Question

A block weighing `14 kg` is on a plane with an incline of `pi/6` and friction coefficient of `1/5`. How much force, if any, is necessary to keep the block from sliding down?

Answer 1

In order for the object to not slide down, a force of `F_0=45.75` Newtons must be applied upon it.

Explanation:

Let us visualise the situation:

To analyze the motion of the block, we must know about the forces which act on it.

First of all, the gravity force `G=mg` which we can break down into `G_x` and `G_y`, where `G_x` is parallel to the plane and `G_y` is perpendicular to it. Second of all, the normal force `N` and finally the force of friction `F_f`.

We ask: what force `F_0` must we apply on the block such that it remains motionless? We can assume this force to be on the same direction as `G_x` and `F_f`, parallel to the plane, and then apply Newton's Second Law:

`F=ma`

Where `F` is the sum of all the forces applied on a certain direction.

If the plane has an inclined `theta` and since the triangle formed by the vectors `vec G`, `vec G_y` and `vec G_x` is similar to the original triangle, we get the following relations:

`{(G_x = mgsintheta),(G_y = mgcostheta) :}`

Now, on the parallel direction to the plane, we have:

`F_("parallel")=-F_0 + G_x - F_f=ma`

The sign of the force is determined by whether it acts for or against motion. In our case, `F_0` must be opposing motion.

In order for the object to not move, the accerelation `a` must be zero, hence

`F_0=G_x - F_f`

Finding `F_f`; The friction force is calculated by the formula `F_f=muN`, where `mu` is the coefficient of friction. Since the object is not moving perpendiculary, it means that

`F_("perpendicular") = 0 = N-G_y=> N = G_y = mgcostheta`

If we take `g` to be `10`:

`N = 140cos(pi/6) =70sqrt3~~121.24 " N"`

`=> F_f = 1/5*121.24 ~~ 24.25 " N"`

Finding `G_x`;

`G_x = mgsintheta=140sin(pi/6) = 70 " N"`

Therefore, for the block to not move, we must apply a force of:

`F_0 =70-24.25=45.75 " N"`