kg lying on a horizontal surface is pulled by a force F acting at an angle 30 ° , with horizontal. For μ s = 0 . 25 , the block will just start to move for the value of F : [Given g = 10 m · s – 2 ]

Step-by-step solution

The free body diagram for the given scenario is shown below- From the equilibrium of the vertical component of forces, it can be written that N = m g - F   sin   30 ° = m g - F 2 = 100 - F 2 = 200 - F 2 . . . . . . . . . . . . . . . . . . . . . . . . ( 1 ) From the equilibrium of horizontal component of forces, it can be written that, F   cos   30 ° =   μ N . . . . . . . . . . . . . . . . . . . . . . ( 2 ) Substitute the expression for the normal reaction force from equation