Equilibrium under parallel forces

[th4450]

Three identical bricks A , B and C are placed on the edge of a table as shown. Each brick is of length 24 cm and C overhangs the table by x cm. What is the maximum value of x so that the system can still be in equilibrium?

Any idea?

 

[Delphi51]

Welcome to PF!
We aren't supposed to offer help until you post an attempt, but can make an exception on your first post. When it just begins to fall, C will turn ever so slightly with the edge of the table as fulcum. In turn, B will tip ever so slightly with its left end as fulcrum. You need to figure out the torques on B and C, which are related by the force B exerts on C.
Setting the torque for C equal to zero (just at the point of rotating), solve for x.

 

[th4450]

Oops thanks.. Sorry for not following the template.

So, first take moment at the bottom-left corner of B.
_CF_B(18) - mg_B(12) = 0
_CF_B = _BF_C = (2/3)(mg_B) (Action-reaction pair)

Then take moment at the edge of the table.
_BF_C(24-x) - mg_C(x-12) = 0
(2/3)(mg_B)(24-x) = mg_C(x-12)
∵mg_B = mg_C
∴x_max = 16.8

But is it necessary to consider the horizontal force acting on A by C? As C slightly overturns, it will push A slightly towards the left.

 

[Delphi51]

Looks good! The question doesn't mention a coefficient of friction, so I doubt if you are expected to include the forces of the brick pushing sideways.