A block of wood weights 10 N and is resting on an inclined plank. The coefficient of friction is 0.7. The frictional forces that acts on the block when the plank is 30° inclined with the horizontal is

See fig. of the question,From figure, R=10cos⁡30∘=53Nand μR=10sin⁡30∘=5N(Note that μR=0⋅7×53=6⋅062NThis is maximum attainable frictional force. When the inclination of the plank is raised, then 10 sin θ increases. The block will remain in rest till 10 sin θ attains a value 6.062 N. After that the box will start sliding down the plank)