Chapter 14, Problem 98P

a.

To determine

The value of coefficient of static friction between box and floor.

Explanation of Solution

Given:

Period of oscillation is $0.80\text{s}$ .

Box started to slip when amplitude reaches $10\text{cm}$ .

Formula used:

For equilibrium:

Write expression for force on x-direction.

  ${\displaystyle \sum f_x}=0$

Substitute $f_s-m_2{a}_{\max }$ for ${\displaystyle \sum f_x}$ in above expression.

  $f_s-m_2{a}_{\max }=0$  ......(1)

Here, $f_s$ is the maximum friction on box and $a_{\max }$ is the maximum acceleration.

Write expression for force on y-direction.

  ${\displaystyle \sum f_y}=0$

Substitute $F_n-m_{2}g$ for ${\displaystyle \sum f_y}$ in above expression.

  $F_n-m_{2}g=0$  ......(2)

Here, $F_n$ is the normal force on the box and $g$ is the acceleration due to gravity.

Write expression for maximum acceleration related to angular frequency and amplitude.

  $a_{\max }=A{\omega }^2$  ......(3)

Here, $A$ is the amplitude and $\omega$ is the angular frequency.

Substitute ${\mu }_s{F}_n$ for $f_s$ in equation (1) and solve for $F_n$ .

  $\begin{array}{c}{\mu }_s{F}_n-m_2{a}_{\max }=0\\ F_n=\frac{m_2{a}_{\max }}{{\mu }_s}\end{array}$

Substitute $\frac{m_2{a}_{\max }}{{\mu }_s}$ for $F_n$ in equation (2) and solve for $a_{\max }$ .

  $\begin{array}{c}\frac{m_2{a}_{\max }}{{\mu }_s}-m_{2}g=0\\ \frac{m_2{a}_{\max }}{{\mu }_s}=m_{2}g\\ \frac{a_{\max }}{{\mu }_s}=g\\ a_{\max }={\mu }_{s}g\end{array}$

Substitute ${\mu }_{s}g$ for $a_{\max }$ and in equation (3) and solve for ${\mu }_s$ .

  $\begin{array}{c}{\mu }_{s}g=A{\omega }^2\\ {\mu }_s=\frac{A{\omega }^2}{g}\end{array}$

Substitute $\frac{2\pi }{T}$ for $\omega$ in above expression.

  ${\mu }_s=\frac{4{\pi }^{2}A}{T^{2}g}$  ......(4)

Calculation:

Substitute $10\text{cm}$ for $A$ , $0.80\text{s}$ for $T$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in equation (4).

  $\begin{array}{c}{\mu }_s=\frac{4{\pi }^2\left(10\text{cm}\right)}{{\left(0.80\text{s}\right)}^2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =0.63\end{array}$

Conclusion:

Thus, the value of coefficient of static friction between box and floor is $0.63$ .

(b)

To determine

The maximum amplitude of vibration for given value of ${\mu }_s$ .

Explanation of Solution

Given:

The value of ${\mu }_s$ is $0.40$ .

Formula used:

For equilibrium:

Write expression for force on x-direction.

  ${\displaystyle \sum f_x}=0$

Substitute $f_s-m_2{a}_{\max }$ for ${\displaystyle \sum f_x}$ in above expression.

  $f_s-m_2{a}_{\max }=0$  ......(1)

Here, $f_s$ is the maximum friction on box and $a_{\max }$ is the maximum acceleration.

Write expression for force on y-direction.

  ${\displaystyle \sum f_y}=0$

Substitute $F_n-m_{2}g$ for ${\displaystyle \sum f_y}$ in above expression.

  $F_n-m_{2}g=0$  ......(2)

Here, $F_n$ is the normal force on the box and $g$ is the acceleration due to gravity.

Write expression for maximum acceleration related to angular frequency and amplitude.

  $a_{\max }=A{\omega }^2$  ......(3)

Here, $A$ is the amplitude and $\omega$ is the angular frequency.

Substitute ${\mu }_s{F}_n$ for $f_s$ in equation (1) and solve for $F_n$ .

  $\begin{array}{c}{\mu }_s{F}_n-m_2{a}_{\max }=0\\ F_n=\frac{m_2{a}_{\max }}{{\mu }_s}\end{array}$

Substitute $\frac{m_2{a}_{\max }}{{\mu }_s}$ for $F_n$ in equation (2) and solve for $a_{\max }$ .

  $\begin{array}{c}\frac{m_2{a}_{\max }}{{\mu }_s}-m_{2}g=0\\ \frac{m_2{a}_{\max }}{{\mu }_s}=m_{2}g\\ \frac{a_{\max }}{{\mu }_s}=g\\ a_{\max }={\mu }_{s}g\end{array}$

Substitute ${\mu }_{s}g$ for $a_{\max }$ and in equation (3) and solve for $A$ .

  $\begin{array}{c}{\mu }_{s}g=A{\omega }^2\\ A=\frac{{\mu }_{s}g}{{\omega }^2}\end{array}$

Substitute $\frac{2\pi }{T}$ for $\omega$ in above expression.

  $A=\frac{{\mu }_{s}g{T}^2}{4{\pi }^2}$   ...... (4)

Calculation:

Substitute $0.40$ for ${\mu }_s$ , $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $0.80\text{s}$ for $T$

  $\begin{array}{c}A=\frac{\left(0.40\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right){\left(0.80\text{s}\right)}^2}{4{\pi }^2}\\ =0.63\text{m}\end{array}$

Conclusion:

Thus, the maximum value of amplitude is $0.63\text{m}$ .