# Is there a possible solution here to who carries more weight? [closed]

I am usually skeptical of riddles online that have to do with Physics but this one struck me as they had an oddly specific answer to it. So my question is for the following picture:

I simplify a bit with placing the two men at the very ends and the middle man at the exact center of the log in line with its center of mass(considering log to have weight $W$). I know I have to find the normal reaction force on the middle man, suppose it to be $R$ and I know by solving moments that the normal reaction for two men on the sides are equal, both equal to suppose $F$. Now even if I solve using both resolving vertically and using moments I end with the same result of : $$2F+R=W$$ Even if I consider the two men at the ends to be not exactly at the ends but at a length $l$ from either side, we get using moments $2Fl+Rl=Wl$ which again ends up as $2F+R=W$.

So my question is, how did they get the result of the middle man having 25% more normal reaction force and if I'm missing any results here. Also, is it really possible to compare the normal reaction forces between the three men?

• The problem is not fully specified unless you say something about (a) the rigidity of the log, (b) the give in the log/shoulder meeting and (c) the heights of the bearers. But for the usual introductory classroom case the center bearer can carry anywhere from 0 to 100% of the weight. Or even pull down and make is partners carry more than the full weight of the log. Jul 10 '18 at 19:13
• This can't be solved with the given data. The beam is balancing on the middle man, so it only needs a tiny poke at the ends to stay balanced. Force $F$ could be tiny and $R$ large. Like when a crane carries a metal bar and you gently push the end of it. Or vice versa, the middle man could duck and let go, and he would do 0 % force and the other two take over and do 50 % each. You need more info that couples the three mens' efforts together before you can solve it. Jul 10 '18 at 19:14
• Possible duplicate of A simple (?) problem of static equilibrium Jul 10 '18 at 19:18
• Thank you for clearing this up, but just out of curiosity if we had to rigorously model the situation, how would that be possible? (Assume the heights of the bearers are the same and they all meet at the shoulder at around the same area) Jul 10 '18 at 20:06
• @dmckee, for that matter, the two on the ends can conspire to make the man in the middle carry more than 100%. Jul 10 '18 at 20:10

This is a problem of Mechanics of solids. You are right in using the concept of moment equilibrium to get the equation 2F+R=W. But the thing is that this is a statically indeterminate problem. I'll tell you why, in the above equation you have two unknowns F&R. So for two unknowns you need two distinct equations, but here if you apply the other static equation of equilibrium (sum of all downward forces = sum of all upward forces) , you get the same equation W =2F + R. Statically indeterminate problems are solved by basically two methods: force method and displacement method. I have a similar problem in my book solved by Castigliano's theorem. I will present the solution briefly. Let the length of wooden log be 2a. We can consider the situation in riddle as a beam of negligible self weight, under the action of the same weight W but with it being uniformly distributed over length 2a, having load intensity w, such that w*2a = W. Also the three men are replaced in beam analogy by three rigid(non-deformable) supports.

Now, under the condition that the two men on the ends are both totally extreme i.e 'a' distance away from the middle man and that they both exert equal force F on the log, then the man in the middle will have to exert a force R= 5/4 wa = 5/8 W; since w*2a =W.

F will be equal to 3/8 wa = 3/16 W. Now, 2F=3/8 W & R-2F = 2/8 W= 0.25 W. Thus, the force applied by middle man exceeds the combined effort of the other two men by 0.25W or in other words the effort applied by middle man in lifting load W is 25% more than the effort of other two combined.

• +1 For the great insight, but could you please go into detail what load intensity here means and how R=5/4wa. It would be helpful to explain in such a way that involves the basics so I can approach more problems like this. Jul 12 '18 at 13:03
• Load intensity simply means force per unit length, so that w(force per unit length) * 2a (total length) gives us the total force/load W. Jul 12 '18 at 13:05
• Then, how does that result in R being equal to 5/4 wa? Jul 12 '18 at 13:06
• The R=5/4 wa is a bit technical and i simply wrote the result, so as to not burden you with technical information. Jul 12 '18 at 13:06
• I have made a pdf of the solution upto R= 5/8 W. Here's the link drive.google.com/open?id=1IO9uzvyr--PG8Gdnx-Vo2oh-Wm9c0gzf Jul 13 '18 at 15:39

Unless there is more to the puzzle than the image suggests, it seems to me fairly straightforward to show that the two outside men carry an equal amount of weight and the inside man carries any weight not carried by the outside men. If either outside man carries more weight than the other, the log tips over. The middle man can carry any amount of weight (he could even not be there).

• From my personal experience with railway sleepers, its easier to be in the middle. Jul 10 '18 at 19:20
• I understand now it is because in the middle you have the freedom of applying less or more force. Just out of curiosity, could you elaborate a bit that could you apply more force if you wanted in a practical situation? Jul 10 '18 at 20:05
• Not sure I understand the question. Are you asking how the middle person could apply varying amounts of force? Jul 10 '18 at 20:13

This riddle is wrong because a log is far stiffer than a person, so the appropriate assumption is that the log does not bend, but people are squished down by an amount proportional to the weight they carry (replace people with springs).

If the log remains horizontal and does not deflect, then each person is squished by the same amount, and thus each carries $$1/3$$ of the total weight.