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?

We're asked to find the necessary force (if such a force is necessary) needed to keep an object on an incline at rest.

![ds055uzetaobb.cloudfront.net]()

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"#.