Chapter 19.1, Problem 19.14P

(a)

To determine

The frequency $\left(f_n\right)$, amplitude $\left(x_m\right)$, and the maximum velocity $\left(v_m\right)$ of the resulting motion.

Answer to Problem 19.14P

The frequency $\left(f_n\right)$, amplitude $\left(x_m\right)$ and the maximum velocity $\left(v_m\right)$ of the resulting motion are $\underset{\_}{5.81\text{Hz}}$, $\underset{\_}{4.91\text{mm}}$, and $\underset{\_}{0.1791\text{m/s}}$ respectively.

Explanation of Solution

Given Information:

The mass $\left(m_1\right)$ of the electromagnet is 150 kg.

The mass $\left(m_2\right)$ of the scrap steel is 100 kg.

The spring constant (k) is $200\text{kN/m}$.

The value of acceleration due to gravity (g) is $9.81{\text{m/s}}^{\text{2}}$.

Calculation:

Show the electromagnet with cable and crane arrangement as in Figure (1).

Refer Figure (1), when the electromagnet is off, the tension in the cable is equal to the force due to the mass of the electromagnet.

Express the force balance equation for the first case.

$T_0=m_{1}g$ (1)

Here, $T_0$ is tension in the cable and $m_1$ is the mass of electromagnet.

Calculate the natural circular frequency $\left({\omega }_n\right)$ using the relation:

${\omega }_n=\sqrt{\frac{k}{m_1}}$

Substitute $\text{200}\text{kN}/\text{m}$ for k and 150 kg for $m_1$.

$\begin{array}{c}{\omega }_n=\sqrt{\frac{\text{200}\text{kN}/\text{m}}{150\text{kg}}}\\ =\sqrt{1,333.33}\\ =\text{36}\text{.515}\text{rad}/\text{s}\end{array}$

Calculate the natural frequency $\left(f_n\right)$ using the relation:

$f_n=\frac{{\omega }_n}{2\pi }$

Substitute $\text{36}\text{.515}\text{rad}/\text{s}$ for ${\omega }_n$.

$\begin{array}{c}{f}_n=\frac{\text{36}\text{.515}\text{rad}/\text{s}}{2\pi }\\ =5.81\text{Hz}\end{array}$

By referring the Figure 1, when the electromagnet is on, the tension in the cable is equal to the force due to the mass of the electromagnet and that due to mass of the scrap steel.

Express the force balance equation for the second case.

$T_0+k{x}_m=m_{1}g+m_{2}g$ (2)

Here, $x_m$ is amplitude of vibration.

Substitute Equation (1) in Equation (2).

$\begin{array}{l}{T}_0+k{x}_m=m_{1}g+m_{2}g\\ T_0+k{x}_m=T_0+m_{2}g\\ k{x}_m=m_{2}g\end{array}$ (3)

Calculate the amplitude $\left(x_m\right)$ of vibration:

Substitute $\text{200}\text{kN}/\text{m}$ for k, 100 kg for $m_2$, and $\text{9}\text{.81}\text{m}/{\text{s}}^{\text{2}}$ for g in Equation (3).

$\begin{array}{l}k{x}_m=m_{2}g\\ x_m=\frac{m_{2}g}{k}\\ x_m=\frac{\left(100\text{kg}\right)\left(\text{9}\text{.81}\text{m}/{\text{s}}^{\text{2}}\right)}{\text{200}\text{kN}/\text{m}}\\ x_m=4.905\times {10}^{-3}\text{m}\\ x_m=4.905\times {10}^{-3}\left(\text{m}\times \frac{{10}^3\text{mm}}{1\text{m}}\right)\\ x_m=4.91\text{mm}\end{array}$

Calculate the maximum velocity $\left(v_m\right)$ using the relation:

$v_m=x_m{\omega }_n$

Substitute $\text{36}\text{.515}\text{rad}/\text{s}$ for ${\omega }_n$ and $4.905\times {10}^{-3}\text{m}$ for $x_m$.

$\begin{array}{c}{v}_m=\left(4.905\times {10}^{-3}\text{m}\right)\left(\text{36}\text{.515}\text{rad}/\text{s}\right)\\ =0.1791\text{m}/\text{s}\end{array}$

Therefore, the frequency $\left(f_n\right)$, amplitude $\left(x_m\right)$, and the maximum velocity $\left(v_m\right)$ of the resulting motion are $\underset{\_}{5.81\text{Hz}}$, $\underset{\_}{4.91\text{mm}}$, and $\underset{\_}{0.1791\text{m/s}}$ respectively.

(b)

To determine

The minimum tension $\left(T_{\min }\right)$ which will occur in the cable during the motion.

Answer to Problem 19.14P

The minimum tension $\left(T_{\min }\right)$ which will occur in the cable during the motion is $\underset{\_}{491\text{N}}$.

Explanation of Solution

Given Information:

The mass $\left(m_1\right)$ of the electromagnet is 150 kg.

The mass $\left(m_2\right)$ of the scrap steel is 100 kg.

The spring constant (k) is $200\text{kN/m}$.

The value of acceleration due to gravity (g) is $9.81{\text{m/s}}^{\text{2}}$.

Calculation:

The minimum value of tension occurs when the displacement (x) is maximum at upward direction $\left(-x_m\right)$.

Express the minimum tension $\left(T_{\min }\right)$ in cable.

$T_{\min }=T_0-k{x}_m$ (4)

Substitute Equations (1) and (3) in Equation (4).

$\begin{array}{c}{T}_{\min }=m_{1}g-m_{2}g\\ =\left(m_1-m_2\right)g\end{array}$ (5)

Calculate the minimum tension $\left(T_{\min }\right)$ which will occur in the cable during the motion:

Substitute 150 kg for $m_1$, 100 kg for $m_2$, and $\text{9}\text{.81}\text{m}/{\text{s}}^{\text{2}}$ for g in equation (5).

$\begin{array}{c}{T}_{\min }=\left(m_1-m_2\right)g\\ =\left(\left(150\text{kg}\right)-\left(100\text{kg}\right)\right)\left(\text{9}\text{.81}\text{m}/{\text{s}}^{\text{2}}\right)\\ =\left(50\right)\left(9.81\right)\\ =491\text{N}\end{array}$

Therefore, the minimum tension $\left(T_{\min }\right)$ which will occur in the cable during the motion is $\underset{\_}{491\text{N}}$.

(c)

To determine

The velocity $\left(\dot{x}\right)$ of the magnet 0.03 sec after the current is turned off.

Answer to Problem 19.14P

The velocity $\left(\dot{x}\right)$ of the magnet 0.03 sec after the current is turned off is $\underset{\_}{0.1592\text{m/s}(\uparrow )}$.

Explanation of Solution

Given Information:

The mass $\left(m_1\right)$ of the electromagnet is 150 kg.

The mass $\left(m_2\right)$ of the scrap steel is 100 kg.

The spring constant (k) is $200\text{kN/m}$.

The value of acceleration due to gravity (g) is $9.81{\text{m/s}}^{\text{2}}$.

Calculation:

Express the displacement (x) of the simple harmonic motion at any instant.

$x=x_m\sin \left({\omega }_{n}t+\varphi \right)$

Here, $\varphi$ is phase angle.

When time (t) is 0, the initial displacement is $-x_m$.

Substitute 0 for t and $-x_m$ for x.

$\begin{array}{l}-x_m=x_m\sin \left({\omega }_n\left(0\right)+\varphi \right)\\ -x_m=x_m\sin \varphi \end{array}$

For the above equation to satisfy the value of $\sin \varphi$ must be –1, which implies that the value of phase angle must be $-\pi /2$. Thus, the value of phase angle is $-\pi /2$.

Calculate the velocity $\left(\dot{x}\right)$ of the simple harmonic motion at any instant.

$\dot{x}={\omega }_n{x}_m\cos \left({\omega }_{n}t+\varphi \right)$

Substitute $\text{36}\text{.515}\text{rad}/\text{s}$ for ${\omega }_n$, $4.905\times {10}^{-3}\text{m}$ for $x_m$, 0.03 s for t, and $-\pi /2$ for $\varphi$.

$\begin{array}{c}\dot{x}=\left(\text{36}\text{.515}\text{rad}/\text{s}\right)\left(4.905\times {10}^{-3}\text{m}\right)\cos \left(\left(\text{36}\text{.515}\text{rad}/\text{s}\right)\left(0.03\text{s}\right)+\left(-\frac{\pi }{2}\right)\right)\\ =\left(\text{36}\text{.515}\right)\left(4.905\times {10}^{-3}\right)\cos \left(1.095-\frac{\pi }{2}\right)\\ =\left(0.179\right)\cos \left(-0.47535\text{rad}\right)\\ =\left(0.179\right)\cos \left(-0.47535\left(\text{rad}\times \frac{180°}{\pi \text{rad}}\right)\right)\\ =\left(0.179\right)\cos \left(27.236°\right)\\ =\left(0.179\right)\left(0.88913\right)\\ =\text{0}\text{.1592}\text{m}/\text{s}\end{array}$

Therefore, the velocity $\left(\dot{x}\right)$ of the magnet 0.03 sec after the current is turned off  is $\underset{\_}{0.1592\text{m/s}(\uparrow )}$.