Equilibrium under parallel forces

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Homework Help Overview

The problem involves three identical bricks placed on the edge of a table, with a focus on determining the maximum overhang of one brick (C) while maintaining equilibrium. The subject area relates to mechanics, specifically the analysis of forces and torques in static equilibrium situations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss calculating torques for the bricks and the conditions for equilibrium. There is an exploration of moments about different points and the relationships between the forces acting on the bricks. Questions arise regarding the necessity of considering horizontal forces acting on the bricks.

Discussion Status

The discussion is active, with participants providing insights into the mechanics of the problem. Some guidance has been offered regarding the calculation of torques and the conditions for equilibrium, while questions about the relevance of additional forces are being explored.

Contextual Notes

There is a mention of the absence of a coefficient of friction in the problem statement, which may influence the considerations of forces acting on the bricks.

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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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