Equilibrium under parallel forces

AI Thread Summary
The discussion focuses on determining the maximum overhang, x, of brick C on a table while maintaining equilibrium for three identical bricks A, B, and C. The analysis involves calculating torques around specific points, particularly the edge of the table and the bottom-left corner of brick B. The derived equations suggest that the maximum value of x is 16.8 cm. There is a consideration of horizontal forces acting on brick A due to brick C's slight overturning, although friction is not mentioned in the problem. The solution emphasizes the importance of torque balance for maintaining equilibrium in this scenario.
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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?

Y-1.jpg



Any idea?
 
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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.
 
Oops thanks.. Sorry for not following the template. :redface:

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.
 
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.
 

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