91Ó°ÊÓ

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, $t=1.5×{10}^{-3}\mathrm{s}$

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,

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

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