# A block weighing 4 kg is on a plane with an incline of (pi)/2 and friction coefficient of 1. How much force, if any, is necessary to keep the block from sliding down?

Aug 10, 2017

${F}_{\text{applied" >= color(blue)(39.24color(white)(l)"N}}$ directed up the incline.

#### Explanation:

We're asked to find the necessary force (if such a force is necessary) needed to keep an object on an incline at rest. If the object is at rest, then it is in equilibrium, and the net force on it is zero.

Let's take a look at the forces acting on the block parallel to the ramp (which I'll call the $x$-axis):

• gravitational force (acting downward), equal to $m g \sin \theta$

• friction force (directed upward because it opposes the direction of sliding), equal to

$f = \mu n = \mu m g \cos \theta$

The net force equation for the block is

ul(sumF_x = overbrace(mumgcostheta)^"upward force" - overbrace(mgsintheta)^"downward force"

We're given

• $m = 4$ $\text{kg}$

• $\theta = \frac{\pi}{2}$

• $\mu = 1$

• and $g = 9.81$ ${\text{m/s}}^{2}$

Plugging these in:

$\sum {F}_{x} = \left(1\right) \left(4 \textcolor{w h i t e}{l} {\text{kg")(9.81color(white)(l)"m/s"^2)cos[pi/2] - (4color(white)(l)"kg")(9.81color(white)(l)"m/s}}^{2}\right) \sin \left[\frac{\pi}{2}\right]$

= color(red)(ul(-39.24color(white)(l)"N"

That is, the net force acting on the object is color(red)(39.24color(white)(l)"newtons" directed downward.

So, to prevent the object from sliding (or falling in this case, because the angle is ${90}^{\text{o}}$), we must exert an applied force directed upward with magnitude color(blue)(39.24color(white)(l)"N".