You want to lower a heavy object over the edge of a cliff on a rope. You intend to lower it smoothly, at constant velocity, and pull on the rope to prevent it from accelerating. In order to reduce the force which you have to exert, you wrap the rope several times round a smooth cylindrical object which is fastened so that it can neither move nor rotate (Let's say, the handrail at a viewpoint). The rope will tighten around the cylinder, and the ensuing friction will reduce the force.

How large is the force with which you have to pull at your end of the rope, depending on the load, the coefficient of friction and the number of turns of the rope round the cylinder? The number of turns should be treated as a real number, since you can wrap the rope 3/4 of the way 'round.

If you don't know where to start, try to compute the friction force when the rope only just touches the cylinder, with the direction of your pull nearly opposite to the load force. Then build up the turns of the rope from these infinitesimal changes in direction.

Let the weight be W1, the force you pull on the rope W2.  If you sum differential forces about a very small differential angle, in both the radial and tangential directions (relative to the cylinder), you end up with (after integration, and remembering that friction force = coefficient of friction times normal force, per Tristan)

W1 = W2 * e^(mu * angle),  where:  e = 2.718...    mu = the static coefficient of friction between the rope and cylinder (the rope is not moving...) and   angle = the total wrap angle in raidians.

This physical technique is used for climbing and rescueing.

Edited on September 26, 2006, 5:53 pm

Posted by Kenny M on 2006-09-26 17:51:22