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A friend offered me a brain teaser to which the solution involves a $195$ pound man juggling two $3$-pound balls to traverse a bridge having a maximum capacity of only $200$ pounds. He explained that since the man only ever holds one $3$-pound object at a time, the maximum combined weight at any given moment is only $195 + 3=198$ pounds, and the bridge would hold.

I corrected him by explaining that the acts of throwing up and catching the ball temporarily make you 'heavier' (an additional force is exerted by the ball to me and by me onto the bridge due to the change in momentum when throwing up or catching the ball), but admitted that gentle tosses/catches (less acceleration) might offer a situation in which the force on the bridge never reaches the combined weight of the man and both balls.

Can the bridge withstand the man and his balls?

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    $\begingroup$ You have to factor in the force of throwing a ball upward. $\endgroup$
    – Steve Mucci
    Commented Sep 16, 2020 at 18:57

8 Answers 8

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Suppose you throw the ball upwards at some speed $v$. Then the time it spends in the air is simply:

$$ t_{\text{air}} = 2 \frac{v}{g} $$

where $g$ is the acceleration due to gravity. When you catch the ball you have it in your hand for a time $t_{\text{hand}}$ and during this time you have to apply enough acceleration to it to slow the ball from it's descent velocity of $v$ downwards and throw it back up with a velocity $v$ upwards:

$$ t_{\text{hand}} = 2 \frac{v}{a - g} $$

Note that I've written the acceleration as $a - g$ because you have to apply at least an acceleration of $g$ to stop the ball accelerating downwards. The acceleration $a$ you have to apply is $g$ plus the extra acceleration to accelerate the ball upwards.

You want the time in the hand to be as long as possible so you can use as little acceleration as possible. However $t_{\text{hand}}$ can't be greater than $t_{\text{air}}$ otherwise there would be some time during which you were holding both balls. If you want to make sure you are only ever holding one ball at a time the best you can do is make $t_{\text{hand}}$ = $t_{\text{air}}$. If we substitute the expressions for $t_{\text{hand}}$ and $t_{\text{air}}$ from above and set them equal we get:

$$ 2 \frac{v}{g} = 2 \frac{v}{a - g} $$

which simplifies to:

$$ a = 2g $$

So while you are holding one 3kg ball you are applying an acceleration of $2g$ to it, and therefore the force you're applying to the ball is $2 \times 3 = 6$ kg.

In other words the force on the bridge when you're juggling the two balls (with the minimum possible force) is exactly the same as if you just walked across the bridge holding the two balls, and you're likely to get wet!

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    $\begingroup$ Am I the only person who noticed you converted the OPs pounds to kg using a 1:1 ratio? Were you assuming that g~=22m/s/s? $\endgroup$
    – Dan Is Fiddling By Firelight
    Commented Sep 12, 2012 at 20:40
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    $\begingroup$ I didn't want to put the force in Newtons because I suspected that might confuse the OP. I also didn't want to put the units in as kgf for the same reason. I used kg as the unit because outside of us physics geeks your normal public doesn't make a distinction between mass and weight/force. $\endgroup$
    – John Rennie
    Commented Sep 13, 2012 at 6:15
  • $\begingroup$ After thinking about this, I don't buy the $t_{\text{hand}}$ = $t_{\text{air}}$ thing. If I launched the balls a mile into the air with a cannon, acceleration is much greater, but the time in the air could still equal the time in the cannon... $\endgroup$
    – adamdport
    Commented Sep 16, 2012 at 1:22
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    $\begingroup$ @adamdport: when you fire the cannon the force on the bridge will be much higher than the weight of the ball. The lowest acceleration, so the lowest force on the bridge, is got by accelerating the ball uniformly for as long as possible i.e. you spread the acceleration out. $\endgroup$
    – John Rennie
    Commented Sep 16, 2012 at 7:46
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    $\begingroup$ Looking back, if the weights were left in pounds like the OP said, there wouldn't have been any confusion about using it as a unit for force and mass at the same time. Also, I wouldn't be caught wondering what it would look like for a 195-kg man to be juggling. $\endgroup$
    – krs013
    Commented Apr 1, 2013 at 12:57
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I love this class of problem as a fantastic physical example of the mean value theorem. Allow me to describe a specific case that fits the following conditions:

  • The man plus the balls has a total weight of $m$
  • The entire system (man+balls) starts at rest and ends at rest

From these relatively simple assumptions, I will claim that the average normal force (the force the ground exerts upward) is equal to the weight of the the system. In other words, for a given period of time of length $T$ we have this:

$$ m g = \frac{1}{T} \int_0^T \vec{F}(t) \cdot \vec{n} dt $$

This is a spectacular claim actually. To simplify the notation, consider that $\vec{F}(t) \cdot \vec{n}$ is just equal to the weight a scale would read (this isn't a bad assumption, depending on the scale). Imagine the man is juggling, standing on a scale, and the scale reads a value that depends on time, $w(t)$. The average value the scale reads will be equal to gravity times his mass, including everything he's holding or wearing.

In the story of the man walking across the bridge juggling balls, the total weight is $201 lb$. For every second he weighs $200 lb$, he spends one second weighing $202 lb$ or something similar. The point is that the average value is the same.

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    $\begingroup$ I just realized - the dot product with the normal vector isn't actually needed, provided you write $g$ as a vector too. $\endgroup$
    – Alan Rominger
    Commented Sep 11, 2012 at 21:50
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    $\begingroup$ "The entire system (man+balls) starts at rest and ends at rest" - this assumption is vital for this solution, otherwise it should be possible to reduce the weight on the bridge by either catching off the bridge or throwing when off the bridge $\endgroup$
    – Casebash
    Commented Sep 12, 2012 at 16:05
  • $\begingroup$ Originally I though, we needed to also assume the man doesn't end up with the balls in a lower position or lower his center of gravity as per Bobbi's comment. Just realised that this assumption isn't necessary if we assume the system ends at rest. $\endgroup$
    – Casebash
    Commented Sep 12, 2012 at 16:27
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    $\begingroup$ But if you're allowed to do your first throw and last catch off either end of the bridge, aren't you better off doing a spectacular throw of one of the balls and then dashing across with the other? $\endgroup$
    – Emilio Pisanty
    Commented Sep 12, 2012 at 16:52
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    $\begingroup$ @EmilioPisanty That's right, either way it violates the assumption of starting and ending at rest. The average weight may be less if the balls are moving up upon entering the bridge, or moving down upon exiting. This applies for the cases that you catch them in the middle or don't. There is another answer where there is significant discussion on this point, although I don't think it's been mathematically addressed yet. $\endgroup$
    – Alan Rominger
    Commented Sep 12, 2012 at 17:29
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Put one ball down. Walk the other across. Go back, get the second ball.

Or, roll the two balls across, then run after them.

Or, the juggler takes off his shoes and walks across barefoot.

This is solved as a "nonlinear thinking" problem, not with "juggling is anti-gravity". The ball-man system must be accelerated downwards with an average of 1 lb of force or the bridge will break. Otherwise you could build a perpetual motion machine from two jugglers on a see-saw who take turns juggling.

(Also, running is like juggling in that the weight is up in the air much of the time--if this could work, you could also just hold the balls and run.)

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    $\begingroup$ +1 good point about the jugglers on a seesaw making a perpetual motion machine. $\endgroup$
    – John Rennie
    Commented Sep 12, 2012 at 8:59
  • $\begingroup$ The original riddle actually stated that you had to do it in one pass (no second trip), a 3-pound cell phone and its 3-pound battery (no rolling), and that you were naked (no stripping). $\endgroup$
    – adamdport
    Commented Sep 12, 2012 at 12:43
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    $\begingroup$ @adamdport - Better start thinking about making a rope out of your hair, then. Or...have you gone to the bathroom lately? $\endgroup$
    – Rex Kerr
    Commented Sep 12, 2012 at 15:34
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    $\begingroup$ My allegedly "incorrect" answers had been "take a 1-pound plank off the bridge before crossing", "run leaning forward so your body acts like an airfoil" and "wait for high-tide and exploit lunar gravitational influences" ^_^ $\endgroup$
    – adamdport
    Commented Sep 12, 2012 at 16:22
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Imagine for simplicity that the juggler at some instant repeats himself, i.e. that the juggler and balls (with masses $M$ and $2m$, respectively) are in the precise same kinematic state at times $t_1$ and $t_2$.

Consider man + 2 balls as the system, and bridge, etc., as the environment.

Let $p(t)$ be (the vertical component of) the total momentum of the system.

Newton's second law applied to the system yields:

$$\tag{1} \dot{p}(t) ~=~ F_n(t) - F_g, $$

where

$$\tag{2} F_g~=~(M+2m)g, $$

and where $F_n(t)$ is the normal force from the bridge, which may vary in time $t$ as the juggler does his routine.$^1$

Because of our simplifying assumption of repeating states, we have

$$\tag{3} 0~=~p(t_2)-p(t_1)~=~ \int_{t_1}^{t_2} F_n(t)dt - (t_2-t_1)F_g, $$

or

$$\tag{4} F_g ~=~ \frac{1}{t_2-t_1} \int_{t_1}^{t_2} F_n(t)dt ~=~\langle F_n \rangle. $$

But if the average $\langle F_n \rangle$ is $F_g$, then clearly at at least one instance $t_3\in [t_1,t_2]$, one must have$^2$

$$\tag{5} F_n(t_3)\geq F_g.$$

In other words, the bridge collapses.

$^1$ The juggler is allowed to do whatever motion he thinks would benefit his case. Whether he wants to jump with both feet leaving the bridge, or lower his center-of-mass, or fall down, is up to him. It seems physically reasonable to assume that the normal force $F_n(t)$ is a piecewise continuous function of time $t\in [t_1,t_2]$, with only finitely many discontinuity points. In that case the integral $\int_{t_1}^{t_2} F_n(t)dt$ can be defined using the Riemann integral without involving the technically more complicated Lebesgue integral. (Also note that the mean value theorem does not apply for discontinuous functions, and from a mathematical purist point of view, the mean value theorem is not needed, i.e., the crucial ineq.(5) may be established with considerations that are even more elementary.)

$^2$ Indirect proof of eq.(5): Assume

$$\tag{6} \forall t\in [t_1,t_2]:~ F_n(t)~<~ F_g.$$

Then

$$\tag{7} \int_{t_1}^{t_2} F_n(t)dt ~<~ (t_2-t_1)F_g,$$

if we assume piecewise continuity $t\mapsto F_n(t)$. But eq.(7) is inconsistent with eq.(3). QED.

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    $\begingroup$ ... unless the period is greater than the time it takes for him to cross the bridge, i.e. he could throw the balls in the air, cross the bridge, then catch them on the other side ;) $\endgroup$
    – UncleZeiv
    Commented Sep 12, 2012 at 15:10
  • $\begingroup$ I'm only interested in an idealized version of the puzzle (v1) that deals with whether the man can stay on the bridge juggling rather than whether he can make it across the bridge. $\endgroup$
    – Qmechanic ♦
    Commented Sep 12, 2012 at 18:19
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It depends on how long his arms are!! (and how long the bridge is) If he starts in first position, arms held high, and imparts -0.17G to his balls while crossing, he will make it. Oops. I did the math wrong in my comment.

Besides, he can do a juggler's trick, and !gradually lower his center of gravity! as he walks across the bridge. The juggling is optional, a distraction from what they are really doing. He only has to accelerate at G*(1/201) to have the bridge bear, not 201 lbs (195+6), but 200 lbs. If he can crouch down to 2 ft, I get 5 seconds to cross the bridge.

1/2 ( 0.16 ft / s^2 ) t^2 = 2 ft

t = sqrt[ 4ft/(0.16ft) sec^2 ]
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  • $\begingroup$ +1 if you can work out how long his arms would have to be hahaha $\endgroup$
    – adamdport
    Commented Sep 12, 2012 at 13:13
  • $\begingroup$ 1/2 A t^2 = drop (he goes from arms up to arms down, so l=0.25At^2), but we do not know how fast the juggler walks. A is 32 ft/sec^2 * (1-1/6), apx 27 ft/sec^2. In one second, he needs 7 ft arms. $\endgroup$
    – Bobbi Bennett
    Commented Sep 12, 2012 at 14:51
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It thinks that its reasonable to assume: "The entire system (man+balls) starts at rest and ends at rest". Then we can completely avoid integrals and dealing with time. For the moment, let's just consider the speeds of the balls and pretend his arms have unlimited length. We can only provide 5 pounds of force per second => an acceleration of 5/3 g, although this can be divided between two balls. The balls experience a downward acceleration of g each or 2g overall. Therefore, the total acceleration downwards (possibly divided between the two balls) is g/3 and we can't end up with them both at rest. The only way we could end up with them both at rest, is if we we were allowed 6 pounds of weight instead of 5 pounds (ie. same as carrying)

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I think it might be possible if the man first throws one of the balls into the air before he steps on the bridge. In that case the man could apply 4 pounds of force upwards on one ball initially, then step on the bridge. At that point the bridge would be holding 198 pounds. The man can then accelerate the other ball upwards with 4 pounds of force before the other ball lands. This would mean that the bridge would be holding 199 pounds at that point. When both balls are in the air the bridge would be holding 195 pounds. Then the first ball would land in the man's hand, and the man would need to apply 4 pounds of force to decelerate it to rest. During deceleration the bridge would be holding 199 pounds. After deceleration the bridge would be holding 198 pounds. Then the man can repeat accelerating the ball upwards with 4 pounds of force as he crosses the bridge.

It may also be possible to do this if the balls had a large volume and you counted air resistance, in which case the air would help decelerate the balls as they fell down, but the man would still have to throw one of the balls into the air before he stepped on the bridge.

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    $\begingroup$ No he can't, unless he throws the ball across the bridge, see Alan SE's answer. No downvote for now, but please delete this. $\endgroup$
    – Ron Maimon
    Commented Sep 11, 2012 at 20:49
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    $\begingroup$ AlanSE's answer assumes that the whole system starts at rest, which is not an assumption my answer makes. $\endgroup$
    – Thomas
    Commented Sep 11, 2012 at 21:04
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    $\begingroup$ @RonMaimon The upward force on your control volume won't balance gravity by the initial and/or final upward/downward momentum. If the ball is thrown up right before entering the bridge, the man-ball system has upward momentum. $\endgroup$
    – Alan Rominger
    Commented Sep 11, 2012 at 21:37
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    $\begingroup$ @RonMaimon Yes, the momentum would be decreasing. If the man-ball starts with an upward momentum at the moment he enters the bridge, there will be a negative average rate of vertical momentum change over time. The proposition is to make it a non-steady state problem. $\endgroup$
    – Alan Rominger
    Commented Sep 12, 2012 at 12:56
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    $\begingroup$ @AlanSE: Ok, I see. Then you could do it, but the center of mass of the three balls would have to be effectively falling at 1/3 the acceleration of gravity all along the bridge. $\endgroup$
    – Ron Maimon
    Commented Sep 12, 2012 at 18:24
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Objectively solving.

As the juggler-ball system is pulled down by 201 pound-force. There has to be atleast 201 'pounds of force' acting upwards. Otherwise the centre of mass of the system would accelerate downwards.

Throwing the ball would not create any net force on the system. And the only thing I see, that can produce an appreciable force upwards on the system is the bridge.

So...I say it's a no, if not for the creative ways like the ones that are mentioned in other answer.

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