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The power of friction (Posted on 2006-09-26) Difficulty: 3 of 5

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.

  Submitted by vswitchs    
Rating: 4.0000 (1 votes)
Solution: (Hide)

Physical arguments

Before computing the solution rigorously, let's look at the problem with the eyes of a physicist and see what we can learn. First, the force your hand has to exert FH should be proportional to the load force FL. There is no other force given in this business on which it might depend. So it remains to determine the factor f:=FH/FL, which will depend on the number of times the rope is wrapped round the railing, n. The second observation is that two turns equal twice one turn, so

f(n=2) = f(n=1)^2

and similarly for any factor between two values of n:

f(n1) = f(n2)^(n1/n2)

This can only be fulfilled if f is an exponential function of n. However, we cannot find out by physical arguments which coefficient n will have, though we might guess that it is proportional to the coefficient of friction between rope and cylinder. To quantify that, we need the

Formal derivation

The method to derive the solution formally is to work things out for an infinitesimal n and then integrate. n being infinitesimal means that the rope only just touches the cylinder and the direction of your pull is nearly opposite to the direction of the load, differing only by an angle 2 pi n << 1. We build a parallelogram of forces in which we neglect the infinitesimal difference between FL and FH and get the normal Force FN with which the rope presses against the cylinder:

FN = 2 FL sin(2 pi n/2) = FL 2 pi n

The difference between FL and FH is the friction force:

FL - FH = u FN = u 2 pi n FL

where u is the coefficient of friction. It is the sliding coefficient as you are lowering the load; if you were merely supporting it, it would be the static friction coefficient. The right-hand side of this equation contains the infinitesimal n, and the left-hand side is an infinitesimal reduction in the force in the rope. We will now call this force just F and obtain the differential equation of an exponential function:

dF/dn = - u 2 pi F

Integrating up, we get the result:

FH = FL exp(-2 pi u n)

A Knot site has a nice article on this effect, with a sketch and a statement of this result.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionsolutionKenny M2006-09-26 17:51:22
Some ThoughtsThe easy partTristan2006-09-26 13:50:41
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