2 Engineering Example 1

2.1 Pulley belt tension

Problem

Consider that a belt is partially wound around a pulley so that there is a difference in the tension either side of the pulley (see Figure 4). The pulley will be stationary as long as the friction between belt and pulley is sufficient. The frictional force on the pulley will depend on the extent of the contact between belt and pulley i.e. on the angle $\theta$ shown in Figure 4. Given that the tensions on either side of the belt are $T_2$ and $T_1$ and that the coefficient of friction between belt and pulley is $\mu$ , find an expression for $T_2$ in terms of $T_1$ , $\mu$ and $\theta$ .

Solution

Consider a small element of the belt, at angle $\theta$ where the tension is $T$ . Changing the angle by a small amount $\Delta \theta$ changes the tension from $T$ to $T+\Delta T$ .

Figure 4

Take moments about the centre of the pulley, denoting the radius of the pulley by $R$ and assuming that the frictional force is $\mu T$ per unit length. For the pulley to remain stationary,

$R\Delta \theta \mu T=R\left(T+\Delta T\right)-RT$ or $\Delta \theta =\frac{\Delta T}{\mu T}.$

Using integration as the limit of a sum,

$\theta ={\int \; <!--nolimits\; -->}_{T_1}^{T_2}\frac{dT}{\mu T}=\frac{1}{\mu }{\left[lnT\right]}_{T_1}^{T_2}=\frac{1}{\mu }ln\left(\frac{T_2}{T_1}\right).$ So $T_2=T_1{e}^{\mu \theta }.$

Exercises

1. Find the area under $y=x+1$ from $x=0$ to $x=10$ using the limit of a sum.
2. Find the area under $y=3x^2$ from $x=0$ to $x=2$ using the limit of a sum.
3. Write down, but do not evaluate, the integral defined by the limit as $\delta x\to 0$ , or $\delta t\to 0$ of the following sums:
   1.   ${\sum }_{x=0}^{x=1}{x}^3\delta x,$
   2.   ${\sum }_{x=0}^{x=4}4\pi x^2\delta x,$
   3.   ${\sum }_{t=0}^{t=1}{t}^3\delta t,$
   4.   ${\sum }_{x=0}^{x=1}6m{x}^2\delta x$ .

Answer