July 17, 1981, Kansas City: The newly opened Hyatt Regency is packed with people listening and dancing to a band playing favorites from the 1940s. Many of the people are crowded onto the walkways that hang like bridges across the wide atrium. Suddenly two of the walkways collapse, falling onto the merrymakers on the main floor.

The walkways were suspended one above another on vertical rods and held in place by nuts threaded onto the rods. In the original design, only two long rods were to be used, each extending through all three walkways (Fig. 5-24a). If each walkway and the merrymakers on it have a combined mass of M, what is the total mass supported by the threads and two nuts on (a) the lowest walkway and (b) the highest walkway?

Apparently, someone responsible for the actual construction realized that threading nuts on a rod is impossible except at the ends, so the design was changed: Instead, six rods were used, each connecting two walkways (Fig. 5-24b). What now is the total mass supported by the threads and two nuts on (c) the lowest walkway, (d) the upper side of the highest walkway, and (e) the lower side of the highest walkway? It was this design that failed on that tragic

night—a simple engineering error.

Step by step solution

01

Given information

$\mathrm{Mass}\mathrm{of}\mathrm{each}\mathrm{walkway}=\mathrm{M}$

02

To understand the concept

The problem is based on the concept of equilibrium of forces. A force has both a magnitude and a direction. If the magnitude and direction of the forces acting on an object are exactly balanced, then there is no net force acting on the object and the object is said to be in equilibrium.

Formula:

$\sum \mathbf{F}\mathbf{=}\mathbf{0}$

03

(a) To find the total mass supported by the threads and two nuts on the lowest walkway

Each nut supports$\frac{\mathrm{M}}{2}$ mass, hence total mass supported will be M .

04

(b) To find the total mass supported by the threads and two nuts on the highest walkway

Same thing here. Each nut supports mass $\frac{\mathrm{M}}{2}$independently, hence total mass supported will be M.

Free body diagrams of each walkway

05

(c) To find the total mass supported by the threads and two nuts on the lowest walkway after changing the design

From FBD I

${\mathrm{T}}_3+{\mathrm{T}}_3-\mathrm{Mg}=0$

Hence,

${\mathrm{T}}_3=\frac{\mathrm{Mg}}{2}$

$$\begin{gathered} \mathrm{Total}\mathrm{Weight}\mathrm{supported}={\mathrm{T}}_3+{\mathrm{T}}_3 \\ =\mathrm{Mg} \end{gathered}$$

Therefore,

localid="1656997385642" $\mathrm{Total}\mathrm{mass}\mathrm{supported}=\mathrm{M}$

06

(d) To find the total mass supported by the threads and two nuts on the upper side of the highest walkway

From FBD II

$2{\mathrm{T}}_2-2{\mathrm{T}}_3-\mathrm{Mg}=0$

By solving the above equation:

${\mathrm{T}}_2=\mathrm{Mg}$

localid="1656997266335" $$\begin{gathered} \mathrm{Total}\mathrm{Weigt}\mathrm{supported}={\mathrm{T}}_2+{\mathrm{T}}_2 \\ =2\mathrm{Mg} \end{gathered}$$

Therefore,

$\mathrm{Total}\mathrm{mass}\mathrm{supported}=2\mathrm{M}$

07

(e) To find total mass supported by the threads and two nuts on The lower side of the highest walkway

From FBD II

$2{\mathrm{T}}_1-2{\mathrm{T}}_2-\mathrm{Mg}=0$

By solving the above equation:

${\mathrm{T}}_1=\frac{3\mathrm{Mg}}{2}$

$$\begin{gathered} \mathrm{Total}\mathrm{Weight}\mathrm{supported}={\mathrm{T}}_1+{\mathrm{T}}_1 \\ =3\mathrm{Mg} \end{gathered}$$

Therefore,

$\mathrm{Total}\mathrm{mass}\mathrm{supported}=3\mathrm{M}$

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