Chapter 8.3, Problem 8.97P

Solve Prob. 8.93 assuming that the normal force per unit area between the disk and the floor varies linearly from a maximum at the center to zero at the circumference of the disk.

8.93 A 50-lb electric floor polisher is operated on a surface for which the coefficient of kinetic friction is 0.25. Assuming that the normal force per unit area between the disk and the floor is uniformly distributed, determine the magnitude Q of the horizontal forces required to prevent motion of the machine.

Fig. P8.93

To determine

Find the magnitude Q of the horizontal force required to prevent motion of the machine.

Answer to Problem 8.97P

The magnitude Q of the horizontal force is $\underset{\_}{2.8125\text{lb}}$.

Explanation of Solution

Given information:

The diameter of the electric floor polisher is $D=18\text{in}\text{.}$.

The weight of the electric floor polisher is $W=50\text{-lb}$.

The coefficient of kinetic friction is ${\mu }_k=0.25$.

Calculation:

Show the free-body diagram of the thrust bearing as in Figure 1.

It has been assumed that the normal force per unit area is inversely proportional to r.

$\Delta A=r\Delta \theta \Delta r$

Here, the distance between the point to the axis of the shaft is r, the change is angle is $\Delta \theta$, and the change in distance is $\Delta r$.

Find the normal force exerted $\Delta N$ due to the element of area $\Delta A$ using the relation;

$\Delta N=k\left(1-\frac{r}{R}\right)\Delta A$

Substitute $r\Delta \theta \Delta r$ for $\Delta A$.

$\Delta N=k\left(1-\frac{r}{R}\right)r\Delta \theta \Delta r$ (1)

Find the value of k using the relation.

$\begin{array}{c}P={\displaystyle \sum \Delta N={\displaystyle \int dN}}\\ ={\displaystyle {\int }_0^{2\pi }{\displaystyle {\int }_0^R\Delta N}}\end{array}$

Substitute $k\left(1-\frac{r}{R}\right)r\Delta \theta \Delta r$ for $\Delta N$.

$\begin{array}{c}P={\displaystyle {\int }_0^{2\pi }{\displaystyle {\int }_0^{R}k\left(1-\frac{r}{R}\right)r\Delta \theta \Delta r}}\\ =k{\left[\frac{r^2}{2}-\frac{r^3}{3R}\right]}_0^R{\left[\theta \right]}_0^{2\pi }\\ =k\left[\frac{R^2}{2}-\frac{R^3}{3R}\right]\left[2\pi -0\right]\\ =2\pi k\left[\frac{3R^2-2R^2}{6}\right]\end{array}$

$\begin{array}{c}P=2\pi k\left(\frac{R^2}{3}\right)\\ k=\frac{3P}{\pi R^2}\end{array}$

Substitute $\frac{3P}{\pi R^2}$ for k in Equation (1).

$\Delta N=\frac{3P}{\pi R^2}\left(1-\frac{r}{R}\right)r\Delta \theta \Delta r$

Find the change in friction force $\Delta F$ using the relation.

$\Delta F={\mu }_k\Delta N$

Substitute $\frac{3P}{\pi R^2}\left(1-\frac{r}{R}\right)r\Delta \theta \Delta r$ for $\Delta N$.

$\Delta F={\mu }_k\frac{3P}{\pi R^2}\left(1-\frac{r}{R}\right)r\Delta \theta \Delta r$

Find the change in moment $\Delta M$ using the relation.

$\Delta M=r\Delta F$

Substitute ${\mu }_k\frac{3P}{\pi R^2}\left(1-\frac{r}{R}\right)r\Delta \theta \Delta r$ for $\Delta F$.

$\Delta M=r{\mu }_k\frac{3P}{\pi R^2}\left(1-\frac{r}{R}\right)r\Delta \theta \Delta r$

Integrate the equation to find the couple M.

$M={\displaystyle {\int }_0^{2\pi }{\displaystyle {\int }_0^R\Delta M}}$

Substitute $r{\mu }_k\frac{3P}{\pi R^2}\left(1-\frac{r}{R}\right)r\Delta \theta \Delta r$ for $\Delta M$ and integrate.

$\begin{array}{c}M={\displaystyle {\int }_0^{2\pi }{\displaystyle {\int }_0^{R}r{\mu }_k\frac{3P}{\pi R^2}\left(1-\frac{r}{R}\right)r\Delta \theta \Delta r}}\\ ={\mu }_k\frac{3P}{\pi R^2}{\left[\frac{r^3}{3}-\frac{r^4}{4R}\right]}_0^R{\left[\theta \right]}_0^{2\pi }\\ ={\mu }_k\frac{3P}{\pi R^2}\left[\frac{R^3}{3}-\frac{R^4}{4R}\right]\left[2\pi -0\right]\\ ={\mu }_k\frac{3P}{\pi R^2}\times 2\pi \times \left[\frac{4R^3}{12}-\frac{3R^3}{12}\right]\end{array}$

$\begin{array}{c}M={\mu }_k\frac{3P}{R^2}\times 2\times \frac{R^3}{12}\\ =\frac{1}{2}{\mu }_{k}PR\end{array}$

Substitute 0.25 for ${\mu }_k$, 50 lb for P, and $\left(\frac{18}{2}\right)\text{in}\text{.}$ for R.

$\begin{array}{l}M=\frac{1}{2}\times 0.25\times 50\times \frac{18}{2}\\ M=56.25\text{lb}\cdot \text{in}.\end{array}$

Refer to the given figure;

Resolve the moment component in y-axis as follows;

$\begin{array}{l}{\displaystyle \sum M_y=0}\\ M-Q\left(10\right)-Q\left(10\right)=0\\ 20Q=M\end{array}$

Substitute $56.25\text{lb}\cdot \text{in}.$ for M.

$\begin{array}{l}20Q=56.25\\ Q=2.8125\text{lb}\end{array}$

Therefore, the magnitude Q of the horizontal force is $\underset{\_}{2.8125\text{lb}}$.