Chapter 17, Problem 17.137RP

To determine

(a)

The impact speed when the hammer hits the test specimen.

Answer to Problem 17.137RP

The impact speed when the hammer hits the test specimen is $18.2121\text{ft/s}$.

Explanation of Solution

Given information:

$\begin{array}{l}OA=15.25in\\ OH=36.48in\\ M_{arm}=19.5lbs\\ M_{hammer}=71.2lbs\\ I_{arm}=47.1slug.i{n}^2=1515.398lb.i{n}^2\\ I_{hammer}=20.9slug.in=672.4976lb.i{n}^2\\ g=32.174ft/s^2=386.088in/s^2\end{array}$

Initial energy of the system is calculated as follows:

$\begin{array}{c}{E}_i=m_{A}g\left[OH+OA\sin 3g\right]+m_{h}g\left[OH+OH\sin 3g\right]\\ =\left[19.5\left[36.48+15.25\sin 3g\right]+1.2\left[36.48+36.48\sin 3g\right]\right]\\ =\left[898.504+4231.98\right]\left(386.088\right)\\ =1980809.748lb.i{n}^2/s^2\end{array}$

Final energy of the system is calculated as follows:

$\begin{array}{c}{E}_f=m_{A}g\left[OH-OA\right]+\frac{1}{2}I{\omega }^2\\ =19.5\left(36.48-15.25\right)386.088+\frac{1}{2}I{\omega }^2\\ =159834.64+\frac{1}{2}I{\omega }^2\end{array}$

Apply the conservation of energy as follows:

$\begin{array}{c}{E}_i=E_f\\ 1980809.748=159834.6407+\frac{1}{2}\left(101475.08\right){\omega }^2\\ \omega =5.990834\text{rad/s}\end{array}$

The impact speed when the hammer hits the test specimen is calculated as follows:

$\begin{array}{c}V=\omega \left(OH\right)\\ V=5.990834\times 36.48\\ V=218.5458\text{in/s}\\ V=18.2121\text{ft/s}\end{array}$

Thus, the impact speed when the hammer hits the test specimen is $18.2121\text{ft/s}$.

Conclusion:

The impact speed when the hammer hits the test specimen is $18.2121\text{ft/s}$.

To determine

(b)

Find the force on the pin O

Answer to Problem 17.137RP

Force at the point O before impact is $268.87\text{lb}$.

Explanation of Solution

Moment of inertia about “O” is calculated as follows:

$\begin{array}{l}{I}_o=\left[I_{arm}+M_a{\left(OA\right)}^2\right]+\left[I_{hammer}+M_h{\left(OH\right)}^2\right]\\ I_o=\left[1515.398+\left(19.5\right){\left(15.25\right)}^2\right]+\left[672.4376+71.2{\left(36.48\right)}^2\right]\\ I_o=101475.0808\text{lb}{\text{.in}}^{\text{2}}\end{array}$

Free body diagram of the system is shown below:

Applying newtons second law in rotation as follows:

$\begin{array}{c}{m}_{A}g\times 15.25\cos \theta +m_{H}g\times 36.48\cos \theta =I_o\alpha \\ 19.5\times 15.25\cos 39°+71.2\times 36.48\cos 39°=\frac{101475.0808}{12}\times \alpha \\ \alpha =0.266rad/s^2\end{array}$

Center of gravity of whole system is calculated as follows:

$\begin{array}{c}r=\frac{m_H{r}_1+m_A{r}_2}{m_H+m_A}\\ r=\frac{71.2\times 36.48+19.5\times 15.25}{71.2+19.5}\\ r=31.915\text{in}\end{array}$

Apply Newton’s second law of motion before impact in horizontal direction as follows:

$\begin{array}{l}{R}_H=\left(m_H+m_A\right)r\alpha \\ R_H=\left(\frac{71.2}{32.2}+\frac{19.5}{32.2}\right)\times \frac{31.915}{12}\times 0.266\\ R_H=2\text{lb}\end{array}$

Apply Newton’s second law of motion before impact in vertical direction as follows:

$\begin{array}{l}{R}_V=\left(m_H+m_A\right)r\times {\omega }^2\\ R_V=\left(\frac{71.2}{32.2}+\frac{19.5}{32.2}\right)\times \frac{31.915}{12}\times {\left(5.990834\right)}^2\\ R_V=268.86\text{lb}\end{array}$

Force at the point O before impact is calculated as follows:

$\begin{array}{l}R=\sqrt{R_H{}^2+R_V{}^2}\\ R=\sqrt{{\left(2\right)}^2+{\left(268.86\right)}^2}\\ R=268.87\text{lb}\end{array}$

Thus, Force at the point O before impact is $268.87\text{lb}$.

To determine

(c)

The amount of energy that test specimen absorb.

Answer to Problem 17.137RP

Amount of energy that test specimen absorb is $1978089.57\text{lb}{\text{.in}}^{\text{2}}{\text{/s}}^{\text{2}}$.

Explanation of Solution

Initial energy of the system is

$\begin{array}{c}{E}_i=m_{A}g\left[OH+OA\sin 39°\right]+m_{h}g\left[OH+OH\sin 39°\right]\\ =\left[19.5\left[36.48+15.25\sin 39°\right]+71.2\left[36.48+36.48\sin 39°\right]\right]\\ =\left[898.504+4231.98\right]\left(386.088\right)\\ =1980809.748\text{lb}{\text{.in}}^{\text{2}}{\text{/s}}^{\text{2}}\end{array}$

Final energy of the system is calculated as follows:

$\begin{array}{l}{E}_f=m_{A}g\left[OA\sin 70°\right]+m_{h}g\left[OH\sin 70°\right]\\ E_f=\left[19.5\left[15.25\sin 70°\right]+71.2\left[36.48\sin 70°\right]\right]\\ E_f=\left(279.44+2440.735\right)\\ E_f=2720.175\text{lb}{\text{.in}}^{\text{2}}{\text{/s}}^{\text{2}}\end{array}$

Amount of energy that test specimen absorb is calculated as follows:

$\begin{array}{l}E=E_i-E_f\\ E=1980809.748-2720.175\\ E=1978089.57\text{lb}{\text{.in}}^{\text{2}}{\text{/s}}^{\text{2}}\end{array}$

Thus, amount of energy that test specimen absorb is $1978089.57\text{lb}{\text{.in}}^{\text{2}}{\text{/s}}^{\text{2}}$.