From: relativity@paulba.no   
      
   Den 24.01.2026 13:08, skrev Maciej Woźniak:   
   > On 1/24/2026 12:34 PM, Paul B. Andersen wrote:   
   >> Den 24.01.2026 07:51, skrev Maciej Woźniak:   
   >>> On 1/23/2026 8:32 PM, Paul B. Andersen wrote:   
   >>>> Den 22.01.2026 22:31, skrev Maciej Woźniak:   
   >>>>> On 1/22/2026 9:57 PM, Paul B. Andersen wrote:   
   >>>>>> Den 21.01.2026 16:11, skrev Maciej Woźniak:   
   >>>>>>> On 1/21/2026 3:31 PM, Paul B. Andersen wrote:   
   >>>>>>>>   
   >>>>>>>> You are swinging an object around you in a string.   
   >>>>>>>> Is the object accelerating towards you, or away   
   >>>>>>>> from you? Or is it not accelerating at all?   
   >>>>>>>   
   >>>>>>> Towards me.   
   >>>>>>   
   >>>>>> Right, well done.   
   >>>>>>   
   >>>>>> So you have understood that the object is accelerating   
   >>>>>> in the direction the force acting on it is pulling it.   
   >>>>>> This direction is along the string towards your hand.   
   >>>>>> The acceleration is:   
   >>>>>>    a = F/m   
   >>>>>> where F is the tension in the string and m is the mass   
   >>>>>> of the object.   
   >>>>>>   
   >>>>>> Let us look at a concrete example.   
   >>>>>> The length of the string is L = 1.4142 m, and the mass m is 1 kg.   
   >>>>>> You swing the object such that it takes t = two seconds to make   
   >>>>>> a full turn.   
   >>>>>> You observe that the angle of the string to the ground is 45⁰.   
   >>>>>> That means that the radius of the circle the object is moving   
   >>>>>> along is:   
   >>>>>>   r = L/√(2) = 1 m   
   >>>>>> It is then easy to calculate that the speed of the object is   
   >>>>>>   v = 2⋅π⋅r/t = 3.14 m/s   
   >>>>>> The horizontal centripetal acceleration is:   
   >>>>>>   aₕ = v²/r = 9.8 m/s²  (towards centre of circle)   
   >>>>>> The horizontal component of the tension in the string is:   
   >>>>>>   Fₕ = m⋅aₕ = 9.8 N   
   >>>>   
   >>>> The horizontal component of the force acting on the object is:   
   >>>>    Fₕ = 9.8 N  (constant towards centre of circle))   
   >>>>   
   >>>>>> Since the angle of the string to the ground is 45⁰, the vertical   
   >>>>>> component of of the tension in the string is:   
   >>>>>>   Fᵥ = Fₕ   
   >>>>   
   >>>> The vertical component of the force acting on the object is:   
   >>>>    Fᵥ = 9.8 N  (constant upwards)   
   >>>   
   >>> So what?   
   >>   
   >> Do you have a reading comprehension problem?   
   >>   
   >> So according to Newton's law F = ma the acceleration   
   >> of the object is a = F/m = 9.8 m/s²  (upwards)   
      
      
   > Do you have a reading comprehension problem,   
   > poor trash?   
   > Of course you do, otherwise you would know   
   > that Newton's law is  for unbalanced (or,   
   > if you prefer, net) force.   
   >   
      
   The net vertical component of the force acting on   
   the object is:  Fᵥ = 9.8 N  (constant upwards)   
      
   You can feel the net force acting on _you_   
   right now, and you can measure it precisely.   
   If you sit on a weight, you can measure the net   
   force the weight is acting on your butt.   
   It will be F = 9.8⋅m N  where m is your mass.   
   This upwards force is the net unbalanced force.   
   If the force were balanced it would be zero,   
   and the weight would show that you were weightless.   
      
   The force pushing your butt will give you a proper   
   acceleration 9.8 m/s² upwards.   
      
   >>> The vertical component of the velocity   
   >>> remains constant and 0. No vertical acceleration,   
   >>> sorry, poor trash.   
      
   Right above you claim that since   
   "the vertical component of the [object's] speed   
   remains constant and 0.   
   [Then it will be} No vertical acceleration."   
      
      
   >> I see. You think that if an object is stationary   
   >> in its rest frame, then its acceleration must be zero.   
      
   This statement of mine is just a re-statement   
   of your own words.   
   This is what you claim to be thinking.   
      
   >   
   > No, I don't - yet another lie from a well known   
   > piece of lying shit.   
      
   Don't you mean what you said?   
   Then you were lying when you said:   
   " No vertical acceleration. "   
      
   Is the truth that you mean it will be   
   a vertical acceleration?   
      
   --   
   Paul   
      
   https://paulba.no/   
      
   --- SoupGate-Win32 v1.05   
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