Great Pulley Problem

[bottomscraper]

For those interested in this problem and others like it might I suggest taking a look at this:

http://www.physicsclassroom.com/Class/v ... ectoc.html

The math is just vector math which you probably already know for solving navigation problems.

[Chris Reinke]

Carl - I believe you have captured the true understanding of the question, and possed the best explanation as to an answer. I would like to offer a simple summary I was offered many years ago in sailing school :

A stationary block represents a change in direction, while a moving block represents mechanical advantage. I am sure someone will find fault in this simple explanation......but it works for me.

Now to add fuel to the fire - How is the weight secured to the line? Most knots used in sailing have an inner bend that is less than 2x the diameter of the line. In doing so, the outer edge of the bent line is weighted with the majority of the load, while the inner edge of the line is slack. This is the fundamental reason for recommended block diameters being more then twice the diameter of the line running over them. In the case of a bowline the breaking strength of the knot is as low as 60% of the line tensile strength. Is the line secured with a thimble, or does John need to stay awake nights trying to include a reduced breaking strength component in his theory?

[Neil Gordon]

Now to add fuel to the fire - How is the weight secured to the line?

True, knots will reduce the breaking strength of the line. By the way, in John's example, the line would have failed because he/we didn't factor in the weight of the line itself. I suggest eliminating the weights and the knots and instead just start with 20 lbs of rope, i.e., 10 lbs on each side of the pulley.

[John Vigor]

Thank you, Carl. You have clearly described what I have always regarded as magic--the fact that a line able to bear a tension of only 10 pounds will support 20 pounds without breaking because of the way it sheds the load to the sheave or peg. The tension in the line is no more than 10 pounds at any one point. As you say, the line will not break. This is obviously a very difficult concept for some people to absorb.

Perhaps we need to think about anchor chain now. The principle seems to be exactly the same. An upright link of chain has two rods bearing the tension vertically, but it has only one rod going horizontally across the top--yet that one rod can bear the same force as the two rods combined. Once again, it's magic as far as I'm concerned.

Cheers,

John V.

[Neil Gordon]

Perhaps we need to think about anchor chain now. The principle seems to be exactly the same. An upright link of chain has two rods bearing the tension vertically, but it has only one rod going horizontally across the top--yet that one rod can bear the same force as the two rods combined.

The difference being that the advertised breaking strength for chain already takes that into consideration.

On the other hand, if you set two anchors from a single chain looped over a windlass, can you then cut the chain size (i.e., breaking strength) in half?

[Carl Thunberg]

I haven't taken the time to fully consider this, but my first inclination is that our analogy breaks down here. Rope is flexible and chain is made of solid links. The links themselves cannot deform. The two rods are in tension, but where the two links bear on each other, there's a significant shear component. There are different stresses at work here than in rope. And, as Neil said, the rating on the chain takes this into account. My GUESS is, the strength in tension is greater than the shear strength, so shear would govern. Strictly a guess, but an educated guess.

[Ed Haley]

Yup, Carl, you and Didereaux cleared it up for me. I guess when you're retired for a while you shouldn't shoot from the hip w/o making sure your belt is tight.

Thanks.

[Neil Gordon]

... magic--the fact that a line able to bear a tension of only 10 pounds will support 20 pounds without breaking because of the way it sheds the load to the sheave or peg. The tension in the line is no more than 10 pounds at any one point.

Wait... I think it's what Didereaux said. The peg/sheave doesn't absorb anything; all it does is facilitate the change in direction of the force.

Okay, try this one, as a way to eliminate the effect of the pulley and the change in direction. Let's make the exercise horizontal. I exert 10 lbs of force on the line and you exert 10 lbs in the other direction. What's the force on the line?

10 lbs <---------------10 or 20?----------------> 10 lbs

[s-dupuis]

As long as the line is free, that is, not fixed to anything.

Picture a scale attached to each end of the line, the type you might use to weigh a fish. The spring inside the scale needs 10 pounds of force for it to accurately read 10 lbs. If both scales indicate 10 lbs. then there must be a total of 20 lbs. being exerted on the rope.

I'm also guessing that the 20 lbs. of force is spread out along the entire length of the rope.
Wouldn't this beg the question that the breaking strength of a line would need to be based on a specific length of line?

Steve

[Ed Haley]

Suppose, as in Neil's example, you attach a spring scale (like you weigh fish with) to a hook on a wall. You pull on the other end of the spring with a force of 10 pounds. The spring scale is reading 10 pounds and the wall is applying a force of 10 pounds against you (through the material in the wall).

If you remove the spring scale from the wall and have someone pull against your 10 pound force with a 10 pound force, the scale would still read 10 pounds.

[Neil Gordon]

The spring inside the scale needs 10 pounds of force for it to accurately read 10 lbs. If both scales indicate 10 lbs. then there must be a total of 20 lbs. being exerted on the rope.

I'm also guessing that the 20 lbs. of force is spread out along the entire length of the rope.
Wouldn't this beg the question that the breaking strength of a line would need to be based on a specific length of line?

Do it vertically and the scales will both read 10 lbs if you hang a 10 lb weight. The lower scale weighs the 10 lb weight. The upper scale weights everything under it which, if you ignore the lower scale and the rope, is the same 10 lb weight.

Tension in the line is the same along the whole line and so doesn't depend on length. If you exceed the breaking strength, the line will fail, probably at the manufacturing weak point (and excluding weaker connectors and/or knots) if all else is equal, since 100% consistency along the line seems impossible.

[Neil Gordon]

Suppose, as in Neil's example, you attach a spring scale (like you weigh fish with) to a hook on a wall. You pull on the other end of the spring with a force of 10 pounds. The spring scale is reading 10 pounds and the wall is applying a force of 10 pounds against you (through the material in the wall).

If you remove the spring scale from the wall and have someone pull against your 10 pound force with a 10 pound force, the scale would still read 10 pounds.

So if you pass the line over a pulley and change its direction but apply the same force, do you still get 10 lbs no matter where you put the scales?