The Naked Scientists

The Naked Scientists Forum

Author Topic: What happens when an opposing force is removed?  (Read 2307 times)

Offline Thebox

  • Neilep Level Member
  • ******
  • Posts: 3164
  • Thanked: 47 times
    • View Profile
What happens when an opposing force is removed?
« on: 23/09/2015 12:15:12 »
If I was to apply a force on the northerly point of a stationary basket ball , creating pressure that contorts the shape of the basketball into an obloid, and fixed this pressure to be set, then I glazed the surface of the ball with icing, and let the icing set,



what will happen to the icing when I remove the fixed force?




« Last Edit: 23/09/2015 22:40:24 by chris »


 

Offline ProjectSailor

  • Full Member
  • ***
  • Posts: 83
  • Thanked: 1 times
    • View Profile
Re: A question of the future
« Reply #1 on: 23/09/2015 13:47:48 »
First .. pedantry makes me point out that you will need to apply an equal force on the southerly point as well otherwise it will move to the south!

The ball will exert force on the icing, due to the pressure inside focused at the northerly and southerly points.. the icing will be put under a strain at these points that depending on the tensile strength and elasticity will cause the icing to want to stretch to form a shape where this strain is no longer significant enough for the icing to stretch.

IF you are talking 'theoretical icing' this works.. in reality all the icing everywhere will shatter off due to the properties of sugar icing and propagation of cracks.. 
 

Offline Craig W. Thomson

  • Sr. Member
  • ****
  • Posts: 370
  • Thanked: 4 times
    • View Profile
Re: A question of the future
« Reply #2 on: 23/09/2015 19:51:24 »
If I was to apply a force on the northerly point of a stationary basket ball , creating pressure that contorts the shape of the basketball into an obloid, and fixed this pressure to be set, then I glazed the surface of the ball with icing, and let the icing set,
what will happen to the icing when I remove the fixed force?
That depends on the flavor of the icing. If it's chocolate, it will exert isotropic pressure on my stomach lining.

Let's do some calculations and see if we can't answer your question more precisely:

http://farm5.staticflickr.com/4151/4981227274_7eb32e7736.jpg

 

Offline chris

  • Neilep Level Member
  • ******
  • Posts: 5341
  • Thanked: 65 times
  • The Naked Scientist
    • View Profile
    • The Naked Scientists
Re: What happens when an opposing force is removed?
« Reply #3 on: 23/09/2015 22:43:04 »
Removal of the force applied by the ball means that the icing will experience a force in one direction unopposed by the outward force of the ball. If the icing is sufficiently stiff, and tough, that the force imbalance will not break it, then the outcome will be no change for the icing. But if the icing is not sufficiently tough then it will try to deform and probably break.
 

Offline Thebox

  • Neilep Level Member
  • ******
  • Posts: 3164
  • Thanked: 47 times
    • View Profile
Re: What happens when an opposing force is removed?
« Reply #4 on: 24/09/2015 02:18:38 »
Would this be the correct algebra explanation?




ΔF[y]/t=Δ[dy]=ΔF
  • /t= Δ[dx]=[dy]>[dy]=[dx]<[dx]□










« Last Edit: 24/09/2015 19:40:05 by Thebox »
 

Offline Colin2B

  • Global Moderator
  • Neilep Level Member
  • *****
  • Posts: 1921
  • Thanked: 125 times
    • View Profile
Re: What happens when an opposing force is removed?
« Reply #5 on: 25/09/2015 10:56:17 »
Would this be the correct algebra explanation?




ΔF[y]/t=Δ[dy]=ΔF
  • /t= Δ[dx]=[dy]>[dy]=[dx]<[dx]□
Reasonable first try, but some thoughts:
F[y1]= -F[y2] because these have to be equal and opposite.
Any force in the x direction would also need to be equal and opposite, but if you are talking about applied force, then in your example this would be 0.
I don't understand what you are trying to say with g+av, use words.
Also, what is r? Usually it stands for radius, but a force cannot = a distance. This is the equivalent in English of saying that cat=rock, makes no sense.
Neither does this ΔF[y]/t=Δ[dy]=ΔF
  • /t= Δ[dx]=[dy]>[dy]=[dx]<[dx].

You have to try to make sense otherwise people will stop responding to you. Notice how they do it with others who make silly statements, they are not being sent to Coventry, it just becomes more effort than it's worth. The only reason some of us persist with you is that we believe you have made real progress, please don't regress.



 

Offline Thebox

  • Neilep Level Member
  • ******
  • Posts: 3164
  • Thanked: 47 times
    • View Profile
Re: What happens when an opposing force is removed?
« Reply #6 on: 25/09/2015 14:14:54 »
Would this be the correct algebra explanation?




ΔF[y]/t=Δ[dy]=ΔF
  • /t= Δ[dx]=[dy]>[dy]=[dx]<[dx]□
Reasonable first try, but some thoughts:
F[y1]= -F[y2] because these have to be equal and opposite.
Any force in the x direction would also need to be equal and opposite, but if you are talking about applied force, then in your example this would be 0.
I don't understand what you are trying to say with g+av, use words.
Also, what is r? Usually it stands for radius, but a force cannot = a distance. This is the equivalent in English of saying that cat=rock, makes no sense.
Neither does this ΔF[y]/t=Δ[dy]=ΔF
  • /t= Δ[dx]=[dy]>[dy]=[dx]<[dx].

You have to try to make sense otherwise people will stop responding to you. Notice how they do it with others who make silly statements, they are not being sent to Coventry, it just becomes more effort than it's worth. The only reason some of us persist with you is that we believe you have made real progress, please don't regress.

 r is radius Colin which I would think everyone knows from Pi,


ΔF[y]/t=Δ[dy]=ΔF
  • /t= Δ[dx]=[dy]>[dy]=[dx]<[dx]□



Change of force in the Y axis over time=change of distance of Y axis=change of force of the X axis over time=change of distance x axis=distance Y being greater than distance Y=distance x being less than distance x

As the vertical axis expands the horizontal access will contract.


av is acceleration velocity.


av=0 but also at the same time = 9.81m/s and at the same time = an unknown based on 1035 mph

the combination of the two make 0.

stationary = 0

centripetal = 9.81m/s2

centrifugal = 9.81m/s2?






« Last Edit: 25/09/2015 14:25:31 by Thebox »
 

The Naked Scientists Forum

Re: What happens when an opposing force is removed?
« Reply #6 on: 25/09/2015 14:14:54 »

 

SMF 2.0.10 | SMF © 2015, Simple Machines
SMFAds for Free Forums