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Interesting riddle/puzzle


Tibbz2
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Assuming equal dimensions of the two balls and amount and densities of fluid etc the one on the left should drop

 

Make sure you take tensions/buoyancy and any other possible forces into account ;)

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Guest Kheldar

:thumbsup:

 

But why? :P

 

You want me to spoil it for everyone? :D

 

It's not the ping pong ball at all as I previously thought - suffice to say the steel ball is having an effect that no (here or SB) has taken into account yet :)

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@Tibbzs............ It tips to the right = the left side goes up?

 

 

Tension in the ping pong ball string

 

Yes overall the right container will push the scales down, but it's not due to the tension in the left string, it's to do with the tension in the right string :)

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You want me to spoil it for everyone? :D

 

It's not the ping pong ball at all as I previously thought - suffice to say the steel ball is having an effect that no (here or SB) has taken into account yet :)

 

Indeed, the ping pong ball is just there as a distraction in the end :D

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Guest Kheldar

Bugger, yes. I've imagined replacing the steel ball with another ping pong ball, it would float, the steel one stays submerged. Right?

 

You're there I think :) In my mind I replaced the water with mercury ... same conclusion :)

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Bugger, yes. I've imagined replacing the steel ball with another ping pong ball, it would float, the steel one stays submerged. Right?

 

Yup... Just think if you were to push that ping pong ball down into the water though, and the reason it drops is to do with forces when that happens (apart from in this system instead of you pushing, the steel's weight is 'pushing' it under)

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Initially i thought "balance" as the volume of water was equal, then i thought right would drop as the mass of the beaker on the right is more, then i thought maybe left would drop as the steel ball wasn't exerting much force on the base of the beaker, then thought right would fall as the buoyancy of the ping pong would do something, then had no idea.

Fascinating.

Jd

Edited by dawson
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Hmmmm... I would have gone for the left one dropping; reasoning -

The volume of liquid displaced on each side is equal, irrespective of what it is being displaced by. So there is no change in the scales/container/liquid system. The steel ball mass does not impinge on the liquid/ container as it is independently supported. The Ping pong ball does impinge on the liquid/container as it has mass (albeit minimal) and increases the mass on that site by a very small amount, even though it appears to be trying to lift that side. The only other factor I can see is that the steel ball, being separately supported, would act as a damper in any movement, but it wouldn't influence which way the movement was.

Would like to see a proof of the solution?

Alan

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Guest ecopley

What actually happens depends on loss of factors and we haven't been given enough information to model the situation fully.

If one assumes the strings have no volume or mass and that the outer fisheye of the ping-pong ball is the same as the diameter of the steel ball (and that the ping-pong ball has no mass) then the right hand side will fall.

The two breakers contain equal amounts of water so they are balanced in terms of mass. The ping-pong ball just represents empty space and exerts no force (it doesn't pull up - or at least the resultant force is zero because any upward force is balanced by downward force or it would be moving, which it can't).

The right side took a while to work out. If the steel ball is submerged then, even though it would sink, there is done upward force on it due to the displacement of the water. Of there is upward force on the steel then there must be downward force on the beaker.

The tricky thing to get your head around is that there can be upward force even though the ball is held up. If it were a ping-pong ball being held down it would be much easier to see. So the upward force due to the displacement if the water is matched by a downward force on the beaker, which will move downward.

I think that's right. It took a while to think it through but, of the assumptions I made are right (of they aren't then if need lots more information to answer the question) then I think that's what happens.

Now, a man is driving a van full of budgies over a bridge. The van full of budgies weighs 6 tonnes. The budgies weigh one tonne. The bridge has a capacity of five tonnes. Before he drives over her bands on the side of the van and every single budgie takes flight. He drives over the bridge while the budgies are in flight. Will the bridge hold? That is, does the van weigh less while the budgies are flying?

Happy skies...

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