Question 13.154: In order to test the resistance of a chain to impact, the ch...

\begin{aligned}T_{1} & =0 \quad V_{1}=W h=(60 \mathrm{\ lb})(5 \mathrm{\ ft})=300 \mathrm{\ lb} \cdot \mathrm{ft} \\T_{2} & =\frac{1}{2} m \nu^{2} \quad V_{2}=0 \\T_{1}+V_{1} & =T_{2}+V_{2} \\0+300 & =\frac{1}{2}\left\lgroup\frac{60}{g}\right\rgroup \nu^{2} \\\nu & =\sqrt{\frac{(600)(32.2)}{60}} \\& =17.94 \mathrm{\ ft} / \mathrm{s}\end{aligned}

(a) Rigid columns:

\begin{gathered}+↑ -m \nu+F \Delta t=0 \quad\left\lgroup\frac{60}{g}\right\rgroup(17.94)=F \Delta t \\F \Delta t=33.43 \mathrm{\ lb} \cdot \mathrm{s} \uparrow \text { on the block. } && F \Delta t=33.4 \mathrm{\ lb} \cdot \mathrm{s}\blacktriangleleft\end{gathered}

All of the kinetic energy of the block is absorbed by the chain.

\begin{aligned}T & =\frac{1}{2}\left\lgroup\frac{60}{g}\right\rgroup(17.94)^{2} \\& =300 \mathrm{\ ft} \cdot \mathrm{lb} && E=300 \mathrm{\ ft} \cdot \mathrm{lb}\blacktriangleleft\end{aligned}

(b) Elastic columns:

Momentum of system of block and beam is conserved.

\begin{aligned}& m \nu=(M+m) \nu^{\prime} \\& \nu^{\prime}=\frac{m}{(m+M)} \nu=\frac{60}{300}(17.94 \mathrm{\ ft} / \mathrm{s}) && \nu^{\prime}=3.59 \mathrm{\ ft} / \mathrm{s}\end{aligned}

Referring to figure in part (a),

\begin{aligned}-m \nu+F \Delta t & =-m \nu^{\prime} \\F \Delta t & =m\left(\nu-\nu^{\prime}\right) \\& =\left\lgroup\frac{60}{g}\right\rgroup(17.94-3.59) && F \Delta t=26.7\mathrm{\ lb}⋅\mathrm{s}\blacktriangleleft\\E & =\frac{1}{2} m \nu^{2}-\frac{1}{2} m \nu^{\prime 2}-\frac{1}{2} M \nu^{\prime 2} \\& =\frac{60}{2 g}\left[(17.94)^{2}-(3.59)^{2}\right]-\frac{240}{2 g}(3.59)^{2} &&\quad E=240\mathrm{\ ft}⋅\mathrm{lb}\blacktriangleleft\end{aligned}