A block of mass 5 kg is placed on a rough inclined plane which makes an angle of 37^ with the horizontal. A purely horizontal force F (parallel to the ground) pushes the block towards the incline. The coefficient of static friction between the block and the incline is 0.5 . What is the minimum horizontal force F required to just start moving the block up the incline? [Take g = 10 m/s ^2 , 37^ = 3 5 , 37^ = 4 5 ]

Step-by-step solution

Let us resolve the forces parallel and perpendicular to the inclined plane. The weight mg has a component mg 37^ down the incline and mg 37^ perpendicular to the incline (downwards). The horizontal force F has a component F 37^ up the incline and F 37^ perpendicular to the incline (downwards, pressing the block into the surface). The normal reaction N is: N = mg 37^ + F 37^ N = (5 10 0.8) + F(0.6) = 40 + 0.6F The net driving force pushing the block up the incline is: F_ up = F 37^ - mg 37^ F_ up = F(0.8) - (5 10 0.