Chapter 17, Problem 1P

Example 17-2 resulted in selection of a 10-in-wide A-3 polyamide flat belt. Show that the value of F₁ restoring f 0.80 is

$F_1=\frac{\left(\Delta F+F_c\right)\exp f\varphi -F_c}{\exp f\varphi -1}$

and compare the initial tensions.

To determine

Whether the value of $F_1$ restoring $f$ to $0.80$.

Explanation of Solution

Write the expression for tight side in belt.

$F_1=F_c+\Delta F\left[\frac{e^{f\phi }}{e^{f\phi }-1}\right]$ (I)

Here, the centrifugal tension is $F_c$, the angle of overlap is $\phi$, the coefficient of the friction is $f$, the difference between tight side and slack side of the belt is $\Delta F$.

Write the expression for slack side in belt.

$F_2=F_1-\Delta F$ (II)

Write the expression for initial tension in belt.

$F_i=\frac{F_1+F_2}{2}-F_c$ (III)

Write the expression for coefficient of friction of belt material.

$f^{\text{'}}=\frac{1}{\phi }\ln \left(\frac{F_1-F_c}{F_2-F_c}\right)$ (IV)

Now add and subtract $F_c\left[\frac{e^{f\phi }}{e^{f\phi }-1}\right]$ in Equation (I).

$\begin{array}{c}{F}_1=F_c+\Delta F\left[\frac{e^{f\phi }}{e^{f\phi }-1}\right]+F_c\left[\frac{e^{f\phi }}{e^{f\phi }-1}\right]-F_c\left[\frac{e^{f\phi }}{e^{f\phi }-1}\right]\\ =\left(F_c+\Delta F\right)\left[\frac{e^{f\phi }}{e^{f\phi }-1}\right]+F_c-F_c\left[\frac{e^{f\phi }}{e^{f\phi }-1}\right]\\ =\left(F_c+\Delta F\right)\left[\frac{e^{f\phi }}{e^{f\phi }-1}\right]-\left[\frac{F_c}{e^{f\phi }-1}\right]\\ =\frac{\left(F_c+\Delta F\right)e^{f\phi }-F_c}{e^{f\phi }-1}\end{array}$

Conclusion:

Substitute $3.037\text{rad}$ for $\phi$ and $664\text{lbf}$ for $\Delta F$, $0.80$ for $f$ and $73.4\text{lbf}$ for $F_c$ in Equation (I).

$\begin{array}{c}{F}_1=73.4\text{lbf}+664\text{lbf}\left[\frac{e^{0.80\times 3.037\text{rad}}}{e^{0.80\times 3.037\text{rad}}-1}\right]\\ =73.4\text{lbf}+728.12\text{lbf}\\ =801.52\text{lbf}\end{array}$

Substitute $664\text{lbf}$ for $\Delta F$, and $801.52\text{lbf}$ for $F_1$ in Equation (II).

$\begin{array}{c}{F}_2=801.52\text{lbf}-664\text{lbf}\\ =137.5\text{lbf}\end{array}$

Substitute $801.52\text{lbf}$ for $F_1$, $137.5\text{lbf}$ for $F_2$, $73.4\text{lbf}$ for $F_c$ in Equation (III).

$\begin{array}{c}{F}_i=\frac{801.52\text{lbf}+137.5\text{lbf}}{2}-73.4\text{lbf}\\ =469.51\text{lbf}-73.4\text{lbf}\\ =396.11\text{lbf}\end{array}$

Substitute $801.52\text{lbf}$ for $F_1$, $137.5\text{lbf}$ for $F_2$, $3.037\text{rad}$ for $\phi$, and $73.4\text{lbf}$ for $F_c$ in Equation (IV).

$\begin{array}{c}{f}^{\text{'}}=\frac{1}{3.037\text{rad}}\ln \left(\frac{801.52\text{lbf}-73.4\text{lbf}}{137.5\text{lbf}-73.4\text{lbf}}\right)\\ =\frac{1}{3.037\text{rad}}\times 2.43\\ =0.80\end{array}$

Thus, the value of $F_1$ restoring $f$ to $0.80$.