Pancake collapse of a tall building. In the section of a tallbuilding shown in Fig. 9-71a, the infrastructureof any given floor Kmust support the weight Wof allhigher floors. Normally the infrastructureis constructed with asafety factor sso that it can withstandan even greater downward force of sW. If, however, the support columns between Kand Lsuddenly collapse and allow the higher floors to free-fall together onto floorK(Fig. 9-71b), the force in the collision can exceed sWand, after a brief pause, cause Kto collapse onto floor J, which collapses on floor I, and so on until the ground is reached. Assume that the floors are separated by d=4.0 mand have the same mass. Also assume that when the floors above Kfree-fall onto K, the collision last 1.5 ms. Under these simplified conditions, what value must the safety factor sexceed to prevent pancake collapse of the building?

Weight of the higher floors is W.

Safety factor of the infrastructure is s.

Distance through which floors are separated is,$d=4.0\mathrm{m}$

Collision time is, $t=1.5\times {10}^{-3}\mathrm{s}$

Using the formula for impulse in terms of velocity, average force in terms of impulse and the law of conservation of energy we can find the average force exerted on lower floors due to higher floors. Then comparing it with the given force in terms of s we can find the value of s.

Formula:

$$\begin{gathered} E=\mathrm{constant} \\ \mathrm{J}=\mathrm{P} \\ {\mathrm{F}}_{\mathrm{avg}}=\frac{\mathrm{J}}{\mathrm{t}} \end{gathered}$$

Let’s assume the mass of the higher floors than K is m. Therefore, its weight is,

$W=mg$

Impulse when they collapse is,

$$\begin{gathered} J=P \\ =mv \\ =m\left(v_f-v_i\right) \\ J=m{v}_f \\ =mv \end{gathered}$$

According to the law of conservation of energy,

$$\begin{gathered} E=\mathrm{constant} \\ \mathrm{mgd}=\frac{1}{2}{\mathrm{mv}}^2 \\ {\mathrm{v}}^2=2\mathrm{gd} \\ \mathrm{v}=\sqrt{2\mathrm{gd}} \end{gathered}$$

Hence, impulse duringtheimpact is

$J=m\sqrt{2gd}$

The average force acting on the lower floor due to these higher floors is,

$$\begin{gathered} F_{avg}=\frac{J}{t} \\ {F}_{avg}=\frac{m\sqrt{2gd}}{t} \end{gathered}$$

Let’s multiply and divide it by g,

$$\begin{gathered} F_{avg}=\frac{mg\sqrt{2d}}{gt} \\ {F}_{avg}=\frac{W\sqrt{{\displaystyle \frac{2d}{g}}}}{t} \end{gathered}$$

We are given that

$F_{avg}=sW$

Thus,

$$\begin{gathered} sW=\frac{W\sqrt{{\displaystyle \frac{2d}{g}}}}{t} \\ s=\frac{\sqrt{{\displaystyle \frac{2d}{g}}}}{t} \\ =\frac{\sqrt{{\displaystyle \frac{2\left(4\right)}{9.8}}}}{1.5\times {10}^{-3}} \\ =602 \\ =6.0\times {10}^2 \end{gathered}$$

Therefore, the value that safety factor s must exceed to prevent pancake collapse of the building is equal to $6.0\times {10}^2$