A 10 kg block lies on a smooth plane inclined at 51°. What force parallel to the incline would prevent the block from slipping?

Answer 1

Hi Khushi,

The force parallel to the inclined is a component of the gravitational force that is acting straight downward. A simple google search (parallel force on inclined plane) will help you see that a little right triangle trigonometry is all that is needed to show that the parallel force is equal to mg sinθ, where m is the mass in kg, g is the acceleration due to gravity which is 9.80 m/s^2, and theta is the angle.

With this equation, the force needed to keep the block from slipping is a friction force acting up the plane with a magnitude of mg sinθ = (10 kg)(9.80 m/s^2) * sin 51 = 65.7 kg m/s^2 or 65.7 N (Newtons).

Hope this helps,

Jim