Question #3abf0
1 Answer
Explanation:
We're asked to find the acceleration of the elevator necessary for the tile to break loose, and in what direction (up or down) the elevator must do so.
Let's examine the forces acting on the loose tile:

the gravitational force (directed downward), equal to its apparent weight
#mg# 
the static friction force (directed upward, opposite to eventual motion) exerted by all four adjacent tiles, equal to
#f_s = mu_sn#
where

#mu_s# is given as#0.500# 
#n# is the normal force, given as#2.00# #"N"#
Plugging these in:
#f_s = (0.500)(2.00color(white)(l)"N") = 1.00color(white)(l)"N"#
Since there are four tiles that each exert this retarding friction force, the equivalent friction force exerted on the loose tile is
#1.00# #"N"# #xx 4 = 4.00color(white)(l)"N"#
The equation for the net force on the loose tile is given by
#sumF = overbrace(f_s)^"upward"  overbrace(mg)^"downward" = ma#
The mass is given as
#sumF = 4.00color(white)(l)"N"  (0.400color(white)(l)"kg")(9.81color(white)(l)"m/s"^2) = color(red)(ul(0.076color(white)(l)"N"# directed upward.
This makes sense, because this tells us that if the elevator is not accelerating, the tile remains in place because the static friction force is greater than the downward gravitational force.
The elevator must accelerate in the opposite direction as the gravitational force (upward) for the tile to break loose, because that increases the normal reaction force as part of its apparent weight. The minimal force needed is seen by
#f_s = m(g+a)#
That is to say, the friction force is equal to the apparent weight
Plugging in values, we have
#4.00color(white)(l)"N" = (0.400color(white)(l)"kg")(9.81color(white)(l)"m/s"^2 + a)#
#color(blue)(ulbar(stackrel(" ")(" "a = 0.19color(white)(l)"m/s"^2" "))# directed upward.