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B Does Normal force always HAVE to be perpendicular to plane ?

  1. Nov 29, 2016 #1
    Say i Have
    1. A perfectly spherical and perfectly rigid Ball
    2. A perfectly flat and perfectly rigid wall

    Say i throw the ball at my FLAT wall also giving the ball a linear velocity
    The only contact with the wall and the ball is a point... a POINT !!
    So, I can as well forget about the rest of the wall except the POINT ( Forget the wall but keep in mind of the point that the ball collides with. I did that since the existence of the rest of the wall has no effect on the after-collision motion of the ball. Here, I'm also assuming the wall and the ball are perfectly rigid so they meet at a single point only )
    The point would have resisted the motion of the ball in the opposite direction to the velocity of the ball( makes sense, as the motion is an example of 1D collision ) And the ball also move back on the same path but in opposite direction like stuff in 1D motion do.
    The rigid body of the ball is made of a set of points at a certain fixed distance from each other
    The point of contact of the ball and the point of contact of the wall are in 1D motion
    The point of contact of the ball which has initial velocity v now has a velocity that is in opposite direction, V ( at the instant of collision)
    If the other points continue to be in v, the distance between the points change...
    I see only two thing the body could do to stay rigid; for all the set of points to instantly have the velocity, V after collision or for the body to rotate
    ( or does it want to stay rigid at all? )Anyways.....
    If it does move in V all at once, the ball will retrace it's original path !
    so.... no matter how the wall was aligned as long as the ball makes a point sized contact,... the force may or may not have been perpendicular and the ball WILL HAVE traveled the same path in reverse
    That's my intuition .. but do correct me if i'm wrong
    .... or is the whole problem is wrong to the base? ( don't say that my assumption of a perfectly rigid/perfectly flat/ perfectly spherical object is impossible... it's fun when such stuff are involved in calculations)
    What's your predictions of what will happen after collision?
     
    Last edited: Nov 29, 2016
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  3. Nov 29, 2016 #2

    Drakkith

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    Certainly. The force on the ball during the impact (considered instantaneous in this case since we are dealing with perfectly rigid bodies) is the normal force. The word "normal" just means that the force acts perpendicularly to the surface.
     
  4. Nov 29, 2016 #3
    Which way do you think the ball will move after the impact... with FLAT wall
    I mean the only contact with the wall and the ball is a point... a POINT !!
    So, I can as well forget about the rest of the wall without the POINT ( Forget the wall but keep the point that the ball is in contact with )
    I a Ball with a velocity hit this point, the point would have hit it against the way the ball is moving and the ball will move away from the point but in the same path that it took to approach the point ( in simple words, the motion is in 1D )
    so.... no matter how the wall was,... the force may or may not have been perpendicular and the ball WILL HAVE travelled the same path in reverse\
    Thats my intuition .. but do correct me if im wrong
     
  5. Nov 29, 2016 #4

    Drakkith

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    Well, if the force acts perpendicularly to the wall, and the ball is striking it at an angle of 90 degrees, which way do you think it will move?

    Edit: The above post was edited subsequent to my reply. I will respond to it in a new post.
     
    Last edited: Nov 29, 2016
  6. Nov 29, 2016 #5
    And.. here i forgot about the rest of of the part of the wall, because i assumed everything to be rigid
     
  7. Nov 29, 2016 #6

    Drakkith

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    I'm afraid not. If the force points in a different direction, either because the ball is striking the wall at a different angle, or because of some esoteric mechanism, the ball will not bounce back along its original path.
     
  8. Nov 29, 2016 #7
    I mean the only contact with the wall and the ball is a point... a POINT !!
    So, I can as well forget about the rest of the wall except the POINT ( Forget the wall but keep the point that the ball is in contact with )
    I threw the Ball with a velocity and it hit this point. The point would have hit it against the way the ball is moving and the ball will move away from the point but in the same path that it took to approach the point ( in simple words, the motion is in 1D collision )
    so.... no matter how the wall was aligned as long as the ball makes a point contact,... the force may or may not have been perpendicular and the ball WILL HAVE travelled the same path in reverse
    Thats my intuition .. but do correct me if im wrong
    Why not...
    The ball contacts the wall at a point
    forget the wall but keep in mind of the point
    the ball goes and hits the point
    returns like a 1D collision should happen
    Even though the situation is given in 3D, i reduced it to 1D

    I forgot about the rest of the wall as their presence wouldn't have made much difference to the motion of the ball as they did not even have contact with the ball
    I think you must remember i said that the wall and the ball are both perfectly rigid and that the ball is perfectly spherical and the wall is perfectly flat
    Is my intuition correct
    if i am correct, what about the normal force in this situation
     
  9. Nov 29, 2016 #8
    Well. maybe the normal force requires an area of contact to exist...
    i'm just guessing... am not sure if it's right
    so, what IS happening in this case??
     
  10. Nov 29, 2016 #9

    Drakkith

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    The fact that the point of contact is, well, a point, is irrelevant. If the ball strikes the wall at an angle then the direction of the normal force is not directly opposite to the motion of the ball. Since the force is not pointing back along the ball's original path the ball will not be accelerated straight backwards.

    Don't get stuck on points and perfectly rigid objects. They don't exist in the real world.
     
  11. Nov 29, 2016 #10
    So the ball will act as if the force acted perpendicular to the plane?
    So, if the normal force was normal,.... the ball would have gone in 2D (or in some case 1D)
    but, the ball has linear velocity and contacts a point in space (not the wall as a whole...only it's point sized thingy) Which give no reason for the ball to move in 2D
    Given that the Ball and the point are the only things involved...
    But still you say that the ball moves in 2D.. why is that
     
  12. Nov 29, 2016 #11

    Drakkith

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    Of course there was a reason. The reason is that the force points at an angle. Draw a force diagram and you can see this yourself. Again, the fact that the point of contact is a single point makes no difference. The rules work out to be the same.
     
  13. Nov 29, 2016 #12
    So,.. you mean that.. NORMAL FORCE WILL BE NORMAL TO THE PLANE... NO MATTER WHAT>>>PERIOD!
    Okay, i will work out the diagrams and calculations on this....( will take time...lots! )
    Thank you!
    hmm! wonder what went wrong with the way I explained it...
     
  14. Nov 29, 2016 #13

    Drakkith

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    As far as I know, yes.
     
  15. Nov 29, 2016 #14
    Thank you!
     
  16. Nov 29, 2016 #15

    jbriggs444

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    The definition of normal force is that it is the component of the contact force that is perpendicular to the surface that is contacted. In this context, "normal" is a synonym for "perpendicular".

    Contact forces can have a component parallel to the surface. For instance, surface friction acts parallel to the surface.

    If a spherically symmetric object approaches a surface dead-on, and the surface is symmetric about the point of collision then symmetry demands that the contact force during the collision can have no parallel component (there being no preferred direction for a parallel component to point). The collision force must therefore be completely perpendicular. That is, it must be a "normal" force.
     
  17. Nov 29, 2016 #16

    ZapperZ

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    This is a rather odd question. It is like asking "If vector A is perpendicular to vector B, does A have to be perpendicular to B?"

    As jbriggs has stated, the word "normal" means perpendicular to that surface. It is its definition. If it is not, we won't call it the "normal force".

    There are other cases where the word "normal" is used, such as "normal incidence", "normal angle", etc... and they all mean perpendicular to something. So you might as well get used to this definition.

    Zz.
     
  18. Nov 29, 2016 #17
    What about my prediction of the ball retracing it's own path after collision?
     
  19. Nov 29, 2016 #18

    ZapperZ

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    I don't see the problem. It simply means that the force acts parallel to the initial direction of motion, and thus, the ball bounces back along the same path it came from.

    Try this: what if the ball hits another fixed ball but off-center? It still, in the ideal case, hits the "wall" only at one point, the same as your scenario above. But what is the direction of the normal force, and consequently, what is the direction of its motion after collision? Do you think it'll trace the same path that it came from?

    Zz.
     
  20. Nov 29, 2016 #19
    I had the same next question in mind ;-)
    The ball could either rotate as well as move back the same path
    or it xan just move back, anyway
    that what I think
    thanks for bringing this up

    [Post undeleted by moderator. Please don't delete posts that already have replies]
     
    Last edited by a moderator: Nov 30, 2016
  21. Nov 29, 2016 #20

    ZapperZ

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    Wait, you actually think that it is possible for the ball hitting, say, a wall that is at an angle, and for it to move back along the same path that it came from? You've never played pool/billiards, or haven't seen anyone playing that before?

    Throw a ball towards the ground at an angle and tell me what you see.

    Zz.
     
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