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I Gravity question

  1. Feb 1, 2017 #1

    JLT

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    If acceleration involves a change in velocity - dv/dt, what velocity is changing for an object at rest with gravitational force? Example - an apple on a table, mg down = the normal force ma up, but neither a nor g seem to involve a dv? the forces involved do not involve changing any velocities, and yet there are still accelerations there... so what is acceleration if there are no (V2-V1)/(t2-t1)'s going on? Did that make sense?
     
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  3. Feb 1, 2017 #2

    Drakkith

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    Staff: Mentor

    What do you mean? What is accelerating in your example of an apple sitting on a table?
     
  4. Feb 1, 2017 #3

    JLT

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    Draw the FBD of it - there are forces in static systems. F = ma, there are "a"'s in static systems, even without any dv's.
     
  5. Feb 1, 2017 #4

    Drakkith

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    As you said, acceleration is defined as the change in velocity over the change in time: ##a=\frac{dv}{dt}##
    If you substitute this into the force equation you can see that ##dv## is indeed present and is zero in the case of balanced forces.
     
  6. Feb 1, 2017 #5

    JLT

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    example; F=kx=ma is not zero, a = kx/m, a positive value even though dv/dt = 0 if the system is in equilibrium. It just seems like there is some other acceleration going on, or some other way to think of it - were a is not dv/dt in all cases?
     
  7. Feb 1, 2017 #6

    Drakkith

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    The F in F=ma is the net force on an object unless you're explicitly breaking force and acceleration down into their different components. So you can certainly calculate that the force from a spring should give an acceleration to a block, but unless you account for all the forces acting on the block then you're doing things wrong.
     
  8. Feb 1, 2017 #7

    Dale

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    F is the net force. It is 0 in your example.
     
  9. Feb 1, 2017 #8

    JLT

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    let's say we are breaking the net force up, and just looking at each individual component. we call each component a force, it could be mg, or uN or kx or F=kq1q2/r^2, or whatever the case may be. If you let go of the system each of those individual forces can turn into a change in velocity for something, but it does not have to change the velocity of something to be considered a force. You can squeeze an apple, and squeeze it more and more with larger and larger forces without changing the position or velocity of the apple - forces have changed, the velocity has not changed. Is force not always equal to ma? not always equal to mdv/dt?
     
  10. Feb 1, 2017 #9

    JLT

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    So the "net" force is always ma, but the individual components of the forces are not equal to ma?
     
  11. Feb 1, 2017 #10

    Dale

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    Correct.
     
  12. Feb 1, 2017 #11

    JLT

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    kg m/s^2... if the individual component is not mdv/dt, then what is it?
     
  13. Feb 1, 2017 #12

    Dale

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    It is a force. Why should it be anything else.
     
  14. Feb 1, 2017 #13

    JLT

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    I guess I'm having a hard time understanding what a force really is if it does not have to involve dv.
    Thanks for all the replies!
     
  15. Feb 1, 2017 #14

    Dale

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    You can think of pressure as being momentum flux. Then force is the integral of the momentum flux over a surface. Force is then just an idealization of momentum flux when you are not interested in the details of the surface.

    But I don't think that is helpful.
     
  16. Feb 1, 2017 #15

    JLT

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    pressure involves a bunch of dv's, I can visualize that one...
    but gravity, GmM/r^2, no dv's in that one... it exists without anything changing velocity? What is the physical sig of G? 6.67 × 10-11 m3kg-1s-2 ... what do those units even mean?
     
  17. Feb 1, 2017 #16

    Dale

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    Not in general. Fields exert pressure even without any "dv"s. Not everything is a gas

    Looking for what something "really is" usually just leads to confusion. Force is the thing which behaves as described by Newton's laws. If there are no forces the acceleration is 0. Forces are vectors quantities that can be added together and if you add up all the forces acting on an object then they are equal to the mass times the acceleration of the center of mass. And when an object exerts a force on another object then the other object also exerts an equal and opposite force on the first object. That is what force really is.
     
  18. Feb 1, 2017 #17

    JLT

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    So science only describes interactions, but never actually explains the thing itself? We'll always be standing on the outside looking at the surface of things? That's disappointing....
     
  19. Feb 1, 2017 #18

    Dale

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    Sorry that you are disappointed, but it isn't too bad. If you could predict how a person would behave in every interaction then you would probably feel like you know them very well, even if it is all from the outside looking at the surface.
     
  20. Feb 1, 2017 #19

    JLT

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    Knowing "how/what" is different than knowing "why"... without knowing the why's it doesn't feel like knowing much. Thanks again.
     
  21. Feb 1, 2017 #20

    Nugatory

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    You can't ever "really explain" anything. Every explanation you come up with will generate a new question needing a new explanation. For example, mankind spent millennia looking for an explanation of the motion of the planets in the night sky before Copernicus, Galileo, Kepler, and Newton figured it out: the planets are bodies orbiting the sun, the earth is one of those bodies, and the force between them is given by Newton's ##F=Gm_1m_2/r^2##. Now that is a truly great explanation, and I cannot imagine anyone being even slightly disappointed by it.

    But have "really explained" planetary motion? No. We just have a new problem, namely explaining why the force is given by that formula and not some other. Why ##r^2## instead of, for example, ##r^3##? Eventually we have to end up with something for which the only explanation is "because that's the way the universe we live in works".
     
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