...or, would they be lifted all togheter? mmmh (I'd say A too, at first sight) - Claudio Cicali

assuming no friction? - Alex Scrivener

liked so I get the answer later.... - WoH: Minding her Botts

Alex, that's the first thing I was thinking. - Rah-PM 2012

all of them. - Guy

A would rise first if there was a surface for the other two to rest upon, and since it looks like they're all on resting on the same plane as the man's feet. A, is the answer. - Jimminy, CoG of FF

Don't we need to know how much force the gentleman is pulling at? - Alonzo Hayden

Nope force does not matter Alonzo. But good stalling tactic. - SteVe C

guess B - shâhin

Just so you guys know, I have no idea what the answer is. I feel like there isn't enough information, but that's probably me overthinking it. - Rah-PM 2012

++ Rah 8) - Zu from AOD

I'm going to say that C is the answer. A seems too easy based on weight and I think that the work will transfer first to the set of pulleys closest to the puller. - Scoble, Alex Scoble

Rah, you can;t ask a question like this without knowing the answer . . . . can you? - Friar Will (:^)

I'm reminded of the airplane/treadmill problem http://blog.xkcd.com/2008... . - Micah

Unlike the airplane/treadmill problem, this one is unambiguous and has one correct answer. I think that answer is A. - Bruce Lewis from fftogo

But the "This is a man" part is always ambiguous. - Micah

My gut reaction is C for the same reason Scoble says. I was hoping someone had seen this problem somewhere before :) - Rah-PM 2012

I'm guessing "C" - visualizing pulling the rope C has to move before B before A. - WarLord

It's B or A. C is definitely out, because it doesn't matter what's first, but what has less downward force, you're just drawing the rope taught, so the weight with lowest force will rise first. - Jimminy, CoG of FF

What do you mean "I have no idea what the answer is."?? A lot of us will lose tonight's sleep. - Guy

I'm pretty sure it's A. But I haven't been in physics class for a long time. - Heather

Secret answer D: The rope would snap and the man would be sent flying by his own momentum into the inconveniently placed lava pit just out of frame. - Joe "Looptid" Pierce

Where the hell is Mo, he would know the answer. - Jimminy, CoG of FF

I'm told B because 40 is half of 20+60 which are at the extremities. And this comes from a woman. - Guy

I'm now sure it's A. Will explain later. - Bruce Lewis from fftogo

@Jimminy Here's Moe http://www.herbanmedia.com/shoppin... - Alonzo Hayden

Oh didn't guess yet. Well all those years of college physics say I forget. Normal rule of thumb is a pulley divides work by half. Based on that rule I'd go B. but my brain isn't turned on today - SteVe C

I found a 6 page forum threat about this exact problem, pretty interesting responses. http://gbxforums.gearboxsoftware.com/showthr... - Chris Topher

The rope is inelastic. So all the weights go up at the same time. - The original Kevin

I have not only the answer, but an explanation that will be persuasive to everyone. Should I post the spoiler here, or put it somewhere else? - Bruce Lewis

Might as well post it on your own feed, Bruce and link to it here. - Scoble, Alex Scoble

hmm, looks like one of the people built a rig to test it. I think that friction might have something to do with his finding, though: http://gbxforums.gearboxsoftware.com/showthr... - Chieze Okoye

It's A :) - HoldMe

Yes, the friction totally threw that experiment off. Cool that he built something, though. His finding is wrong, and my explanation will be persuasive to everyone. Writing it up now. - Bruce Lewis

My explanation: http://friendfeed.com/brlewis... - Bruce Lewis

I think Bruce is right. - Scoble, Alex Scoble

My problem is that I assumed that this wasn't a straightforward question and so looked for the answer that wasn't straightforward. - Scoble, Alex Scoble

C - Outsanity

I can analyze it as long as it's static, but when things start moving I have to get an ME. - Kevin L

Assuming the numbers are weights, then it's A. The mechanical advantage on each weight is 2-fold, but identical across each of them. So the lightest weight lifts first, regardless of where it is. Only when it reaches the top and is blocked from moving will any of the other weights lift, because the man cannot put more than 10 pounds of tension on the line as long as the 20 pound weight can move. Note: Friction changes things. In a real world scenario, if you have *any* friction, then it will be possible to lift the 60 pound weight first, as the downforce provided by the other two weights will then be >30 pounds. - Otto

I'd go with C. - jbrotherlove

Close, Sean. Remember: in stasis, the tension is equal at every point along the rope. - Kevin L

I think Bruce has it. - Chieze Okoye

Define "rope"... just kidding. - Mark J

OK, here's a follow-up question. Assume the man pulls A and B all the way to the top. Finally, C lifts off the ground. How many pounds of force is he exerting on the rope? - Bruce Lewis

C, .. - pyla

Total download force = 60+ 40 + 20 = 120. 3 free axles means at equilibrium, the force in each segment is 120/(3*2) = 20. So for each weight, the total upward force is 2 * 20 = 40... so weight A is the only one that will move up. - Ken Morley

For Bruce's question the answer is 60/2 = 30 :o) - Ken Morley

If we guess backwards, which one moves firstly? My answer is C. - _©onta©t_

My mistake. Actually, I'd assumed all the weights were unsupported, so the total downward force doesn't change when A and B reach the top. The force should still be 120 / 6 = 20, not 30. If the weights are all sitting on the ground, then the forces in the rope will change as each weight leaves the ground, but A will still rise first....but I could be wrong :o) <I am... see below> - Ken Morley

3 free pulleys x 2. Only the movable pulleys multiple the forces. - Ken Morley

Ken, each weight is independent, so the factor of the pulley's benefit is 1/2, the third pulley in the sets are only changing the vector. And the benefit is non cumulative, since it's 1/2 for each weight. - Jimminy, CoG of FF

But the system is in equilibrium, so all the forces must be equal. http://en.wikipedia.org/wiki... - Ken Morley

Ken, the issue with that, is it's an individual weight, distributed, not independent weights. So each weight's force measure, is calculated independently. - Jimminy, CoG of FF

3 free pulleys, 120 total pounds, force = 120/3/2. What am I missing? - Ken Morley

3 free pulleys? Labeled with a number right to left, which ones are free? - Jimminy, CoG of FF

The ones on the weights are the only ones that can move, and so are the only ones that multiply the force. - Ken Morley

But they don't multiply force, they only split the amount of effort required to lift that wait, it doesn't transfer over the entire system. - Jimminy, CoG of FF

"The addition of a fixed pulley to the single pulley system can yield an increase of advantage..." yes, but in that case the end is attached back to the weight, not fixed. - Ken Morley

Okay my brain hurts... I'm going to bed and will reconsider in the morning ;o) - Ken Morley

Ken, if we were talking about one 120 weight attached in the configuration, you would be correct. Independent weights don't work the same way. - Jimminy, CoG of FF

Yes, W / 2 (free pulleys) / 2 = W/4 :o) - Ken Morley

Assuming that the wheels turn equally, the force is equally spread across all the weights. Imagine pulling on the rope with 1lb of pressure - nothing moves, but the rope gets taught. Next, imagine 2 lbs of pressure - still nothing. Imagine increasing the force applied by small increments: at some point, the rope will transfer a force equal to the weight of one of the blocks. At that point, the block lifts off the ground, right? - I like big Botts

If they are attached to the same weight. - Jimminy, CoG of FF

I'm going to go buy some pulleys tomorrow ;o) - Ken Morley

g'nite :o) - Ken Morley

Ken, here is an explanation, somewhat, http://ff.im/q9wKF, of why the diagrams on wikipedia are wrong, in application to this system. Each of these weights are independent and and thus examples of diagram 1, you can see this if you remove the 7 right most pulleys from the diagram, and then only the 3 right most after that. You end up with a system that only contains the 20[unit] weight, and a second that contains only the 20 and 40[unit] weights. - Jimminy, CoG of FF

Computer programmers, you should be able to come up with the answer to my followup question, and with an explanation that's easily persuasive to everyone. Black-box abstraction is the key to simplifying this system. - Bruce Lewis

For non-programmers: Black-box abstraction is enclosing a system in a mental black box, so that you only think about what's going in and/or out of the box, not what's inside. Don't be thrown off by the fancy word "abstraction". It's a way to think less, not think more. Maybe you can get the answer before any programmers do. - Bruce Lewis

Jimminy, the key point is that In order for the starting position to be stable, the weights must all be supported by a surface. As the tension in the system is increased and reaches 10 lbs (I'm assuming the units are lbs), the first weight (A) will rise until it reaches the top (the other two pulleys are acting as fixed pulleys at this point). When the tension reaches 20 lbs the 'B' weight will rise. The 'C' weight will not move until the tension is increased to 30 lbs. - Ken Morley

Just came across this from HN, Ken, the assumption is that the system is stable, due to the fact that the weights are all at the same distance from the pulley arm, also they are apparently at the same level as the mans feet. I also, it's cumulative, the mechanical advantage is only a factor of 2, so it's, 20/2, 10 lbs. to raise A, to raise A and be it's, (20+40)/2, 30 lbs, and to raise A, B, and C you would need 120/2, 60 lbs. Which may have been what you were saying, but I'm not quite sure. - Jimminy, CoG of FF

Jimminy says 60 lbs, Ken says 30. If you're the one who's right, see if you can use black-box abstraction (see my comment above) to persuade the other. - Bruce Lewis

Jimminy, the system can not be stable if the weights are not supported. Imagine if the weights are all sitting on a platform that is gradually lowered. What would happen? In your example above you are correct that 10 lbs of force is required to raise weight A, but when it reaches the top and runs out of rope, it is no longer acting as a pulley. It is as though it were just attached at that point. The tension in the system is still only 10 lbs as the other weighs are still sitting on the ground. As you pull harder and the tension reaches 20 lbs then the second weight raises... etc. - Ken Morley

I'll guess A as well before I look at the 83 comments here and see what the answer is. - Dan owns Comicsforge.com

Hiding the thread, just thinking about having to think about it makes my head hurt :P - Rene, Pro Button Pusher

Ken, considering, that the man is also standing on the same surface as the weights, nothing would change. Also, you still have to support weight A, even once it reaches the top, it is still a downward force, on the line, so you have to support it until you reach 20 lbs, then weight... Woah. It's 40 lb's required, to get all 3 off the ground. You have two parameters that have to be met, that you have more tension on the line than what is required for the weight, and you have an minimum of an equivalent amount of tension to support the weight that lifted up prior. So even though it would only take 30lbs to lift C, you have to raise 40 all the weigh up to the top, or the total mass that you are lifting is 100lbs, and thus requires 50lbs of tension. 40 is the sweet number where you remove the tension required to lift it, as well as meeting the tension requirement for weight C. If you want to raise them all up to the top, you need 60. - Jimminy, CoG of FF

I'd like you guys to hover over the picture to see what company is asking this question on an employment application. I'm pretty sure they wanted a common sense answer that an average person would know, but it's obviously just not that simple of a question :) - Rah-PM 2012

I think Otto has pretty-much nailed it (see comments): http://ourdoings.com/ourdoin... - Ken Morley

I just read Otto's comment more carefully, and I disagree about A and B making 30 lbs of force to help lift C. A and B act as counterweights to each other, not just to C. - Bruce Lewis

I suppose the easiest way to visualize what happens is to imagine them all being lifted to the middle. When you let go of all of them at the same time, C will drop down because it's heavier than A, so A will go up. - Richard

My first guess is A. ***My father is a Mechanical engineer and he says A as well. *****************He says it would take 10lbs of force to lift A 20lbs to lift B and 30lbs to lift C. - Jim