A block of mass 102 kg is prevented from sliding on a plane with an inclination angle 30° with the horizontal by applying a force of 750 N parallel to the inclined plane. Find the frictional force acting on the block, if the coefficients of static friction and kinetic friction between the block and the plane are 0. 4 and 0.3, respectively. 

Let P be the force that is applied to the block to prevent it from sliding down. Now from the FBD, we will be balancing the forces which are parallel and perpendicular to the wedge..
Net force along the wedge

$=\mathrm{P}-\mathrm{mgsin}\theta =750-500=250\mathrm{N}$

'Now this 250" " N force should be balanced by frictional force. But before that we need to check the value of limiting friction.
Here limiting friction
$\mathrm{F}={\mu }_{\mathrm{s}}\mathrm{mgcos}\theta$

$\Rightarrow \mathrm{F}=0.4\times 102\times 9.8\times \cos {30}^{\circ }=346\mathrm{N}$

As net external force is less than limiting friction, therefore friction on the body will be static friction of 250N.