68 + (3x1) = 70?

Cecil's Columns/Staff Reports
21

After some thought, I think you actually COULD get juggling to work…

A juggled coin only imparts force while he is holding it, thus, by maintaining 2 coins in the air he would be able to remain at or under the 70 KG limit.

Let me explain… suppose he starts juggling before crossing the bridge, when juggling, suppose there are always at least two coins in the air, and the rest of the time he is either empty-handed (thus at 68 kg), he is holding the third one (and thus is at 69 kg) or he is throwing the third one (assume he throws with force that is less than that which would make his effective weight exceed 70 kg, or less than 19.62 N).

Thus, provided he has long enough arms to exert this force long enough to gain enough air time to maintain at least two coins in the air at the same, he should be able to stay under the 70 kg limit.
ETA: There would also be a fourth state of catching… obviously, he’d have to be careful to catch in such a way as to slow it down appropriately so as not to exceed an upward slowing force that exceeds the 19.62 N limit.

22

[QUOTE=Blaster Master]
when juggling, suppose there are always at least two coins in the air, and the rest of the time he is either empty-handed (thus at 68 kg), he is holding the third one (and thus is at 69 kg) or he is throwing the third one …
Thus, provided he has long enough arms to exert this force long enough to gain enough air time to maintain at least two coins in the air at the same, he should be able to stay under the 70 kg limit.
[/QUOTE]

Doesn’t work. If he’s catching-throwing a coin 1/3 of the time, the necessary acceleration averages (big surprise here) 3G. If you postulate some gaps, the required acceleration increases. The average downward force on the bridge remains equal to the weight of the magician and his props.

23

[QUOTE=Xema]
Doesn’t work. If he’s catching-throwing a coin 1/3 of the time, the necessary acceleration averages (big surprise here) 3G. If you postulate some gaps, the required acceleration increases. The average downward force on the bridge remains equal to the weight of the magician and his props.
[/QUOTE]

I disagree… I crunched some numbers while I was at the gym.

He has 2kg @ 1G of leeway. This is the same amount of force as 1kg @ 2G. IOW, 2kg exerts **-19.62N ** in -1G, similarly 1kg exerts 19.62N in 2G.

Further, if a projectile has an initial vertical velocity of V and height of H, then V / G time later, it will have a velocity of -V and a height H. That is, if an object starts at a height of H = 0m with a velocity of 9.81m/s, 2s later it will be at a height of H = 0m with a velocity of V = -9.81m/s.

Finally, to maintain 2 coins in the air, he must exert a force in 1/3 of the juggling cycle that will exactly reverse its downward velocity. IOW, 2T[sub]c[/sub] = T[sub]a[/sub]. Where the c refers to when the coin is being caught/thrown and the a refers to when it’s in the air.

Thus, the velocity at which it leaves his hands (let’s make this H = 0) must equal the negative of the velocity at which he catches it so we can say that V[sub]c[/sub] = -V[sub]a[/sub].

Thus, from the information above, we can say the following are true:

F = MA = 2kg * 1G = 1kg * 2G = 19.62N

V[sub]c[/sub] = 2G * T[sub]c[/sub] = 2G * 1/2T[sub]a[/sub] = G * T[sub]a[/sub] = 9.81m/s * T

V[sub]a[/sub] = -G * T[sub]a[/sub] = -9.81m/s * T

IOW, if we apply a force of 2 * M * G on an object for time T that had an initial velocity of -G * T then the object will have a velocity of G * T and remain in the air for time 2 * T after which it will have a velocity of -G * T… on and on.

You will also notice that this will result in each coin receiving a force of exactly the limit of 19.82N for exactly 1/3 of the time, assuming his timing is perfect (that is, he’s catching one coin at the exact moment he’s releasing the last one).

Thus, besides perfect timing, he’d just have to make sure he can exert a constant force over the range of the catch/throw of 19.82N. Plus, with that force and the corresponding speeds and the length of his arms in mind, he’d have to adjust T to be within human constrants. That is, if T is too large, his arms would have to be very long to exert that much force over that much distance for that much time. However, any T > 0 will work, so I’m not going to bother trying to calculate a reasonable one.

Now, I just hope I didn’t mess up anywhere…

24

Here’s another wild possibility. The force of tossing the coins up in the air will only affect the bridge if it gets passed through the magician’s body. But it is possible for force to be absorbed by his body. He could for example compensate for the upward force by absorbing the downward force with his upper body mass and, by bending his knees, not pass any of this downward force into the bridge. The force of throwing a coin up is slight and it would only take a slight knee bending to compensate for each throw. He couldn’t unbend his knees because straightening up would exert the accumulated force back on to the bridge. But a highly skilled contortionist could duck-walk across the bridge with his increasingly bended knees as he juggled the coins. And as an encore, he could walk back across the bridge while playing “Johnny B. Goode” on a Gibson 3x5.

25

Blaster Master, maybe it would help to think of it in energy terms.

The change in energy of a system is defined as work, and the equation for that is force times distance. Let’s use proper units here and say that I weigh 98 Newtons, and the coins weigh one Newton each. The amount of potential energy something has is equal to it’s weight times it’s height.

Imagine that I am jugging on an elevator platform that is going to be raised one meter. Let’s assume that what you say is true, and that I can keep my weight averaged to 100 newtons. It would take 100 joules (remember Work=Force x Distance) to raise the platform one meter. However, since potential energy is equal to weight x height I would have 101 joules of potential energy. Since it’s not possible to just create energy like this, it’s clear that the scenario is physically impossible.

26

Don’t forget that as he throws each coin upwards, it’s not just the upward-moving coin that exerts downward reaction force that eventually gets transmitted down to the bridge, it’s his arm too. Stand on a set of scales and throw nothing into the air, and there will still be a measureable momentary weight increase.

27

[QUOTE=friedo]
In order to keep that bag twirling you have to move your arm back and forth with sufficient force to overcome that gold’s weight. And that causes you to push really hard with your legs in order to maintain your balance, adding to your weight. This may not seem immediately obvious based on your experience with twirling things above your head. But try it with something heavy. Your legs will get quite a work-out.

So all twirling does is confuse matters by moving the gold sideways instead of up and down, but it doesn’t solve the problem.
[/QUOTE]

What if the bridge was simply a log and you had monkey-legs such that your feet would be not on top but sorta on the sides? Wouldn’t that lessen the total downward force?

28

[QUOTE=Blaster Master]
Now, I just hope I didn’t mess up anywhere…
[/QUOTE]

There’s got to be something wrong there, otherwise it amounts to being able to design a self-contained machine that gets lighter when you switch it on.

29

[QUOTE=Moe]
What if the bridge was simply a log and you had monkey-legs such that your feet would be not on top but sorta on the sides? Wouldn’t that lessen the total downward force?
[/QUOTE]

No. You can’t get anything for nothing.

What if it wasn’t a bridge, but a treadmill?

30

[QUOTE=Blaster Master]
IOW, if we apply a force of 2 * M * G on an object for time T that had an initial velocity of -G * T then the object will have a velocity of G * T and remain in the air for time 2 * T
[/QUOTE]

I think this may be where the problem lies. Gravity acts on the coin at all times: while in the air and in your hand; I think your calculations are omitting the force due to gravity during the time the coin is being caught and thrown.

One way to see this would be to think about “catching” a falling ball by applying an upward acceleration of 1G: the ball wouldn’t stop, it would simply continue downward at a constant velocity. If, as you suggest, the upward acceleration is 2G, the path of the ball in your hand will mirror its path in the air (which would require some very long arms).

To get the trajectories you seek, the acceleration will need to be 3G. In the more general case, it will be G * (number of coins). In other words, the average downward force on the bridge can’t be reduced by juggling.

(And, as Mangetout notes, the peak force will be increased - unless the magician has weightless arms.)

31

[QUOTE=Xema]
I think this may be where the problem lies. Gravity acts on the coin at all times: while in the air and in your hand; I think your calculations are omitting the force due to gravity during the time the coin is being caught and thrown.

One way to see this would be to think about “catching” a falling ball by applying an upward acceleration of 1G: the ball wouldn’t stop, it would simply continue downward at a constant velocity. If, as you suggest, the upward acceleration is 2G, the path of the ball in your hand will mirror its path in the air (which would require some very long arms).

To get the trajectories you seek, the acceleration will need to be 3G. In the more general case, it will be G * (number of coins). In other words, the average downward force on the bridge can’t be reduced by juggling.

(And, as Mangetout notes, the peak force will be increased - unless the magician has weightless arms.)
[/QUOTE]

You’re right… this is what I get for crunching numbers in my head… I lose one. I failed to account for gravity while in the “catch/throw” phase.

…Even though my result was counter-intuitive, it just seemed too good, I couldn’t NOT post it. :smack:

32

[QUOTE=Blaster Master]
You’re right… this is what I get for crunching numbers in my head… I lose one. I failed to account for gravity while in the “catch/throw” phase.

…Even though my result was counter-intuitive, it just seemed too good, I couldn’t NOT post it. :smack:
[/QUOTE]
Right. Re-doing my earlier calculation:

  1. Suppose the magician is tossing the coin up in the air so that it leaves his hand with a velocity of V. After it leaves his hand, it slows due to gravitational acceleration (g), so that the velocity at any time is = V - gt. Using that, it must take the coin a time of t = (V/g) to reach the top of its arc (where the velocity = 0), and a total time of t = (2V/g) to return to the magician’s hand.

  2. Likewise, when the magician catches it, he has to accelerate it upward with some acceleration a. For the same reasons, it must take the coin a time of t = (a/g) to reach the bottom of its arc, and a total time of t = (2a/g) to again return to the point where it’s in the air.

  3. While the coin is in the air, its force on the magician is zero, of course. While the coin is in the magician’s hand, the force is F = m(g+a), where m is the mass of the coin. The force is caused by the weight of the coin (mg) plus the additional force required to accelerate the coin upward (ma).

  4. The average force on the magician over time is equal to:
    [(time in air)(force in air) + (time in hand)(force in hand)]/[total time]
    Remembering that the force from the coin while in the air is zero, this is expressed as:
    F = [(2V/a)m(g+a)]/[(2V/g) + (2V/a)]
    Multiply through by ga, and this is:
    F = [2V    m(g+a)g]/[2Va + 2Vg]
    Canceling, the time-averaged force is…
    F = mg
    Which is the weight of the coin.

Note this is independent of the velocity V or acceleration a. What this means is that there no way of tossing the coin that will change the average force, and thus there’s no way of apportioning the time the coin is held to reduce the overall force on the bridge.

33

[QUOTE=Blaster Master]
…Even though my result was counter-intuitive, it just seemed too good, I couldn’t NOT post it.
[/QUOTE]

Apparently, newspapers have the concept of “facts too good to check” (i.e. they fit too well with preconceived notions for skepticism to be applied).

34

[QUOTE=Blaster Master]
You’re right… this is what I get for crunching numbers in my head… I lose one. I failed to account for gravity while in the “catch/throw” phase.

…Even though my result was counter-intuitive, it just seemed too good, I couldn’t NOT post it. :smack:
[/QUOTE]

Hey, it could have been worse - you didn’t set up a website promoting the concept and try to attract investors.