Question

A block of mass m=4kg is placed over a rough inclined plane as shown in figure. The coefficient of friction between the block and the plane is μ=0.6. A force F=10N is applied on the block at an angle of 30∘. The contact force between the block and the plane is

Answer 1

Force acting downwards = mg  

Components of mg = mgsinΘ parallel to the inclined plane and mgcosΘ    perpendicular to the inclined plane

so the Force should be greater than mgsinΘ

f > mgsinΘ

Answer 2

Answer:

Explanation:

Contact force= square root of (f)^2+(N)^2