Question

A 5kg block rests on a 30° incline. The coefficient of static friction between the block and the incline is 0.20. How large a horizontal force must push on the block if the block is to be on the verge of sliding a)up the incline, b)down the incline ?

Answer 1

Let `F` be horizontal force applied on the block as shown above.This force will have two components

(1) `Fcos30^@` that will act upward parallel to inclined plane.

(2) `Fsin30^@` that will act on inclined plane perpendicularly. So it will be added with the component of the weight of the block `5gcos30^@` to enhance the magnitude of Normal reaction `N`
So the frictional force
`f_"fric"=muN=0.2(Fsin30^@+5gcos30^@)`,where acceleration due to gravity `(g) =9.8m"/"s^2`

Again the component of the weight of the block of mass 5kg acting downward parallel to the inclined plane is `5gsin30^@`

(a) Considering that the equilibrium of forces when the body is on the verge of sliding up the incline we can write

`Fcos30^@=5gsin30^@+0.2(Fsin30^@+5gcos30^@)`

`=>F=(5g(sin30^@+0.2cos30^@))/(cos30^@-0.2sin30^@)~~43N`

(b) Considering that the equilibrium of forces when the body is on the verge of sliding down the incline, we can write

`Fcos30^@=5gsin30^@-0.2(Fsin30^@+5gcos30^@)`

`=>F=(5g(sin30^@-0.2cos30^@))/(cos30^@+0.2sin30^@)~~16.6N`