Monday's effort

Re: Monday's effort

You are asking me to remember physics 101?!?!

If the second has the person pulling on the rope in the same direction as the first, then parts one and two can be answered using only a single digit.

I took it that the third part had the user of the rope pulling in the same direction as the item is to be moved (i.e., using all three sheaves of each block), in which case the sum of digits of my answer (rounded to the nearest integer) is 23.

Hopefully this answer is opaque enough that, if right, I won't get sent to the box for an early answer.

Whoops!

I'm afraid both my memory of physics 101 and my reasoning ability failed me late last night. I'll try again:

Part 1: Mechanical advantage = 1 (i.e., no force multiplier); in this case the single pulley only changes the direction. I think that is the only one I thought clearly on last night!

Part 2: Six reversals to get a mechanical advantage of 6, assuming that you pull in the same direction as the object is to be moved. If you make another turn around a sheave in the fixed block to change direction as in the first example, then seven reversals.

Part 3: 6.4 tons = 2 tons x 6 pure mechanical advantage x .9^6 surviving friction.

Thanks for the problem. Should have been easy. Oh, well, can't retain everything!

Ding, ding, dooooing!

Alex answered to the point I was trying to make which is that a single pulley doesn't help you move anything to advantage, it only changes direction of the load at the cost of an additional percentage of friction. Remember that the next time you struggle to lift a wet hay bale or a bundle of shingles a couple of stories. Even if it is Physics 101 don't be fooled by a single pulley as an assist. I'm betting that more than just me was under the impression that using a single pulley (1) was a mechanical advantage. Not so.

The second answer is also correct, the second time. There are seven lines between the pulleys (one is the loose end one is attached to the pulley) provided that the anchor is not attached to the moving pulley and all three pulley sheaves are used.

I arrived at a different answer with my friction numbers. My answer is 7 1/2 tons. This is the method I was taught. To make the the adjustment for friction, multiply 12 by 10 = 120. This result I divided by 10 + 6 (six sheaves) or 16. 120 divided by 16 Answer 7.5. As to Alex's response. I'm not familiar with the ^ symbol so my method may be way off base. There is the distinct possibility of a world of hurt between 7.5 and 6.4 tons on a line. My math skills are long ago and suspect so I'm open (if unprepared) for additional discussion.

The symbol

The ^ symbol is used to represent and exponent when you do not have the normal math symbols/structure available. Thus x^2 would be x squared, etc.

one-pulley clarification and friction formula

Just to point out that the number of pulleys is not all-telling, if one wanted to lift a bundle of shingles up two stories, and attached a rope to the bundle and passed that rope through a pulley on a beam two stories up, you would not get any mechanical advantage. But that one pulley CAN give you a mechanical advantage. Tie the rope to the beam and pass it through the pulley attached to the bundle of shingles. Then the guy on the second floor has a 2:1 mechanical advantage in lifting the shingles.

Where do the numbers in your friction formula come from, and what is their rationale? I think the "12" is the lift with no friction losses, and the formula for friction adjustment is (1/f)/((1/f)+#of pulleys)?

I was taking the total mechanical advantage and adjusting for frictional losses with the following rationale:

The output of the first pulley is 90% of the input, with the rest being lost to heat (or partly to sound if it is a squeaky pulley). The second pulley takes the 90% of original work that gets transfered to it, and transfers 90% or that, or 81% of the original, on to the third pulley. The third pulley takes the 81% of original work and transfers 90% of that, or 72.9% of the original, on to the fourth pulley...

But after writing that out, I think it is wrong. It implicitly assumes that all of the work of lifting the object is subject to the loss in all six sheaves. It seems now that the first pulley does "its" share of lifting after only one friction loss, and passes on the remaining work to the other pulleys. If that is the correct view, the total force applied to lift the object will be .9 + .9^2 + .9^3 + ... + .9^6 = 4.217. If that is the case, you can lift 8.43 tons.

But I highly recommend checking with an engineer or physicist before doing so!

Re: Ding, ding, dooooing!

A comment and a question. You can gain mechanical advantage with just one sheave, it just needs to be fixed to the load (of course this was not how you described the set up in question 1). In all multi sheave rigging, for a given number of sheaves, how the sheaves are reeved deteremines the amount of mechanical advantage. If the dead end of the line is fixed to the load end with the force applied in the same direction as the load is moving one gains one addtional "part" 6 sheaves, 7x advantage. If the dead end is fixed to the non moving block, the advantage is eqaul to the number of sheaves with one sheave providing a direction change to the force.

The question I have is where do you obtain the the 10 percent friction figure? That seems awful high, like you could run out of the ablitiy to do work rather quickly. Admittedly all the rigging I've been around has been with wire rope, and now that I think about it it was awful hard to pull 5/8' wire through 12 part sheaves.

glc

Wikipedia has same formula as Gary *LINK*

Anyone have any explanation for that formula?

Wikipedia version 9same formula, just in different layout:


Wikipedia Block and Tackle article

I should check

with Wikipedia first. I wish I had seen that. Actually where I got my original information is tied to Robert Scharff, who was writing about his 1989 book entitled "Workshop Math". Early on I knew I was treading on thin ice for part three when I couldn't get a confirming response from the author, Sterling Publishing, or Popular Science Magazine who all had a part in the book at that time. The web, of course, has multiple pages on pulleys and friction but somehow I never got to Wikipedia which is where I regularly start a search. I ran out of time and Monday's trivia question was all I could come up with. My apologies if the information is inaccurate.

Re: I should check

Well, since your formula, from a different source, agrees with Wikipedia, I'm inclined to think it is probably correct (although we have to allow for the possibility for Wikipedia being the source for the other material). But you've really got my curiosity up, and why we can't get at that formula from first principles. Don't have time today, but tonight will do a little digging if no-one chimes in before.

Friction in a block and tackle *LINK*

Okay, I found a source. My problem is that I was trying to make this too complicated. I was treating the 10% friction as a friction loss in each pulley. That is not it. It is an estimate of the additional load ON THE WHOLE BLOCK AND TACKLE of friction, as a function of number of sheaves. So if you are lifting 1,000 lb with a three-fold purchase (I love these names), you add 10% times 6 pulleys to the load, so treat it like a lift of 1,600 lbs with a frictionless B&T.

For the example in the original question, we are lifting 1.6*w when you include friction, and with a rope pulling 2 tons, that says that 6(mechanical advantage)* 2 tons = 1.6 w, so W = 7.5, as Gary said.

This web site has more than you could ever want to know (okay, at least more than I ever wanted to know!) about blocks and tackles.


Block and Tackle Info

I'm relieved. Thank you Alex.