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?

1 Answer
Jun 10, 2018

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:

https://en.wikipedia.org/wiki/Inclined_plane

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"#