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a second identical juggler (they are twins) tries to cross the bridge by projecting all 5 balls from one side of bridge and then he plans to run (on the bridge) to the other side and catching them. Will he be successful?

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- #1

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a second identical juggler (they are twins) tries to cross the bridge by projecting all 5 balls from one side of bridge and then he plans to run (on the bridge) to the other side and catching them. Will he be successful?

- #2

Nugatory

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What do you think and why?

(As a small hint - the first case is incompletely unspecified)

(As a small hint - the first case is incompletely unspecified)

- #3

A.T.

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What happens when he accelerates the balls upwards, with his hand?I was thinking about a situation where you have an expert juggler who carries 5 balls. He wants to cross a bridge which can only handle a total weight of lets say 50kg. (These values are randomn). And lets say the juggler weighs 45kg. Each ball weighs 2 kg. Can he cross the bridge by juggling such that only ball is in his hands while other 4 are in air?

Can he run fast enough and/or throw strong enough?a second identical juggler (they are twins) tries to cross the bridge by projecting all 5 balls from one side of bridge and then he plans to run (on the bridge) to the other side and catching them. Will he be successful?

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Although it seems to be possible, what I thinks is as he juggles for a given time interval, you can find the change in momentum. Or the average weight that he carries is 45+10 = 55 and hence the bridge will collapse. But I am still not satisfied with my own solution.What do you think and why?

(As a small hint - the first case is incompletely unspecified)

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Nugatory

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A.T.

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Sounds correct to me.Although it seems to be possible, what I thinks is as he juggles for a given time interval, you can find the change in momentum. Or the average weight that he carries is 45+10 = 55 and hence the bridge will collapse. But I am still not satisfied with my own solution.

- #7

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At any instant the total weight is 47kg.

if t time takes for one ball to come back to hand, and if I take a total time of 5t,

since F=##\Delta P/ \Delta t##

For the whole time interval, the change in momentum of each ball will add up and during that interval the average force F is greater than mg. Is this correct?

- #8

jbriggs444

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That is the gravitational downforce on the juggler plus the one ball in his hand, yes.At any instant the total weight is 47kg.

if t time takes for one ball to come back to hand, and if I take a total time of 5t,

since F=##\Delta P/ \Delta t##

For the whole time interval, the change in momentum of each ball will add up and during that interval the average force F is greater than mg. Is this correct?

Yes. In order for the ball to remain (on average) in place, the average upforce due to the hand must be equal to the average downforce due to gravity. If it is only in hand for time t out of total cycle time 5t then the average upforce on the one ball while it is in hand must be equal to five times that ball's weight.

- #9

Meir Achuz

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Draw a free body diagram of the jugular and the five balls..

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- #11

Vanadium 50

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(As a small hint - the first case is incompletely unspecified)

I don't think that is true - or rather, despite being underspecified, the answer is the same.

- #12

jbriggs444

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Who said the first person can cross the bridge?

- #13

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a second identical juggler (they are twins) tries to cross the bridge by projecting all 5 balls from one side of bridge and then he plans to run (on the bridge) to the other side and catching them. Will he be successful?

My intuition says that the first one can not cross the bridge because in order to keep the balls suspended in the air the average force equivalent to the weight of all balls, when you factor in wind resistance, the downward push of each ball from his accelerating it upwards will be at least equal to the force of holding all balls at the same time.

In the second case, if he manages to throw the balls and run fast enough and be able to catch them, yes he will be successful, at least in the sense that the bridge will not collapse.

- #14

Nugatory

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I don't think that is true - or rather, despite being underspecified, the answer is the same.

You're right. I was trying to point out that although the 45 kg juggler could carry 5kg of balls across the bridge, the bridge could fail while juggling less than that. However, we've already specified that we have more than 5 kg worth of balls, so the bridge fails no matter what.

- #15

A.T.

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I think it's the other way around.If first person can cross the bridge, why not the second one?

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as we see the bridge can withstand 50kg wt force ... so the ball which the juggler will be grasping should not put more than 5 kg-wt force on the juggler with its weight already is going to put a 2kg wt force .. so the juggler has to ensure that he should take time in stopping the ball rather than just grasping at it to decrease the force ( as Ft=mv )

now ... if we suppose a ball is thrown into air with V velocity and it stays on air for T time then he should at least throw a ball every T/5 sec ... and his time to stop the incoming ball should be less than T/5 ( or two balls will be on his hand )

for simplification if we think that he throws the ball as soon as he stops it and then another ball comes to his hand then we get

t = T/5

and V = gT/2 so T=2V/g

∑t = T + t = 5t + t = 6t

now as the extra force should be less than or equal 3 kg wt .... if we assume it as 3 kg wt then we have

mV/t = 3mg/2

so.. t =2V/3g ...

so the whole time he takes is 4v/g .... and as long as he has strength and can slow the ball within time 2V/3g he is gonna make it (if there is no air resistance in play )

and for the second case ... i can only say good luck ( as the length of the bridge is not stated ... so he ll have to be either bolt or FLASH to do it in different cases ( FLASH will do it all the time ))

- #17

A.T.

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I don't get it. 4V/g = 2T, so it's it twice the time he needs for 5 balls. How is that the whole time?t = T/5

V = gT/2

...

so the whole time he takes is 4V/g

The simple momentum argument shows that it's impossible for the 1st person to cross the bridge.

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- #18

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nope ... its twice the time for one ball to come to its origins which two balls take collectively .. ( think one ball comes , he reloads , he throws and grasps the other one )4V/g = 2T, so it's it twice the time he needs for 5 balls.

- #19

Nathanael

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I think of it a bit differently and get a different result. I treated the juggler as slowing down a ball and instantly throwing it back up. As soon as he releases one ball, another comes into his hand. Ideally (to transport the maximum weight) he will be always applying only the average force necessary. If we call the time that it takes for him to slow-down-and-throw-up a ball [itex]T[/itex], then he will be constantly applying a force of [itex]\frac{2mV}{T}+mg[/itex], (neglecting air resistance) where [itex]V[/itex] the speed with which the ball is released and m is the mass of the ball. This force can be at most 5g, so the condition for the bridge to not fall is [itex]\frac{2mV}{T}+mg≤5g[/itex]. At best, the time that the ball is in the air is [itex]4T[/itex] and if we neglect air resistance we can relate this time in the air with the speed [itex]V[/itex] by [itex]\frac{2V}{g}=4T[/itex]. Thus [itex]V=2gT[/itex] and our initial constraint [itex]\frac{2mV}{T}+mg=5mg≤5g[/itex] is false (and so the bridge collapses).now i ll try to explain this 1st case mathematically ...

So neglecting air resistance, the max he can juggle across the bridge is five 1 kg balls. Of course, this assumes he is continuously applying exactly the average force, which is unrealistic. His best bet is to carry them across.

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- #20

A.T.

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Maybe you should define your symbols more clearly. But if your result is that the 1st person can cross the bridge, the you have done something wrong.nope ... its twice the time for one ball to come to its origins which two balls take collectively .. ( think one ball comes , he reloads , he throws and grasps the other one )

- #21

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the way u have calculated constant force should be mV/T ..think of it a bit differently and get a different result. I treated the juggler as slowing down a ball and instantly throwing it back up. As soon as he releases one ball, another comes into his hand. Ideally (to have the highest chance of success) he will be always applying only the average force necessary. If we call the time that it takes for him to slow-down-and-throw-up a ballTT, then he will be constantly applying a force of 2mVT\frac{2mV}{T}, (neglecting air resistance) whereVV the speed with which the ball is released and m is the mass of the ball. This force can be at most 5g, so the condition for the bridge to not fall is 2mVT≤5g\frac{2mV}{T}≤5g. At best, the time that the ball is in the air is 4T4T and if we neglect air resistance we can relate this time in the air with the speedVV by 2Vg=4T\frac{2V}{g}=4T. ThusV=2gTV=2gT and our initial constraint 2mVT=8g≤5g\frac{2mV}{T}=8g≤5g is false (and so the bridge collapses).

Interestingly though, this implies that if the mass of the ball was less than 1.25kg then the juggler would make it across (assuming he starts juggling before he steps on the bridge) even though the total weight is slightly greater than that of 50kg. I'm not sure if this is correct.

Edit:

I found my mistake:

2mVT\frac{2mV}{T} would be the net force on the ball, not the force that the juggler applies... He will actually be applying 2mVT+mg\frac{2mV}{T}+mg

Therefore, ignoring air resistance, the max he can juggle is indeed 1kg (which is a very idealized case).

so we get mv/T =4g which is greater than 3g ( 5-2 ) ... yes the bridge will collapse ...

my bad .. i miscalculated t=T/5 instead of t =T/4

- #22

Nathanael

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The ball starts with velocity V down and ends up with velocity V up. The change in velocity is 2V. The average force is [itex]\frac{2mV}{T}[/itex]the way u have calculated constant force should be mV/T ..

so we get mv/T =4g which is greater than 3g ( 5-2 ) ... yes the bridge will collapse ...

my bad .. i miscalculated t=T/5 instead of t =T/4

You should not be comparing it to 3g, you should be comparing it with 5g. He only ever has one ball in his hand at a time... It's not like he's carrying one ball and throwing another one upso we get mv/T =4g which is greater than 3g ( 5-2 )

- #23

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thanx ... i got it now ... now i ve got the same result as u did ... i didnt keep 2 balls at a time ..... i just substracted the mg from the net force to obtain the working forceThe ball starts with velocity V down and ends up with velocity V up. The change in velocity is 2V. The average force is 2mVT\frac{2mV}{T}

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- #25

A.T.

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That basically boils down to the second scenario.

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