Chapter 12, Problem 12.128RP

To determine

(a)

Minimum value of coefficient of static friction between the collar and rod at $\theta ={90}^0$.

Answer to Problem 12.128RP

Minimum value of coefficient of static friction between the collar and rod is 0.454.

Explanation of Solution

Given information:

Mass of the collar C, m=200 gm or 0.2 kg.

Angular rotation of rod, $\omega$ =6 rad/s.

Position of block E, $\theta ={90}^0$

Calculations:

AB is the axis of rotation and collar C slide on semicircular rod. So, the distance of collar C from the axis AB,

$\begin{array}{l}s=r\sin \theta \\ s=0.6\sin 90\\ s=0.6m\end{array}$

Acceleration of the collar,

$\begin{array}{l}a=s{\omega }^2\\ a=0.6\times 6^2\\ a=21.6m/s^2\end{array}$

Equilibrium force along x direction,

$\begin{array}{l}\sum F_x=m_{c}a\\ N=m_{c}a\\ N=0.2\times 21.6\\ N=4.32N\end{array}$

Equilibrium force along y direction,

$\begin{array}{l}\sum F_y=0\\ F=w\\ {\mu }_{k}N=mg\\ {\mu }_k=\frac{mg}{N}\\ {\mu }_k=\frac{0.2\times 9.81}{4.32}\\ {\mu }_k=\frac{1.962}{4.32}\\ {\mu }_k=0.454\end{array}$

To determine

(b)

Minimum value of coefficient of static friction between the collar and rod at $\theta ={75}^0$.

Answer to Problem 12.128RP

Minimum value of coefficient of static friction between the collar and rod is 0.179.

Explanation of Solution

Given information:

Mass of the collar C, m=200 gm or 0.2 kg.

Angular rotation of rod, $\omega$ =6 rad/s.

Position of block E, $\theta ={75}^0$

Calculations:

AB is the axis of rotation and collar C slide on semicircular rod. So, the distance of collar C from the axis AB is

$\begin{array}{l}s=r\sin \theta \\ s=0.6\sin 75\\ s=0.579m\end{array}$

Acceleration of the collar,

$\begin{array}{l}a=s{\omega }^2\\ a=0.579\times 6^2\\ a=20.86m/s^2\end{array}$

For the value of $\theta ={75}^0$, the direction is not known. The motion of collar is upward or downward. Consider both possibilities.

For downward impending motion,

Equilibrium force along y direction,

$\begin{array}{l}\sum F_y=0\\ N\cos \theta +F\sin \theta =W\\ N\cos \theta +{\mu }_{k}N\sin \theta =W\\ N\left(\cos \theta +{\mu }_k\sin \theta \right)=W\\ N=\frac{m_{c}g}{\cos \theta +{\mu }_k\sin \theta }(1)\end{array}$

Equilibrium force along normal direction,

$\begin{array}{l}\sum F_n=m_{c}a\\ N\sin \theta -F\cos \theta =m_{c}a\\ N\sin \theta -{\mu }_{k}N\cos \theta =m_{c}a\\ N\left(\sin \theta -{\mu }_k\cos \theta \right)=m_{c}a\\ N=\frac{m_{c}a}{\sin \theta -{\mu }_k\cos \theta }(2)\end{array}$

Equating equation 1 and equation 2,

$\begin{array}{l}\frac{m_{c}g}{\cos \theta +{\mu }_k\sin \theta }=\frac{m_{c}a}{\sin \theta -{\mu }_k\cos \theta }\\ g\sin \theta -g{\mu }_k\cos \theta =a\cos \theta +a{\mu }_k\sin \theta \\ a{\mu }_k\sin \theta +g{\mu }_k\cos \theta =g\sin \theta -a\cos \theta \\ {\mu }_k\left(a\sin \theta +g\cos \theta \right)=g\sin \theta -a\cos \theta \\ {\mu }_k=\frac{g\sin \theta -a\cos \theta }{a\sin \theta +g\cos \theta }\\ {\mu }_k=\frac{g\tan \theta -a}{a\tan \theta +g}\text{}\text{}\\ {\mu }_k=\frac{9.81\times \tan {75}^0-20.86}{20.86\times \tan {75}^0+9.81}\text{}\text{}\\ {\mu }_k=\frac{9.81\times 3.732-20.86}{20.86\times 3.732+9.81}\text{}\\ {\mu }_k=\frac{36.61-20.86}{77.86+9.81}\text{}\\ {\mu }_k=\frac{15.75}{87.67}\\ {\mu }_k=0.179\\ \text{}\end{array}$

For upward impending motion,

Equilibrium force along y direction,

$\begin{array}{l}\sum F_y=0\\ N\cos \theta -F\sin \theta =W\\ N\cos \theta -{\mu }_{k}N\sin \theta =W\\ N\left(\cos \theta -{\mu }_k\sin \theta \right)=W\\ N=\frac{m_{c}g}{\cos \theta -{\mu }_k\sin \theta }(4)\end{array}$

Equilibrium force along normal direction,

$\begin{array}{l}\sum F_n=m_{c}a\\ N\sin \theta +F\cos \theta =m_{c}a\\ N\sin \theta +{\mu }_{k}N\cos \theta =m_{c}a\\ N\left(\sin \theta +{\mu }_k\cos \theta \right)=m_{c}a\\ N=\frac{m_{c}a}{\sin \theta +{\mu }_k\cos \theta }(5)\end{array}$

Equating equation 4 and equation 5,

$\begin{array}{l}\frac{m_{c}g}{\cos \theta -{\mu }_k\sin \theta }=\frac{m_{c}a}{\sin \theta +{\mu }_k\cos \theta }\\ g\sin \theta +g{\mu }_k\cos \theta =a\cos \theta -a{\mu }_k\sin \theta \\ a{\mu }_k\sin \theta +g{\mu }_k\cos \theta =a\cos \theta -g\sin \theta \\ {\mu }_k\left(a\sin \theta +g\cos \theta \right)=a\cos \theta -g\sin \theta \\ {\mu }_k=\frac{a\cos \theta -g\sin \theta }{a\sin \theta +g\cos \theta }\\ {\mu }_k=\frac{a\tan \theta -g}{a\tan \theta +g}\text{}\text{}\text{}\text{}\\ {\mu }_k=\frac{20.86\times \tan {75}^0-9.81}{20.86\times \tan {75}^0+9.81}\text{}\text{}\text{}\text{}\\ {\mu }_k=\frac{20.86\times 3.732-9.81}{20.86\times 3.732+9.81}\text{}\text{}\text{}\text{}\\ {\mu }_k=\frac{77.86-9.81}{77.86+9.81}\text{}\text{}\text{}\text{}\\ {\mu }_k=\frac{68.05}{87.67}\\ {\mu }_k=0.776\end{array}$

To determine

(c)

Minimum value of coefficient of static friction between the collar and rod at $\theta ={45}^0$.

Answer to Problem 12.128RP

Minimum value of coefficient of static friction between the collar and rod is 0.218.

Explanation of Solution

Given information:

Mass of the collar C, m=200 gm or 0.2 kg.

Angular rotation of rod, $\omega$ =6 rad/s.

Position of block E, $\theta ={45}^0$

Calculations:

AB is the axis of rotation and collar C slide on semicircular rod. So, the distance of collar C from the axis AB,

$\begin{array}{l}s=r\sin \theta \\ s=0.6\sin 45\\ s=0.424m\end{array}$

Acceleration of the collar,

$\begin{array}{l}a=s{\omega }^2\\ a=0.424\times 6^2\\ a=15.27m/s^2\end{array}$

For the value of $\theta ={45}^0$, the direction is not known. The motion of collar is upward or downward. Consider both possibilities.

For downward impending motion,

Equilibrium force along y direction,

$\begin{array}{l}\sum F_y=0\\ N\cos \theta +F\sin \theta =W\\ N\cos \theta +{\mu }_{k}N\sin \theta =W\\ N\left(\cos \theta +{\mu }_k\sin \theta \right)=W\\ N=\frac{m_{c}g}{\cos \theta +{\mu }_k\sin \theta }(1)\end{array}$

Equilibrium force along normal direction,

$\begin{array}{l}\sum F_n=m_{c}a\\ N\sin \theta -F\cos \theta =m_{c}a\\ N\sin \theta -{\mu }_{k}N\cos \theta =m_{c}a\\ N\left(\sin \theta -{\mu }_k\cos \theta \right)=m_{c}a\\ N=\frac{m_{c}a}{\sin \theta -{\mu }_k\cos \theta }(2)\end{array}$

Equating equation 1 and equation 2,

$\begin{array}{l}\frac{m_{c}g}{\cos \theta +{\mu }_k\sin \theta }=\frac{m_{c}a}{\sin \theta -{\mu }_k\cos \theta }\\ g\sin \theta -g{\mu }_k\cos \theta =a\cos \theta +a{\mu }_k\sin \theta \\ a{\mu }_k\sin \theta +g{\mu }_k\cos \theta =g\sin \theta -a\cos \theta \\ {\mu }_k\left(a\sin \theta +g\cos \theta \right)=g\sin \theta -a\cos \theta \\ {\mu }_k=\frac{g\sin \theta -a\cos \theta }{a\sin \theta +g\cos \theta }\\ {\mu }_k=\frac{g\tan \theta -a}{a\tan \theta +g}\text{}\text{}\\ {\mu }_k=\frac{9.81\times \tan {45}^0-15.27}{15.27\times \tan {45}^0+9.81}\text{}\text{}\\ {\mu }_k=\frac{9.81\times 1-15.27}{15.27\times 1+9.81}\text{}\\ {\mu }_k=\frac{9.81-15.27}{15.27+9.81}\text{}\\ {\mu }_k=-\frac{5.46}{25.08}\\ {\mu }_k=-0.218\\ \text{}\end{array}$

For upward impending motion,

Equilibrium force along y direction,

$\sum F_y=0$

$N\cos \theta -F\sin \theta =W$

$N\cos \theta -{\mu }_{k}N\sin \theta =W$

$N\left(\cos \theta -{\mu }_k\sin \theta \right)=W$

$N=\frac{m_{c}g}{\cos \theta -{\mu }_k\sin \theta }(4)$

Equilibrium force along normal direction,

$\begin{array}{l}\sum F_n=m_{c}a\\ N\sin \theta +F\cos \theta =m_{c}a\\ N\sin \theta +{\mu }_{k}N\cos \theta =m_{c}a\\ N\left(\sin \theta +{\mu }_k\cos \theta \right)=m_{c}a\\ N=\frac{m_{c}a}{\sin \theta +{\mu }_k\cos \theta }(5)\end{array}$

Equating equation 4 and equation 5,

$\begin{array}{l}\frac{m_{c}g}{\cos \theta -{\mu }_k\sin \theta }=\frac{m_{c}a}{\sin \theta +{\mu }_k\cos \theta }\\ g\sin \theta +g{\mu }_k\cos \theta =a\cos \theta -a{\mu }_k\sin \theta \\ a{\mu }_k\sin \theta +g{\mu }_k\cos \theta =a\cos \theta -g\sin \theta \\ {\mu }_k\left(a\sin \theta +g\cos \theta \right)=a\cos \theta -g\sin \theta \\ {\mu }_k=\frac{a\cos \theta -g\sin \theta }{a\sin \theta +g\cos \theta }\\ {\mu }_k=\frac{a\tan \theta -g}{a\tan \theta +g}\text{}\text{}\text{}\text{}\\ {\mu }_k=\frac{15.27\times \tan {45}^0-9.81}{15.27\times \tan {45}^0+9.81}\text{}\text{}\text{}\text{}\\ {\mu }_k=\frac{15.27\times 1-9.81}{15.27\times 1+9.81}\text{}\text{}\text{}\text{}\\ {\mu }_k=\frac{15.27-9.81}{15.27+9.81}\text{}\text{}\text{}\text{}\\ {\mu }_k=\frac{5.46}{25.08}\\ {\mu }_k=0.218\end{array}$