Chapter 8.4, Problem 8.121P

8.121 and 8.123 A cable is placed around three parallel pipes. Two of the pipes are fixed and do not rotate; the third pipe is slowly rotated. Knowing that the coefficients of friction are μ_S = 0.25 and μ_k = 0.20, determine the largest weight W that can be raised (a) if only pipe A is rotated counterclockwise, (b) if only pipe C is rotated clockwise.

Fig. P8.120 and P8.121

To determine

Find the largest weight W that can be raised only when the pipe A is rotated counter clockwise.

Answer to Problem 8.121P

The largest weight W that can be raised only when the pipe A is rotated counter clockwise is $\underset{\_}{28.9\text{lb}}$.

Explanation of Solution

Given information:

The coefficient of static friction is ${\mu }_s=0.25$.

The coefficient of kinetic friction is ${\mu }_k=0.20$.

Calculation:

Show the free-body diagram of the pipes A, B, and C as in Figure 1.

Find the angle of the cable wounded around the pipe as follows;

$\begin{array}{c}{\beta }_A=\frac{1}{4}\times 2\pi \\ =\frac{\pi }{2}\end{array}$

$\begin{array}{c}{\beta }_B=\frac{1}{2}\times 2\pi \\ =\pi \end{array}$

$\begin{array}{c}{\beta }_C=\frac{1}{4}\times 2\pi \\ =\frac{\pi }{2}\end{array}$

Find the largest weight W using the equation.

$\begin{array}{l}{\left(\frac{T_2}{T_1}\right)}_A{\left(\frac{T_2}{T_1}\right)}_B{\left(\frac{T_2}{T_1}\right)}_C=\left(e^{{\mu }_s{\beta }_A}\right)\left(e^{{\mu }_k{\beta }_B}\right)\left(e^{{\mu }_k{\beta }_C}\right)\\ \frac{T_{AB}}{50}\times \frac{T_{AB}}{T_{BC}}\times \frac{T_{BC}}{W}=\left(e^{{\mu }_s{\beta }_A}\right)\left(e^{{\mu }_k{\beta }_B}\right)\left(e^{{\mu }_k{\beta }_C}\right)\\ \frac{T_{AB}}{50}\times \frac{T_{BC}}{T_{AB}}\times \frac{W}{T_{BC}}=\left(e^{{\mu }_s{\beta }_A}\right)\left(e^{-{\mu }_k{\beta }_B}\right)\left(e^{-{\mu }_k{\beta }_C}\right)\end{array}$

Substitute 0.25 for ${\mu }_s$, $\frac{\pi }{2}$ for ${\beta }_A$, $\pi$ for ${\beta }_B$, and $\frac{\pi }{2}$ for ${\beta }_C$.

$\begin{array}{c}\frac{T_{AB}}{50}\times \frac{T_{BC}}{T_{AB}}\times \frac{W}{T_{BC}}=\left(e^{0.25\times \frac{\pi }{2}}\right)\left(e^{-0.20\times \pi }\right)\left(e^{-0.20\times \frac{\pi }{2}}\right)\\ \frac{W}{50}=1.48097\times 0.53349\times 0.73040\\ W=28.9\text{lb}\end{array}$

Therefore, the largest weight W that can be raised only when the pipe A is rotated counter clockwise is $\underset{\_}{28.9\text{lb}}$.

To determine

Find the largest weight W that can be raised only when the pipe C is rotated clockwise.

Answer to Problem 8.121P

The largest weight W that can be raised only when the pipe C is rotated clockwise is $\underset{\_}{28.9\text{lb}}$.

Explanation of Solution

Given information:

The coefficient of static friction is ${\mu }_s=0.25$.

The coefficient of kinetic friction is ${\mu }_k=0.20$.

Calculation:

Show the free-body diagram of the pipes A, B, and C as in Figure 2.

Find the angle of the cable wounded around the pipe as follows;

$\begin{array}{c}{\beta }_A=\frac{1}{4}\times 2\pi \\ =\frac{\pi }{2}\end{array}$

$\begin{array}{c}{\beta }_B=\frac{1}{2}\times 2\pi \\ =\pi \end{array}$

$\begin{array}{c}{\beta }_C=\frac{1}{4}\times 2\pi \\ =\frac{\pi }{2}\end{array}$

Find the largest weight W using the equation.

$\begin{array}{l}{\left(\frac{T_2}{T_1}\right)}_A{\left(\frac{T_2}{T_1}\right)}_B{\left(\frac{T_2}{T_1}\right)}_C=\left(e^{{\mu }_k{\beta }_A}\right)\left(e^{{\mu }_k{\beta }_B}\right)\left(e^{{\mu }_s{\beta }_C}\right)\\ \frac{50}{T_{AB}}\times \frac{T_{AB}}{T_{BC}}\times \frac{W}{T_{BC}}=\left(e^{{\mu }_k{\beta }_A}\right)\left(e^{{\mu }_k{\beta }_B}\right)\left(e^{{\mu }_s{\beta }_C}\right)\\ \frac{50}{T_{AB}}\times \frac{T_{AB}}{T_{BC}}\times \frac{T_{BC}}{W}=\left(e^{{\mu }_k{\beta }_A}\right)\left(e^{{\mu }_k{\beta }_B}\right)\left(e^{-{\mu }_s{\beta }_C}\right)\end{array}$

Substitute 0.25 for ${\mu }_s$, 0.20 for ${\mu }_k$, $\frac{\pi }{2}$ for ${\beta }_A$, $\pi$ for ${\beta }_B$, and $\frac{\pi }{2}$ for ${\beta }_C$.

$\begin{array}{c}\frac{50}{T_{AB}}\times \frac{T_{AB}}{T_{BC}}\times \frac{T_{BC}}{W}=\left(e^{0.20\times \frac{\pi }{2}}\right)\left(e^{0.20\times \pi }\right)\left(e^{-0.25\times \frac{\pi }{2}}\right)\\ \frac{50}{W}=1.36911\times 1.87446\times 0.67523\\ =1.73287\\ W=28.9\text{lb}\end{array}$

Therefore, the largest weight W that can be raised only when the pipe C is rotated clockwise is $\underset{\_}{28.9\text{lb}}$.