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?

1 Answer
Aug 10, 2017

F_"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.

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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 mgsintheta

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

f = mun = mumgcostheta

The net force equation for the block is

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

We're given

  • m = 4 "kg"

  • theta = pi/2

  • mu = 1

  • and g = 9.81 "m/s"^2

Plugging these in:

sumF_x = (1)(4color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)cos[pi/2] - (4color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)sin[pi/2]

= 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^"o"), we must exert an applied force directed upward with magnitude color(blue)(39.24color(white)(l)"N".