As shown in the figure a block of mass 10kg lying on a horizontal...

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

$\begin{cases}N & = & mg - Fsin30 ° \\ & = & mg - \dfrac{F}{2} \\ & = & 100 - \dfrac{F}{2} \\ & = & \dfrac{200 - F}{2} . . . . . . . . . . . . . . . . . . . . . . . . ( 1 )\end{cases}$

From the equilibrium of horizontal component of forces, it can be written that,

Fcos30 ° = \muN . . . . . . . . . . . . . . . . . . . . . . ( 2 )$</p><p>Substitute the expression for the normal reaction force from equation (1) into equation (2) and solve to calculate the value of the applied force.</p><p>$\begin{cases}\sqrt{3}\dfrac{F}{2} & = & 0 . 25 \times \left(\dfrac{200 - F}{2}\right) \\ 4\sqrt{3}F & = & 200 - F \\ F & = & \dfrac{200}{4\sqrt{3} + 1} \\ & \approx & 25 . 2\end{cases}$