# Equilibrium under parallel forces

1. Feb 18, 2012

### 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?

2. Feb 18, 2012

### 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.

3. Feb 18, 2012

### th4450

Oops thanks.. Sorry for not following the template.

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

Then take moment at the edge of the table.
BFC(24-x) - mgC(x-12) = 0
(2/3)(mgB)(24-x) = mgC(x-12)
∵mgB = mgC
∴xmax = 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.

4. Feb 19, 2012

### 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.