Randall D. Knight — Dynamics I: Motion Along a Line: Problem 76

A smaller holding tension \(T_{hold}\) can support a larger load tension \(T_{load}\) if a rope is wrapped partially or completely around a cylinder. In old sailing ships, the enormous tension in a rope holding a sail could be controlled by only a few sailors when the rope was wrapped around a capstan. Similarly, belaying rock climbers can support their weight in case of a fall. FIGURE CP6.76 shows a rope wrapped through angle \(\phi\) around a cylinder of radius \(R\) whose coefficient of static friction is \(\mu_s\). The minimum tension needed to hold the load occurs when the rope is on the verge of slipping. The inset shows a small segment of rope spanning angle \(d\phi\). Assume that the tension on the load side is infinitesimally larger than on the holding side. The static friction force is tangent to the rope and thus perpendicular to \(\vec{n}\). However, the small curvature of this segment means that the two tension forces act at angles \(\frac{1}{2} d\phi\) to the tangent line. a. Apply Newton’s second law to this segment of the rope to get an equation relating \(dT\), the increase in tension across this little segment, to \(d\phi\), then integrate from \(T = T_{hold}\) at \(\phi = 0\) to \(T = T_{load}\) at angle \(\phi\). You will need to use the small-angle approximations \(\sin \theta \approx \theta\) and \(\cos \theta \approx 1\) when \(\theta\), in radians, is very small. Give your result as an equation for \(T_{hold}\), the minimum tension force needed to support load \(T_{load}\). This result is known as the capstan equation. b. A rock climber who weighs \(670 \text{ N}\) is dangling from a rope. His partner uses a belaying device in which the rope is wrapped around a cylinder with \(\mu_s = 0.40\). What is the magnitude of the tension needed to support the climber if the rope is wrapped through an angle of (i) \(90^{\circ}\), (ii) \(180^{\circ}\), and (iii) \(360^{\circ}\)?