Question

A belt is wrapped around a pulley as shown in the figure. The coefficient of friction between the belt and pulley is \(0.3\). If there is no slippage between the belt and pulley, the angle of wrap (\(\theta\)), in degrees, is . (Rounded off to two decimal places)

Hint:

Use the belt friction (capstan) equation relating the tight side and slack side tensions to the coefficient of friction and the angle of wrap.

Answer 1

Correct Answer: 132.38

Rather than quoting the belt friction formula directly, let's build it from the equilibrium of a tiny piece of belt in contact with the pulley, then use it to find the angle of wrap.

Take a small element of belt that subtends an angle $d\theta$ at the pulley centre. The tension on one side of this element is $T$ and on the other side is $T+dT$, since friction increases the tension gradually as we move around the wrap. The belt also presses on the pulley with a small normal force $dN$.

Balancing forces along the direction of pull and perpendicular to it for this tiny element gives $dN = T\,d\theta$ (for small $d\theta$) and, at the point where slipping is about to start, the friction force equals $\mu\,dN$. This friction force is what raises the tension across the element, so $dT = \mu\,dN = \mu T\,d\theta$.

Separate the variables and integrate this over the whole wrap, from the slack side tension $T_2$ up to the tight side tension $T_1$, as $\theta$ runs from $0$ to the full wrap angle:

\[ \int_{T_2}^{T_1} \frac{dT}{T} = \int_0^{\theta} \mu\,d\theta \]\[ \ln\left(\frac{T_1}{T_2}\right) = \mu\theta \]

This is the same capstan relation, now derived rather than assumed. Rearranging for the angle:

\[ \theta = \frac{1}{\mu}\ln\left(\frac{T_1}{T_2}\right) \]

From the figure, the two belt pulls are $5.0$ kN and $2.5$ kN, with the larger one being the tight side tension $T_1$ and the smaller one the slack side tension $T_2$. The coefficient of friction is $\mu = 0.3$.

\[ \theta = \frac{1}{0.3}\ln\left(\frac{5.0}{2.5}\right) = \frac{\ln 2}{0.3} = \frac{0.6931}{0.3} = 2.3105 \text{ rad} \]

Change this to degrees by multiplying by $180/\pi$:

\[ \theta = 2.3105 \times \frac{180}{\pi} = 132.38^{\circ} \]

Let's summarize:

• Friction on a belt element gives $dT = \mu T\,d\theta$, which integrates to the capstan equation $\ln(T_1/T_2) = \mu\theta$.
• The tight side pull is $5.0$ kN and the slack side pull is $2.5$ kN, giving a tension ratio of $2$.
• Solving for $\theta$ and converting to degrees gives $132.38^{\circ}$.

So the angle of wrap of the belt on the pulley is $132.38^{\circ}$.