Chapter 7.5, Problem 66E

Belts and Pulleys A thin belt of length L surrounds two pulleys of radii R and r, as shown in the figure to the right.

1. (a) Show that the angle θ (in rad) where the belt crosses itself satisfies the equation

$\theta +2\cot \frac{\theta }{2}=\frac{L}{R+r}-\pi$

[Hint: Express L in terms of R, r, and θ by adding up the lengths of the curved and straight parts of the belt.]

1. (b) Suppose that R = 2.42 ft, r = 1.21 ft, and L = 27.78 ft. Find θ by solving the equation in part (a) graphically. Express your answer both in radians and in degrees.

To determine

To show: The equation $\theta +2\cot \frac{\theta }{2}=\frac{L}{R+r}-\pi$, where the belt crosses itself with an angle $\theta$.

Explanation of Solution

Given:

The thin belt of length L surrounds two pulleys of radii R and r, as shown in the given figure.

Calculation:

Redraw the given figure for convenient as shown below in Figure 1.

Obtain the angle of $A\widehat{F}B$ as follows.

$\begin{array}{c}\text{angle of }A\widehat{F}B=360°-90-90-\theta \\ =360°-180°-\theta \\ =\pi -\theta \end{array}$

Now, find the angle of $D\widehat{G}C$ as shown below.

$\begin{array}{c}\text{angle of }D\widehat{G}C=360°-90-90-\theta \\ =360°-180°-\theta \\ =\pi -\theta \end{array}$

Note that, the arc length is $2\pi R\frac{C}{360°}$, where C is central angle.

Obtain the arc length of AB as shown below.

$\begin{array}{c}\text{arc length of }AB=2\pi R\frac{\pi -\theta }{360°}\\ =R\left(\pi -\theta \right)\end{array}$

Similarly, obtain the arc length of CD as follows.

$\begin{array}{c}\text{arc length of }CD=2\pi r\frac{\pi -\theta }{360°}\\ =r\left(\pi -\theta \right)\end{array}$

Now, obtain the length of the bell on left side as shown below.

$\begin{array}{c}\text{length of belt on the left side}=2\pi R-R\left(\pi -\theta \right)\\ =2\pi R-R\pi +R\theta \\ =\pi R+R\theta \\ =R\left(\pi +\theta \right)\end{array}$

Similarly, obtain the length on the right side as follows.

$\begin{array}{c}\text{length of belt on the right side}=2\pi r-r\left(\pi -\theta \right)\\ =2\pi r-r\pi +r\theta \\ =\pi r+r\theta \\ =r\left(\pi +\theta \right)\end{array}$

Now consider the triangle AFE,

$\begin{array}{l}\frac{AE}{AF}=\frac{AE}{R}\\ AE=\tan \left(\frac{\pi }{2}-\frac{\theta }{2}\right)\\ AE=R\cot \frac{\theta }{2}\end{array}$

Similarly, $CE=r\cot \frac{\theta }{2}$.

Note that, the length of the belt is denoted by L and it can be calculated as follows.

$\begin{array}{c}L=\text{length of the belt }AB+\text{length of the belt }CD+2AE+2CE\\ =R\left(\pi +\theta \right)+r\left(\pi +\theta \right)+2R\cot \frac{\theta }{2}+2r\cot \frac{\theta }{2}\\ =\left(R+r\right)\left(\pi +\theta \right)+2\cot \frac{\theta }{2}\left(R+r\right)\\ =\left(R+r\right)\left(\pi +\theta +2\cot \frac{\theta }{2}\right)\end{array}$

On further simplification,

$\begin{array}{l}\frac{L}{R+r}=\pi +\theta +2\cot \frac{\theta }{2}\\ \frac{L}{R+r}-\pi =\theta +2\cot \frac{\theta }{2}\\ \theta +2\cot \frac{\theta }{2}=\frac{L}{R+r}-\pi \end{array}$

Hence, it is proved.

To determine

The angle $\theta$, by solving the equation in part (a) graphically.

Answer to Problem 66E

The values of $\theta$ has multiple solutions.

Explanation of Solution

From part (a),note that $\theta +2\cot \frac{\theta }{2}=\frac{L}{R+r}-\pi \left(1\right)$.

Substitute the given values $R=2.42\text{ft},r=1.21\text{ft},L=27.78\text{ft}$ in the equation (1) as shown below.

$\begin{array}{c}\theta +2\cot \frac{\theta }{2}=\frac{L}{R+r}-\pi \\ \theta +2\cot \frac{\theta }{2}=\frac{27.78}{2.42+1.21}-\pi \\ \theta +2\cot \frac{\theta }{2}=4.5113\end{array}$

Use online graph calculator and obtain the graph of the function $f\left(\theta \right)=\theta +2\cot \frac{\theta }{2}$ as shown below in Figure 2.

From Figure, it is observed that the values of $\theta$ are 1.047, 12.048, 18.567… and it has multiple solutions.

Therefore, the values of $\theta$ has multiple solutions.