Four bricks of length L, identical and uniform, are stacked...

Four bricks of length $L,$ identical and uniform, are stacked on top of one another (Fig. $12-71 )$ in such a way that part of each extends beyond the one beneath. Find, in terms of L, the maximum values of (a) $a_{1},(\mathrm{b}) a_{2},(\mathrm{c}) a_{3},(\mathrm{d})$ $a_{4},$ and $(\mathrm{e}) h,$ such that the stack is in equilibrium, on the verge of falling.

Four bricks of length $L,$ identical and
uniform, are stacked on top of one
another (Fig. $12-71 )$ in such a way that
part of each extends beyond the
one beneath. Find, in terms of
L, the maximum values of
(a) $a_{1},(\mathrm{b}) a_{2},(\mathrm{c}) a_{3},(\mathrm{d})$
$a_{4},$ and $(\mathrm{e}) h,$ such
that the stack is in equilibrium, on the verge of falling.

Key Concepts

Sequential Stacking and Series Summation

In problems involving sequential stacking of objects, each additional object alters the system’s center of mass in a cumulative way. The maximum allowable overhang often results in a series summation, such as a harmonic series, which mathematically describes the progressive limits of displacement that maintain equilibrium. This concept ties together the principles of static equilibrium, center of mass, and torque in the context of step-by-step assembly.

Overhang in Stacked Structures

Overhang refers to how far an upper object extends beyond the supporting object below it. Maximizing overhang while preserving equilibrium requires careful consideration of the shifting center of mass and the resulting moments. This concept is a practical application of static equilibrium and center of mass principles in determining how far individual components can extend without compromising stability.

Torque and Moment Balance

Torque is the turning effect produced by a force acting at a distance from a pivot point. Moment balance requires that the sum of all torques about any chosen pivot point must be zero for a system in equilibrium. This principle is essential when analyzing the stability of stacked objects, as each brick’s contribution to the overall moment must be balanced to prevent rotation.

Static Equilibrium

This concept involves ensuring that a system is balanced, meaning that the net force and the net moment (or torque) acting on the system are zero. In the context of stacked objects, achieving static equilibrium means that the sum of forces and the sum of torques about any pivot must cancel out, ensuring that the structure neither translates nor rotates.

Center of Mass

The center of mass is the point where the total mass of the system can be considered to be concentrated. For stacked objects, the location of the combined center of mass relative to the base of support is crucial; if the center of mass lies outside the support area, the arrangement will tip over. This concept is fundamental in determining the limits of overhang in stability problems.

Transcript

00:01 For this problem on the topic of equilibrium and elasticity, we are given four bricks each of length l, which are all identical and uniform stacked on top of each other as shown in the diagram.

00:12 They are stacked in such a way that part of each extends beyond the one beneath.

00:17 We want to find, in terms of l, the maximum values for a1, a2, a3, and a4, and also h, that allows the stack to be in equilibrium and on the verge of falling.

00:31 So for part a, we know the center of mass of the top brick cannot be further to the right with respect to the brick below, which is brick 2, then all over 2.

00:40 Otherwise, its center of gravity is past any point of support and it will fall.

00:45 So we can see simply that a1 is equal to the length of the brick l divided by 2, l over 2.

00:54 And this is the maximum value it can take.

01:02 For part b, with brick 1, the top brick in the maximum situation, can the combined sense.

01:07 Center of mass of brick 1 and brick 2 is halfway between the middle of brick 2 and its right edge.

01:13 That point must be supported, so in the maximum case it is just above the right edge of brick 3.

01:19 And so a2 must equal l over 4.

01:31 Next we want to find the maximum value for a3.

01:37 Now the total center of mass of bricks 1, 2 and 3 is 1 third of the way between the middle of brick 3 and its right edge.

01:44 And we can show this by calculating the center of mass...

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