Chapter 8.8, Problem 107P

If the coefficient of static is, μ_s = 0.3, determine the smallest torque M required to overcome friction.

To determine

The smallest torque required to overcome friction.

Answer to Problem 107P

The smallest torque (M) required to overcome friction is $\underset{\_}{\text{132}\text{N}\cdot \text{m}}$.

Explanation of Solution

Given information:

• The axial force (P) is 5 kN.
• The coefficient of static friction ${\mu }_s$ is $0.30$.
• The outer diameter $\left(D_1\right)$ of the collar is 200 mm.
• The inner diameter $\left(D_2\right)$ of the collar is 150 mm.

Determine the inner radius $R_1$ of the pipe.

$R_1=\frac{D_1}{2}$ (I).

Determine the outer radius $R_2$ of the pipe.

$R_2=\frac{D_2}{2}$ (II).

Determine the smallest torque (M) required to overcome friction using the formula.

$M=\frac{2}{3}{\mu }_{s}P\left(\frac{R_2^3-R_1^3}{R_2^2-R_1^2}\right)$ (III)

Conclusion:

Substitute 150 mm for $D_2$ in Equation (I).

$\begin{array}{c}{R}_1=\left(\frac{150}{\text{2}}\text{mm}\right)\times \left(\frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\right)\\ =0.075\text{m}\end{array}$

Substitute 200 mm for $D_1$ in Equation (II).

$\begin{array}{c}{R}_2=\left(\frac{\text{200}}{\text{2}}\text{mm}\right)\times \left(\frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\right)\\ =0.1\text{m}\end{array}$

Substitute 0.30 for ${\mu }_s$, 5.0 kN for P, 0.1 m for $R_2$, and 0.075 m for $R_1$ in Equation (III).

$\begin{array}{c}M=\frac{2}{3}\left(0.3\right)\left(5\text{kN}\times \frac{\text{1,000N}}{1\text{kN}}\right)\left(\frac{{0.1}^3-{0.075}^3}{{0.1}^2-{0.075}^2}\right)\\ =0.20\times 5,000\times \left(0.1321\right)\\ =132.1\text{N}\cdot \text{m}\\ \approx \text{132}\text{N}\cdot \text{m}\end{array}$

Thus, the smallest torque (M) required to overcome friction is $\underset{\_}{\text{132}\text{N}\cdot \text{m}}$.