Chapter 13.2, Problem 13.69P

To determine

The maximum additional deformation of the spring.

Answer to Problem 13.69P

The maximum additional deflection is $0.174\text{ m}$.

Explanation of Solution

Given:

Mass of the package is $50\text{ kg}$.

Spring stiffness is $30\text{kN/m}$.

Angle of inclination is $20°$.

Initial compression of spring is $50\text{ mm}$.

Velocity of package is $2\text{ m/s}$.

Concept used:

Draw the diagram of the package moving to different positions as shown below:

Consider the package is moving from position (0) to (1).

Write the Expression for the kinetic energy of package at position (0).

$T_0=\frac{1}{2}m{v}_0{}^2$   ...... (1)

Here, $T_0$ is the kinetic energy at position (0), $m$ is the mass of the package and $v_0$ is the initial velocity of the package.

Write the Expression for the kinetic energy of package at position (1).

$T_1=\frac{1}{2}m{v}_1{}^2$   ...... (2)

Here, $T_1$ is the kinetic energy at position (1) and $v_1$ is the initial velocity of the package.

Write the expression for the work done from bringing the package from (0) to (1).

$U_{01}=\left[-\left({\mu }_k{R}_N\right)+mg\sin \theta \right]d$   ...... (3)

Here, ${\mu }_k$ is the coefficient of kinetic friction, $R_N$ is the reaction from the ground, $g$ is the acceleration due to gravity, $\theta$ is the angle of inclination and $d$ is the displacement from (0) to (1).

Write the expression for normal reaction from the ground.

$R_N=mg\cos \theta$   ...... (4)

Write the expression for work energy principle between point (0) and (1).

$T_0+U_{01}=T_1$   ...... (5)

Consider the motion of package from position (1) to (2).

Write the Expression for the kinetic energy of package at position (2).

$T_2=\frac{1}{2}m{v}_1{}^2$   ...... (6)

Here, $T_2$ is the kinetic energy at position (2).

Write the expression for the initial potential energy at position (2).

$V_{2i}=\frac{1}{2}k{x}_1{}^2+mg\left(x_1+x\right)\sin \theta$   ...... (7)

Here, $V_{2i}$ is the initial potential energy, $k$ is the spring constant, $x_1$ is the initial compression of spring and $x$ is the maximum additional deflection.

Write the Expression for the kinetic energy of package at final position.

$T_f=\frac{1}{2}m{v}_f{}^2$   ...... (8)

Here, $T_f$ is the final kinetic energy and $v_f$ is the final velocity of the package.

Write the expression for the final potential energy at position (2).

$V_{2f}=\frac{1}{2}k{\left(x_1+x\right)}^2$   ...... (9)

Here, $V_{2f}$ is the final potential energy at point (2).

Write the expression for the law of conservation between points (2) and the final position.

$T_2+V_{2i}=T_f+V_{2f}$   ...... (10)

Calculation:

Substitute $50\text{ kg}$ for $m$ and $2\text{ m/s}$ for $v_o$ in equation (1).

$\begin{array}{c}{T}_0=\frac{1}{2}\left(50\right){\left(2\right)}^2\\ =100\text{J}\end{array}$

Substitute $50\text{ kg}$ for $m$ in equation (2).

$\begin{array}{c}{T}_1=\frac{1}{2}\left(50\right)v_1{}^2\\ =25v_1{}^2\end{array}$

Substitute $mg\cos \theta$ for $R_N$ in equation (3).

$U_{01}=\left[\left(-{\mu }_{k}mg\cos \theta \right)+mg\sin \theta \right]d$

Substitute $0.2$ for ${\mu }_k$, $50\text{ kg}$ for $m$, $20°$ for $\theta$, $8\text{ m}$ for $d$ and $9.81{\text{ m/s}}^2$ for $g$ in above equation.

$\begin{array}{l}{U}_{01}=\left[-\left(\left(0.2\right)\left(9.81\right)\left(50\right)\cos 20°\right)+\left(50\right)\left(9.81\right)\sin 20°\right]\left(8\right)\\ =604.61\text{}\text{J}\end{array}$

Substitute $100\text{}\text{J}$ for $T_0$, $25v_1{}^2$ for $T_1$ and $604.61\text{ J}$ for $U_{01}$ in equation (5).

$100+604.61=25v_1{}^2$

Simplify the above expression for $v_1$.

$v_1=5.31\text{ m/s}$

Substitute $5.31\text{ m/s}$ for $v_1$ and $50\text{ kg}$ for $m$ in equation (6).

$\begin{array}{c}{T}_2=\frac{1}{2}\left(50\right){\left(5.31\right)}^2\\ =704.9\text{ J}\end{array}$

Substitute $30\text{ kN/m}$ for $k$, $20°$ for $\theta$, $9.81{\text{ m/s}}^2$ for $g$ and $50\text{ mm}$ for $x_1$ in equation (7).

$\begin{array}{c}{V}_{2i}=\left[\begin{array}{l}\frac{1}{2}\left(30\text{ kN/m}\left(\frac{1000\text{ N}}{1\text{ kN}}\right)\right){\left(50\text{ mm}\left(\frac{1\text{ m}}{1000\text{ mm}}\right)\right)}^2\\ +\left(50\right)\left(9.81\right)\left(x_1+x\right)\sin 20°\end{array}\right]\\ =37.5\text{ J + 167}\text{.76}\left(x_1+x\right)\end{array}$

Substitute $0$ for $v_f$ in equation (8).

$\begin{array}{c}{T}_f=\frac{1}{2}m{\left(0\right)}^2\\ =0\text{}\text{J}\end{array}$

Substitute $30\text{ kN/m}$ for $k$ in equation (9).

$\begin{array}{c}{V}_{2f}=\frac{1}{2}\left(30\text{ kN/m}\left(\frac{1000\text{ N}}{1\text{ kN}}\right)\right){\left(x_1+x\right)}^2\\ =15000{\left(x_1+x\right)}^2\end{array}$

Substitute $704.9\text{ J}$ for $T_2$, $37.5\text{ J + 167}\text{.76}\left(x_1+x\right)$ for $V_{2i}$, $0\text{J}$ for $T_f$ and $15000{\left(x_1+x\right)}^2$ for $V_{2f}$ in equation (10).

$704.9+37.5+1667.76\left(x_1+x\right)=15000{\left(x_1+x\right)}^2$

Simplify the above expression for $\left(x_1+x\right)$.

$\left(x_1+x\right)=0.224\text{ m}$

Substitute $50\text{ mm}$ for $x_1$ in above expression.

$50\text{ mm}\left(\frac{1\text{ m}}{1000\text{ mm}}\right)+x=0.224\text{ m}$

Simplify above expression for $x$.

$x=0.174\text{ m}$

The maximum additional deflection is $0.174\text{ m}$.

Conclusion:

Thus, the maximum additional deflection is $0.174\text{ m}$.