From: relativity@paulba.no   
      
   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 the vertical acceleration is:   
   >>   aᵥ = Fᵥ/m = 9.8 m/s²  (upwards)   
   > Nope. The vertical component of the velocity   
   > is constant and 0. No vertical acceleration,   
   > sorry.   
      
   So if a constant horizontal Force F = 9.8 N is acting   
   on an object with mass m, then the acceleration of   
   the object will be constant a = F/m = 9.8 m/s²   
      
   But if a constant vertical Force F = 9.8 N is acting   
   on an object with mass m, then the acceleration of   
   the object will be constant a = 0 m/s²   
      
   So Newton's law F = ma is only valid for horizontal   
   forces. Right?   
      
      
   > Of course - it may be differently in a   
   > completely isolated room without windows   
   > and doors. I've never been in such place   
   > (since such places only exist in sick   
   > delusions of some brainwashed religious   
   > maniacs) - so I can't know.   
   >   
      
   So you think that it may be that Newton's law   
   F = ma only is valid for vertical forces inside   
   a completely isolated room without windows   
   and doors?   
      
   --   
   Paul   
      
   https://paulba.no/   
      
   --- SoupGate-Win32 v1.05   
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